Lilavati Of Bhaskaracarya: Treatise of Mathematics of Vedic Tradition

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THE COSMIC CALCULATOR (A Vedic Mathematics Course for Schools) Kenneth R. Williams & Mark Gaskell The Cosmic Calculator course offers a complete system of mental mathematics which can be taught in (\ holi~tit way. This course consists of three textbooks an Answer Book and a Teacher's Guide. The course is aimed at 11-14-year-oid pupils though some O'l it is very suitable for children from 8 years. Vedic Mathematics is being taught in many schools world-wide with great success: many top mathematics prizes have been won by st'ldents of this system. .

GEOMETRY IN ANCIENT AND MEDIEVAL INDIA

........ilavati - aracarya ( f Vedic Tradition

TA. Saraswati Amma This book is a geometrical survey of the SaJTlskrt and Prakrt scientific and quasi-scientific literature of India beginning with the Vedic literature and ending with t~e early part of the 17th centufV. It deals in detail with the Sulbastltras in the Vedic literatl~re, with the mathematical parts of Jaina Canonical works and of the Hindu Siddhantas and with the contributions to C\ geo,metry II, Sripati, Bhaskara I & II, Sangamagrama Madhava, ParameSvara, Nilakantha, his discipies and a host of others. The works of the mathe~~ticiam Mah'avira, Sridhar?. and Narayal).a Pal)c.lita and the Bakshali Manusc~ pi. have also been studied.

VEDIC MATHEMATICS Jagadguru Swami Sri Bharali Kr~1Ja TirlhajZ MahclTiiJa The book brings to light how great and true knowledge is born of intuition, quite different from modern Western method. The ancient Indian method and its secret techniques are examined and shown to be capable of solving various problems of mathematics. The universe we live in has a basic mathematical structure obeying the rules of mathematical measures and relations. All the subjects in mathematics-Multiplication, Division, Factorization, Equations, Calculus, Analytical Conics, etc. - are dealt with in forty chapters, vividly working out all problems, in the easiest ever method discovered so far.

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hankara Patwardhan Naimpally . Shyam Lal Singh

Ary;,hhatil (h. 476 All) is credited for his piom'crinl( wurk of extra Cling tantrajiis/ra, (i .e. ;Istronomy and mathematics) from Yl'dfl.1 (g<'l hCc~rally known as the oldest 'mhodilllt:nt of knowledge) to the level of COllllnon usage as he appropriately says in his AI),ahhauyam. In 1150 An, Bhaskaradirya (h .ll14 AO). I't: nowncd mathematician and astronomer of V,.tlirlradition' composed I.iliitmtias the fil'st part of this largel' work called Sitldll(inl",-.~imlllani, a comprehensive c:. "r()~iti()n of arithmetic, algebra, geomell)" mensuration, number theory "d relalrd topics. Uiiitlati has been used i'IS . tandard texthook for about 800 years. I~ hU'id, scholarly and literal), presentation has altracted several cultural areas and it was widdy used and commented and translated into several languages of the world . nhaskaracarya himself never gave any derivations of his formulae. N. H. Phadke (1902-1973) worked hard to construct proofs of several mathematical methods and formulae given in original Uliitlali. The present work is an enlargement of his Marii#li work and attempts a thorough math 'lni\l,ical explanation of definitions, 1'I""lIIl1lae. short. cuts and methodology as inwnd by Uhaskara. Stitches are followed lJy literal trans.lation so that the reader c;m enj()y illld appreciate the beauty of ~CC: u"alc and musical presentation in LrlalJQ)l The buuk. is uscfullO school going /1ildrcn. ~ophomorcs, teachers, scholars, hit rian. and lllO~(, working for the cause

r tl'\;'\t;hcrnalics.

LILAvATI OF BHAsKARACARYA

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LILAVATIOF -BHASKARACARYA A'J r'eatise of Mathematics of Vedic Tradition with rationale in terms of modem mathematics largely based on N. H. Phadke's Marathi translation of Liliivati

Translated by KRISHNA]I SHANKARA PATWARDHAN SOMASHEKHARA AMRITA NAIMPALLY SHYAM

i l l SINGH

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Foreword a 'Bhiiskaracarya' and 'UlavaU' are well-known. If a stu,Ii llplays outstanding talent in Mathematics, the school teacher Itlvill ~ ly ca11s the student "Second Bhiiskaraclirya". Many legends AhulIl J'I>,lvatl are in vogue. It is but natural that Indians have a special ItIlOf'rlit ill Bhiiskaracarya and his works, especially the LJlavali IJh"I(lII~clrya was born in Maharashtra. Many translations of the 1." ltvptT lI' well as commentaries on it are available. However, there is 1111 wurk ill Marathi which studies this subject thoroughly from all AlI, lefi ill It modem - twentieth ~entury - fashion. Professor N.H. I·h.tlkt~ hilS laken great pains in the preparation of the present work A N,II' U"hl on Li/avati, thus fi lling a void. r.iahlCCII years ago Professor Phadke wrote a booklet on Indian MlllhcllllltlciiUlS. Therein we find a lot of authentic and interesting 11I""nlllllioll about Bhiiskaracarya and Ulavati. Doubtless the readers tit lIulI "'-Klllet will look forward expectantly at the present treatise. III Ihc Uliivat1, Arithmetic is presented as an enjoyable playful actlvllV . feasor Phadke has admirably succeeded in preserving this Iltlril III Ihe A New Light on LWivali. Since Professor Phadke studied 1IIIIIIy books in Sanskrit, HindI, English and MarathI, he could have , ,, ••Iy l ivclI scholarly notes and references all over. But the author has ....llIled Ihal temptation and has given the relevant supplementary in.... II III

,10 111

v

vi

LT/avall of Bhaskariiciirya

formation in six appendices. So the book is easy to read and the readers enjoy the playful atmosphere. The LT/avall is a book on Arithmetic written in the twelfth century. It was used in India as a textbook for many centuries. Even now it is being used in Sanskrit Schools in some States. Techniques for the solution of problems are simple and easy to use and, moreover, there is a lot of interesting information in the problems presented therein. Here, I would like to narrate my own experience. At the beginning of the Sandhyavandanam (daily prayers), the Maharashtrians repeat twenty-four names of Lord Vi!!.,u but in the version repeated in North India, there are some variations and less than twenty-four names. When I was a Professor at Benares Hindu University, I asked several scholars about the mystery of twenty-four names. None could give a satisfactory explanation. Finally the late Professor Vasudeva Sharan Agrawala of the College of Indology unravelled the mystery. Professor Phadke has given the same explanation in connection with Stanza CCLXX. In the four hands of Lord Vi~.,u there are a conch, a disc, a mace and a lotus. They can be placed in 4x3x2x 1 = 24 ways in the four hands, thus giving rise to 24 different forms with 24 names. Similarly ten weapons can be placed in Lord Siva's ten hands in, 1O! = 36,28,800 ways. But one does not find so many names of Lord Siva in the ancient literature. Reader's attention is specially drawn to various forms of expressions used to address the student in different stanzas. Thus we find: "0, you intelligent girl Li/avall", "0, friend", "My beloved", "Deereyed", "Fickle-eyed" etc. Scholars may draw varying conclusions from these regarding the types of students studying in the "Fun with Arithmetic" classes of Bhiiskaracarya. Yet, one thing is crystal clear to the readers of the LT/avalf. that learning begins with fun and flourishes in wonder. At the end of stanza LXXVIII we find, "if you know Arithmetic well, tell the number" . When such a challenge appears at the ends of an interesting problem, the readers feel that it would have been wonderful to have been the members of Bhaskaracarya's class! In stanza CXXII, the following problem is given: "Suppose relishes are made by mixing I, 2, 3, 4, 5 or 6 at a time from six substances which are respectively sweet, bitter, astringent, sour, salty and hot. 0, ari thmetician, tell me how many different relishes can be pre-

vii

,"I'etl'r' One does not know whether boys and girls prepared such rell.h.1I bllt t he readers relish, time and again, such "gourmet" Arithmetic l l .illl",ol1 is such an ancient Arithmetic! Today we know many more IIdillit111es and results. In this regard, Professor Phadke has been quite 111i~ rt, While explaining stanza CLXXV, he discusses the two cases (i) t~ j¥{() • 0 and (ii) cos«() O. This is an example of a result which was 1101 lcfIown in the times of Bhaskaracarya but well-known now even to "'ollll~i cntious high school students. There is nothing surprising about IlIi·lI. Undergraduate students of Mathematics are now required to I llId y v ral results which Newton did not know. h ) I' those who are interested in both Sanskrit and Mathematics, I.Ilfll'trll is an attractive illustrated arithmetized book of stories. But it w01I11'I not be fair to merely consider A New Light on Li/avall as a " I1I1I1JaI or literary treatise. In the history of Mathematics, LTlavall occupies an honoured place. It shows the extraordinary prowess of Uhftakllrncarya who was not only a top Mathematician but also an ucdknt teacher. Plato says in Republic: "Arithmetic has a very great l iliJ l'IevQling effect, compelling the soul to reason about abstract 11I.lll1hc.!I'!I, and if visible and tangible objects are intruding upon the .r' IIII1l~ lIt, refusing to be satisfied." Ir U I re be any truth in the above statement, then without getting di ll rlH:tcd by interesting problems, poetic fancy and attractive deIMlrl (lti Ofll , 1\ child will be influenced by LT/avall and hopefully a new ilhlakllJ'l\ciuya will emerge. III Ihe history of Mathematics one does fi nd such unusual incidents. 'I'h, wdl -k'nown Malhematician Madame Kowalewski was inspired by pAl•• IIr i\n old Mathematics book which were used as wall paper! I'll nOI h. ' lIrprised if some "Lilavati", inspired by the present treatise A N_w Ught on Li/avatr. will blossom forth into a top-notch re••R:hCr in Modem Mathematics.

'*

Punt Univcn:ity U ·' · 1971

V.V. Narlikar Lokm anya Tilak Professor of Applied Mathematics

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Contents Page liclfcword by V.V. Narlikar I~ rllllun

Transliteration of Devanagari

Trllllsiators' Preface IIhnskariiciirya: His Life and Work I,ilrlvatl

v

xi xiii

xv I

I. Ik fl llitions and Tables

3

2. I.

f' llIl'C Values of Digits

9

Add ition and Subtraction

II

4. I,

M.' thods of Multiplication I livlsion

13

e.

Methods of Finding Squares

19

" S4111lrc R ot . , Method to Find the Cube . , (' lIhc~ Roots

17

23 27 31

10, Hlllht Operations on Fractions II . " .Iclitiun and Subtraction of Fractions

33 39

12. U

Mllltivlication of Fractions

41

UiVI"ioll of Fractions

43

ix

x

LUiivafi of Bhiiskariiciirya

14. Squares. Cubes. Square Roots and Cube Roots of Fractions

45 ,

15. Eight Rules Concerning Zero

47

16. Reverse Process

51

17. To Find an Unknown Quantity

55

18. Method of Transition

63

19. Square Transition

65

Quadratic Equation

71

2l. The Rule of Three

77

Inverse Proportion

81

20. 22.

23. The Rule of Five 24.

Rules for Barter

25. Simple Interest 26. Combinations 27.

Progressions (Series)

83 87 89

Roman Transliteration of Devanagari

101 105

28. Mensuration.

113

29.

Volume

151

30.

Wood Cutting

155

31. Volume of a Heap of Grain

157

32. Shadows 33. Pulverization

161 167,

34. Concatenation (Permutations. Partitions etc.)

177

Index of Verses

183

Subject Index

195

I',,",

t

31

311

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LTiavatl of Bhiiskaraciiryo

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Translators' Preface .I~.~ 1111"

gll'at pleasure in placing before you this book, largely a .1111111'1 III the Liliivati Punardarsana written in Mara~hi by my " .. I (;\1111, the late Professor N.H. Phadke in 1971. He took great 1m III WIIIIIlg this Mara~hi translation with comments and explana-

nil I 'tlll.1 1IIIIIlk Smt. Subhadrabai Phadke for directing me to her eldest hll 'I .N Phadke, Chief Engineer, Metropolitan Railway, Calcutta

II"

IIl1lhllllly to publish this English translation, made by me, of I'hildke's Liliivafi Punardadana. I acknowledge the kindness • WII hy MI . (j.N. Phadke in giving me the authority. It WII~ ""f11l' time in June 1982 that my student Sornashekhar fiif""',t"lly C"xprt·s:-.ed to me his desire to publish the English translation th. I illll'"'' By the grace of the Almighty, I have been able to "ttlll"l' Ih,\ I, "nslation before I completed 71 years of my life. I thank ,.ihllk I1111 I also wish to thank my wife Laxmibai (alias Mai) "hllll' whll\r ~(l(lperation I would not have been able to complete this

..hi."

II I

.,1 .""".

H'

K.S. Patwardhan

v,mlll" ), I !JK2

xiii

. ... 0.,

xiv

LI/avatI of Bhaskaracarya

In 1977, I was inspired to translate Liliivaf; into English. Since I found the task difficult, I requested my esteemed teacher Professor Patwardhan, who not only readily agreed but finished the translation in record time. I went through his version, compared it with the original and made many changes. I have taken the liberty to shorten or simplify some derivations/proofs and to omit those that are found in current textbooks. It gives me a sense of fulfilment to have been able to play this role with my teachers Professors Patwardhan and Phadke who inspired me in my student days. It is a pleasure to express my gratihlde. Thunder Bay June 17, 1983

The LIlavati has been my first interest for the last two decades. I am happy that Professor Som A. Naimpally chose me to join as a translator for this book, and lowe him for the same. It has been my endeavour to narrow down the gap between Sanskrit verses and their English renderings. Although it has been my ample endeavour that the English renderings should not be far away from the literary beauty of the verses, I have preferred to remain deeply faithful to the mathematical message. However, it seems that "faithfulness" and "beautifulness" rarely go together. As a consequence, renderings of certain stanze give only mathematical formulae. I have taken the liberty of making various alterations in the original typescript. On behalf of the authors, I thank Professor K. V. Sarma of Madras for his high appreciation and suggestions to improve upon the original typescript. I record my appreciation and thank Mr. N.P. Jain and Dr. G.P. Bhatt for their interest and personal attention in bringing out this book and sharing some of the editing travails. I believe that LWivati can give pleasure and insight to every class of readers, both from children to senior citizens. Rishikesh 24-11-1991

Bhaskaracarya: His Life and Work

S.A. Naimpally

• one come across a person, at least in Maharashtra, who I'd "hout Bhliskariicarya or Bhaskara It - the great poet 1 11\ 111 lcilUl. His book Siddhantaiiromal)i - especially the ilOWIi as the LIlavati (Slate Mathematics or Arithmetic) IIOWII 1111 over the world. Because of the poetic name the WIhe excellent problems contained therein, Bhaskariicarya UlIl respect of scholars for the last eight hundred years. III lIilleteenth-century German mathematician Weierstrass

ftI

IhulJlllliciiUl, who is also not something of a poet, can

"'I' cmnplete mathematician." I

in the West, many mathematicians who have a flair for

I". QlJlar Khayyam, famous for Rubais and sitting under z

wilio

IlIId

women, was primarily a mathematician .. In the

, WI nllil ill Bhaskaracarya an extraordinary combination of

S.L. Singh xv

LIlavat'i of Bhaskaracarya

xvi

a poet and mathematician. His book amply demonstrates the qualities of head and heart. Bhiiskaracarya was born in 1114 A.D. He has made this clear in the fourth part of the SiddhantaSiroma~ titled 'Goliidhyaya' (Astronomy). In the 58th stanza in the chapter of problems he says: ,(fl!1UI'iuf¥t~i1¥t'lqs1qfl¥tils¥Cq"'''~NRt: I '(,"jUlq~'UJ l'Im ~"idfh~qoft ~: II

I was born in Sakal 1036 (A.D. 1114) and I wrote SiddhantaSiroma~ in Saka 1072 (A.D. 1150) at the age of 36. Bhaskaracarya has given some information about his family background at the end of the chapter on problems in Goladhyaya. From this, it follows that he belonged to the Sfu:l<:lilya lineage and that he lived in Vijjalavida. In the second stanza of the following verses, he talks about his art of writing with great confidence: 3/'~lOCIJ('ft'ii('f~-~ ~~tJ~,wt'" I "'1"''ffWI''~ ~wt('ffU 'iI~~?:l'~'j) fituJ: II 4h'+"rafWtR iiiiI '(ii/i ~ <: 3" C"'Y I'<' y:e t<:: ~: 1 ~ ~'rvl!ft~ ~ II ~ l'lI*,,«~~.!a ~HI4'I..q ~ I

~2A

i!i5f.t1:f
Q

_iII.,'WJ,'· Ifi.~

Life and Work

'"" Imll part of the name ('Vijjalavida') is corrupted into

. "11' thInk that Bhiiskaracarya hailed from Bida. But Bida is rrulll Ahmednagar in Marathawada and is not anywhere

S"hYfl mountain. Besides, there no descendant of i. traceable in Bida. Some opine that Vijjalavida is the .....rll .lIcnr ~Iyderabad. The reason for this goes back to the I'lli/i of Emperor Akbar's court. At the suggestion of Aktrllll!llated Lr/avat'i into Persian. Therein he says that ...lQIirva hailed from Bedara. But Bedara, which is 80 krn far . 11,,111', II not have a single small hill even nearby. Hence, i. 1I(,t Bedara. In Bhiiskaracarya's times, a town Kalyana IIw r by ukya dynasty. There is neither any reference to ( ill Gha karacarya's works nor a reference to him is made "',IUI)' r Cfflukya d~asty. Is Vijjalavida the town of Vijapura? IIJl1r11 ,)oflsider Bhiiskaracarya to be a Vai~l,lavite Brahmin Kanl. t"ka. P.mdit Sudhakar Dwivedi also takes him to be a '"I lhe. basis of his frequent quotes from the Vi~1)u PurliI)a ettlille r ' OhuvlmakoSa' of Goladhyaya. But then Vijapura is at kill Ca lil of' Sahya Mountain. Also, "Vijapura" seems to be ",,'111 " Vidyapura" or "Vijayapura" rather than "Vijjalavida". .lin flllllot be identified with Vijapura. Further, there is an !,hIlt lIhltskaracarya was not from Kamataka. Nrsirp.ba in his '" :\\'(l/1'livarttika " on LTiavat'i writes . ....rv ..

aeo

~: II

Bhiiskaraciirya studied all sciences under the guidance of his father Mahesvara. Mahesvara was a great astrologer. According to late Mr. S.B . Dixit, Mahesvara was born in A.D. 1078. He wrote two books Kara~a-Grantha and JjjJaka- Tika-Grantha. As Bhiiskariicarya' studied 1 under the guidance of such a competent teacher, he became an expert in many branches of learning.

was from Maharashtra. In another commentrary describes Vijjalavida thus:

M lI1I1 II'vara

Location of Vijjalavida

ve quotations that VijjaIavida is situated in met which is north of the river Godavari.

I~

the first stanza quoted above, Bhaskariicarya says that he hailed ' from Villala'lida. But no definite information is available regarding its I

Hindu calendar.

xvii

."11 hul

Kolhapura which is close to the Sabya nut to

_ _ Yijlpur.

IN

north of Godavari. Bida (of Marathawada), way .near the ranges of Sahya Mountain.

xviii

LTlavatI of Bhaskaracarya

MunIsvara wrote his commentary in 1608 which is approximately 500 years after the composition of the Siromal)i. We consider that he must have had a reliable source of information regarding the Bhaskarliclirya's abode. According to him Vijjalavida was situated in between Vidarbha and Sahya Mountain. It was in between Malegaon range of Sahya Mountain and Pitala Valley in the vicinity of the Godavari, as it was about 32 to 48 krn far from this river. Even if these distances are not accurate, Vijjalavida must be near Malegaon, Calisagaon, Jalagaon. This is confirmed by the research of Dr. Bhau Daji Telanga. It appears that the present Patana which is 16 krn from Calisagaon must be the former Vijjalavida. Dr. Telanga discovered tile inscription on land-grant to the temple of Sarajadevi of Patana. He has shown ample proof for this in the publications of the Royal Asiatic Society.' There he has made references to the grants given to the Pandits for the construction of Mathas, schools etc. by the rulers of Suryavanshi Dynasty, viz. Kr~r:Jaraja, Indrarlija, Govana and Siiryadeva. Further, there is a reference to a grant given to Oirigadeva, a grandson of Bhaskaraclirya, to run a school. Nonetheless, no commentator has ever identified Vijjalavida with Patana. Ordinarily, names of towns don't change. Besides, the evidence presented by Dr. Bhau Daji shows that C-angadeva alone belonged to Patana. It is possible that C-angadeva and Anantadeva might have found it more conducive to run the school at Patana rather than at Vijjalavida and so they might have left their original village for good and established themselves at Patana. They might have shifted as per the ruler's orders. This author is of the finn opinion that further research is necessary to establish the location of Vijjalavida. Until then we may accept Dr. Bhau Daji's conclusion. No information is, however, available with us regarding the location of Bhaskaraclirya's school; his patrons in his capacity as an astrologer; or his donors. So it may be concluded that Bhaskaracarya was not attached to any ruler or patron. Names of his descendants for nine generations are known from Copper Plate Inscriptions:

I

Bhau Daji: Journal of the Royal Asiatic Soclery (1 86S), pp. 392-418 and Journal ofthe Bombay Branch of the Royal ASiatic Society. 8 (1868). p. 231.

xix lIftlllc:arabhana,

Govinda,

Prabhakara,

_1Vt.,•.IIhl\lIlolar'!i, Lak!imIdhara, Cingadeva.

Manoratha,

- ••lrtarl'·YMt.lducatioD

11 .....III" u .IIIhIllAllv

III

t lJIention that Bhaskaraclirya must have been an exlI, ent student. He gives his genealogy in his Algebra I of 'Goladhyaya'. In both the places he records that

Mllcation from his father. In Algebra he says:

II is clear that Bhaskaracaya's father was his teacher. towed on him the title of "AClirya" (Preceptor). He . It Al gebra also from his father and mastered it suffi" hlKJk on it. He was well versed in Poetry, Grammar, Imnomy etc. and kept himself up to date. In stanza 'mIT we learn about the subjects one had to master to •Ac!'lrya·. GlUleSa Dai vajiIa calls Bhliskaracarya (A Jewel

It with Bhaskaraclirya's students.

te hi s SiddhantaSiromal)i when he was 36. It is in all. (2) Algebra. (3) Planetary motions. and (4) t• • Til!, LTllfl'llli mainly deals with Arithmetic but also conhfuU' I' tl<'" Trlaonollletry and Algebra. Those days (as it is true ".I.V) ~ C1illcated and uneducated were interested in future . . . . ., A"II lint III lilly every Astronomer had to study Astrology. For ,,",PI1M'I '! noceasary to master Planetary Motions and Astro"/til thl 1.,'fl!TYati provided prerequisites for this study . ....,.' "',V,II wrote his LT/avan by selecting good parts from Itftik!f and MahaVIriiclirya's Gal)itasarasaIiJgraha

Ifh'''''·

11M

E

"·11,.,. ..,' ".,./

I'.

xx

.

LWivatI of Bhaskarac8ryQ

!

and adding material of his own. All the prerequisites for the study of Astronomy are in the LJliivatl. In order to solve problems concerning periods in Planetary Motions, Bbaskaracarya has added solutions of what are known as Diophantine Equations. , Bbaskaracarya's Algebra too is an excellent book on which several' commentaries have been written in several languages. A well-known commentary on Algebra the Nav3iJkura was written in 1612 by Kr~I:Iadaivajna who was an Astrologer in the court of the Empero~ Jehangir. In 1634 a Persian translation of Algebra was made ' by. Ataulla Rasidi, an Astrologer in the court of Shahjahan. Stra~h; translated it into English in 1813 and Khanapur Shastri rendered it, into Mara~hI in 1897. Pandit Sudhakar Dvivedi wrote a Sanskrit com~ ' mentary on Algebra wherein Diophantine Equations, Squares, Pellian' equations are included. " Planetary Motions and Astronomy being more involved compare4 to the first two parts, neither the Eastern nor Western scholars ha'v ' paid any attention. However, they are worth reading by teachers ·and students of Mathematical Astronomy. In his Astronomical book Bbaskariicarya maintains that the earth is stationary.' In his work , J~ Planetary Motions, he considers Lunar motions, Revolutions, EclipSes and such topics which are profound and involved. Besides these, Bbaskaracarya wrote several works: KaralJllkutiihala. Sarvatobhadrayantra. Vasi~(hatulya. Vivahapara1a. It is our misfortune that the original manuscripts are not available. M~- ' harashtrians do not generally have an awareness of the necessity, 01'\ preserving historical artifacts. However, the handwritten copies ~f ! LWivatI with commentaries which are as old as two hundred years are ) still available. . i

'H

Readers ~i.ll notice that LWivatI is a monumental work. According the prevaIling customs, Bbaskaracarya'wrote it in verse form. TOlex, plain a scientific topic like Mathematics in a verse form is indeedli! 'difficult task. But a still more difficult job is to infuse poetic

qualities},

It appean that Bhiskara and the ancient Indian aslrooomm considered earth ItIllon'ary just for observations and subsequent calculations only. '

xxi

_IIIIIeII1IR overcame all these hurdles and wrote the famous lIthe art of printing existed then, several editions of the hnve been published. However, despite the nonof Ihe printing machines, the handwritten-copies of Lilavaf;' rtol'n olle end of India to the other. The Liliivafi replaced most Il.nl were used earlier. The (whole) LWivatI reached other • lillie later, in the sixteenth ce~tury. All those who read this overwhelmed by Bhiiskaraclirya's skill. It seems ."!lnm himself loved his first creation, i.e., Liliivafi more than WOrk•. Beautiful poetic flights of LWivati are not found in m",helllatical or scientific compositions.

worb II is clear that Bhiiskaraciirya had mastered Grammar,

Y"'I of ~ini. Also, like the poets, Bhiiskaraciirya was fond • •110,'1 and metaphors, e.g ..

_ ....nnRllllr.. ~ I I' • • •mfi~ r i .~fas4Idld'tt't

II'

.~: II'

II'

Il1o quite witty. In one problem, a wealthy person is supalvin. one kavacp' in charity to a poor man! We give fur-

..... or TJ,I,'

Valiant Bhaskaracarya

I

R.W: fils Life and Work

RhAskaraclirya's poetical skills by examples from arc vivid descriptions of seasons. It is all beautiful

xxii

poetry. The late

LIllivatf of Bhaskaracary" \ Balalq~l.1a

the following:

Janardhane Mcx:taka rendered into Maratlli '

_""". Hi.,· /.ife and Work

xx ili

valents for numbers. Verily he opened a great store"

l1m'a ) Icnns. If one wants Marathi' equivalents for sci en-

lish, one may easily find them in Bhaskaracarya's ult] be useful to teach Mathematics in Indian lan_.ctll"". his U/avali. has been used as a textbook for the last y ·....s in India. Even now it is used in some places in

«M4&ilt=i "'~«f4~"fA4f ~(iiiQq~qp"s;QilJ4i;;:pg<: I vt;q'llttf~t"i;~l't"il~i ~ ~ ,.vilq'AI: II

( >It¥of'1 1!. .ih.,iUjiQ) In the Hemanta (winter) season the earth looks more enchantin( i because of the plenty of stored grains. The drops of water, in the fonq; of dew, appear like pearls and add to the beauty. The well-fed con! I tented cows and bulls move about in flocks and the farmers enjoy lhil beautiful sight. Teaching Ability In the LWivatf Bhaskaracarya has displayed his various skills. Whil~ teaching Mathematics he also wanted to convey informations on reli! gion, the Vedas, Puriinas, Epics etc. In the stanza 76 the informations! ' . are given regarding a chariot and the fact that Kan:Ja. though a brother of Arjuna, was his enemy. The entire battle-scene is graphically d~­ picted. This stanza can be compared with Kalidasa's Sakunta/am. l Grivabharigabhiramam. The graphic descriptions of a drove of swans ; a flock of elephants; a colony of bees; the attack of a snake by a d~, mesticated peacock; sinking of a lotus in water owing to a strong wind and so on: were to train students in the appreciation of nature and to make mathematics interesting rather than tiresome. Bhaskaracarya haS seldom given proofs or derivations following the contemparory tradi~ tion. Still, he gave a large number of examples. He was a leading Astronomer of his time and had earned the well deserved title of "Jewel among the mathematicians". Bhaskaracarya handled determination of sines of angles much more skilfully than predecessors. Although he took earth to be stationary and the centre of , the universe, it did not adversely affect his calculations. Bhaskaracarya was familiar with many results that were later discoyered by Copernicus and Tayko Brahe. He also had some idea of limi~ which was (500 years) later discovered by Newton. He moulded a difficult language like Sanskrit to express a scientific subject like Mathematics. By using tenns from fine literature, ~e

hi.

l

pl'aelical too as he was of the opinion that if eclipses, • do not occur as given in ancient almanacs, then the I.lO .·h.mged. Orthodox people used to stick to false , '-'10n!! Rahu and Ketu swallow the Moon and the Sun Whal actually happens is that at the time of Lunar Cnlers the shadow of the Earth and so it becomes . carya knew this fact but found it difficult to con0:....1'--'.,- - 1\1 l;Irge. He used to say:

I. Hot a demon. It is the shadow of the earth that MOOn invisible. If you still believe in Rahu, at least ICNd the shadow of the Earth and then swallowed

1II1

Til ~

found a way to reconcile the old and the. new disi'I practical scholar, he was not to compromise with 1'0 was a great Indian Astronomer Lalla (A.D. 768) Xlbook on Mathematics which was used until

- - '• •• UIIIC. That text contained some incorrect propositions his hook on Astronomy Bhaskaracarya corrected I olllilac like the correction of the fonnula for the I ~ surface area of a sphere. He has given this in u h daimed himself to be a "student of Brahmaillllcd to criticize the latter's long or wrong or teOlYlclimcs he also criticized Aryabha~ .ll il conjectured that Bhaskaracarya might have vislOiuC lUlonymous Indian astrologers had visited TurMil1illlto4lUll as honourable guests according to some Araftom Sanskrit.

"

xxiv

LIliivali ofBhaskariiciI'yrA l

If/.f

l.ife and Work

xxv

1-'1

bian works. If this view is acceptable, then there should have beeJt.: j some reference to his · works to the Arabian mathematicians. Hence 1 the above view that Bhaskariicarya had visited the middle-east is nol' ) very convincing. Social Conditions •1

It is rather difficult to draw a picture of the social conditions of the)' Bhaskariicarya's period by way of a book on Arithmetic. IJ11' Bhaskariicarya's times, scholarship was assessed by the study of ten 'l works. All trade was in the hands of Lamanas. The wealthy wore! garments made of silk imported from China. Young women were sold , as slaves. Scholars were patronized by the rulers. Travel at night was I risky. Ni~ka, a silver coin, was the principal currency. Peacocks were ~ common pets. There was abundance of lotus flowers, bees, elephants . , 'I and birds. Siva, Vi~~u, and Ga~eSa were universally worshipped. Tho) RamayaJ.li1 and Mahabharata were studied regularly. On the whole, ' there was prosperity. Gold was expensive but there were plenty of .' food items. Astronomers were more involved with Astrology than its mathematical applications. Although the scholars were generally pa- ,I tronised by rulers, Bhaskaracarya seems to be an exception. He wasI hale and hearty, and died in 1193 at the age of 79. After Bhaskaracarya we don't find well-known learned (traditional); mathematicians in Maharashtra. Though, most students of Indian math- \ ematics merely studied Bhaskaracarya's books still no research was done until seventeenth century. Though a great mathematician, he did not know decimal fractions; which were discovered in Italy in the sixteenth century. It is surprising): that Bhaskaracarya did not discover Newton's Binomial Theorem (in' its general form). Had he discovered logarithms, determination of chord\ , would have been considerably simplified. He knew that the earth attracted; other bodies but he missed Force = Mass x Acceleration. For the' computation of daily motion of planets he introduced the concept 0[. ' instantaneous velocity by dividing the day into large number of smal ~ intervals. The formula &(sin 8) = cos 8 88 is implicitly available in Siromani. He was aware that the differential coefficient vanishes at al'l' . " extreme value of the function. He appears to be the first mathematicianl who could perceive the ideas of differential calculus 500 year~ I

11 ' I

I' I,

IIlId Leibniz. He had little knowledge of ellipse, parho III. Had he proved theorems in Algebra and GeomeII hll\' achieved a greater status as a mathematician. III < 'hievements put him on the top of the list of

'unullatC to have had such a great mathematician in India . " I\( 'ligent that we forgot even to celebrate the SOOth composition of his SiddhantasiromaI.Ji. Let us Bhaskariicarya's eight hundredth

ur

Lilavati (Pali Mathematics)

~~ ~ ift1.ff......{ I ·9'(dl~i~ ....."~~rcft;iiIPJ~ II I II _ _"""'"

~

Sillutations to the elephant-faced Lord who creates Ii I ,by remembering whom all obstacles are der-~ I are revered by the community of gods. Here I l illIe mathematics - LJUivatl - which is loved by hccause of its clarity, brevity as well as its liter-

CHAPTER 1

'cfinitions and Tables 'fIl ~ ~ qUlllq('lu: I

.-.m.

~ ~ ~:II 1111

Jrm make one kakiIJ.I Four

kaki~lis

make one paIJ.a. mll'ke one dramma. And sixteen drammas make one

:'IIIIIIWllclIrya gives the following table concerning the . Ieventh and twelfth centuries: (C wri ) c:: 1 kakil)J (davaifi> . . .iD. I fHI(lO (pa/slf)

......."'.,~ h century. the above coins were more or less used in . Oriiish regime coins pai. ruka and dhabu paisa were ' \joll coins were also in use during the rule of the Petr w possible to get 40 Cowries for one Paisa and U.cd In procure sundries such as hot peppers and corird (h71/1f1 for price has evolved from 'dramma'.

3

4

5

LilavatT ofBhaskaracJ

Measure for gold

• I .'."/11. 24 arigulas = I hasta (forearm). .. I ".~,1t" 2000 da¢as I kroSa or kosa. II ,'I IIIhi " 'aIIlS of barley-com are kept be11ywise close to Ih.' lenl'" is one arigula, i.e., the phalanx of the fingers. . . . . III' , ,"nl.lnnl measures in India. There were many kingi flli'il~II, each having its own weights and measures . ....11 til 11'1'111'" was not the same all over India. However, as~'I¥hl or a man is 31h hastas (foreanns), the above t,t \..' IIl1irunllly IIsed throughout India. Ordinarily I hasta = \ ,'" \• • 2.:'1400 cms. So the height of a man equals 5 ft. 10 A ,'",III,,1'd lilli' of length came into use all over India only .... IItlll'," 1~lIll~. Old measures remained in the literature as III IIItllll )1'11111' transactions. Now miles have given way to • '''"' 1r"...,,1 I II IWO miles but now both kosa and mile will go.

=

~ ~ ~ ~ q~fa~s:ffl ~ 1f a~ I 'Itllolq)fi1C;:Q<4fi1~g~: q<4fi1~it» ~: ~: II III I I

... ,Il"" '''''I ,III'

2 yavas = I guifja (rattl), 3 guffjas = I valla 8 vallas = I dhara~la, 2 dhara1Jlls = I gadyiiIJaka 14 vallas = I dharaka.

Comment: This verse III gives various measures for weighing silvc or gold. A barley-com (with its outer shell) is called a yava '. Nowa days gold and silver are weighed in gms and so guifjiis are not in usc Guifja is a fruit with black, red or blackish red colou~. It is shaped lik, a green lentil but twice its size. Guiija, being tough and durable. has been in use since ancient times. Village goldsmiths use Guifjas fOI weighing silver and gold even now. One guifja weighs 1.2.. troy grain

16

Presently, gadyiiIJa or gadyiiIJaka and dharaka are not in use. But valh or va/a is still in vogue. One dharaka = 42 guiIjas; I gadyiiIJaka =4~ guifjas.

~~~~:~~I q;.N~

tWi

~: ~ ~ 'fjqu':ftSl't II IV II

5 guifjas = I m8$ii. 16 m~

= I kar!f'1

4 kar!f'1s = I pala. Those who know measures for gold call a golden kar!f'1 "suvarlla".

Comment: In the times of Bhaskarliclirya, the principal measure fOi weighing gold was a kan;a. 80 guffjas made one kar!f'1. The modem 3 measure tolii equals 96 guiJjas. I tola = 11- gms. 5

VI \I

,",tftI" - 4 kOo'tn.~. 10 hastas = I bamboo.

"'Iff' "',,., •A",:a

or a square with sides 20 bamboos. A ""lI1bcIII ur a pole is easily available all over for measur'h,' 1'I11ll1lh used 1 Pole (bamboo) 5\h yards. In the fifth

=

.... H"

I',

lIk. n UK unit of length which we can compare with

..." '·,,!!,I,. ' y"lcm, One nivartana is approximatelr two acres • II ,,"'K'J hec.: tftrc).

'u. e';r.ln.ln V lume "","IIull"uA: cmmi Q.,(Cfi1;ftSl't' 'Ii")fa~n ifl'II:H!lI~q)1 ~ I I VII I'

' " "'tllfl1'II "

_lIttllll' 1,1

""I Cube

i a solid which bas twelve edges each of

""' II,IIt,'/ h. volume is the unit. In Magadha state, quantity of

Units of Length q:"'c'h::S~:J('fDI\C+i~: m~: "'~:J~~t.,~: I

~~: ~: ~~dftd~'" ~" V "

"I I •• //!IiJLw·,,'

means a solid with twelve edges and not

•••"" "11'" P. II. Ihe 'alC Khanapurkara Shastri interpreted. A cube ,... h •. ,·,11" '. 1'1", ollly eight comers (vertices). The measure kharT M",""",. 111111' "ave been the standard measure for grains.

6

LIlavatT of Bhaskaracarya

Varanasi. Allahabad. Gaya. Patna which are in southern Uttar Pradesh or Bihar. fOimed an influential part of ancient India. Hence measures from this part of the country were taken as standard. One khiirT equals eight payalfs or 32 kg. 'Sastrodita' must be taken to mean Government order. ~

UT7Jf: mj

~: 'fQ1
~~ UICi4>t

~ 16

..!.4

khiirT

= I drona . • .! 4 drona . = I adhaka. .

adhaka = I prastha.

.

..!.4

prastha = I kudava.

.

l

'./IlIIllons and Tables

1II:lIlIing measures of time etc. are well-known. E.g.: nimi$O. the ~ IIII ullime. measures the time for a wink (of an eye) of a wise man. 1 III't", = 1130 nimi$O and tru~i = 1/100 tatpara 1 1I 1I1"'i~S = I ka$/ha.. 30 kii$/has:::: I kala.. () A"'i \. = I nak$Otra gha~ika.. 2 ghatika<; = 1 k$OIJLl. 60 ghatikas = 1 Day. II 11l;lIc measures of time: I "'11 ( ne complete breath) = Time taken to recite 20 long vowels. 1 pala. 60 palas = 1 gha~ika ) 'lrtll,kas = 1 Day. 30 Days = 1 Month. 12 Months = 1 Year. 1111 i the end of the chapter on terms.

This is what the ancients said. Comment: No reference is made in the verse to the shape of the measures. It appears that they were not cylindrical as they are today. They must be two frustumta of right circular cones joined together. as in the adjoining illustration. At the beginning of this century. adholI. Seer. Quarter Seer were used to measure grain. The words kuqava. a/have. ni~have are still in use in Konkana. The current Government measures are cylindrical and grain is now measured by weight.

m:

q,~cni$fl+"*'I~~: ~S'5I I ~ ~ ~\-Jt=
7

CHAPTER 2

Place ValDes of Digits "'1"'''<j'('i~'kq;f('#iZjf~ I

" "rlt

~ ~a4l"'(ij,qaq>''''d4 II

XI

II

this verse Lord GaI:lpati is invoked again.) I offer homage to that Lord GaI:leSa whose neck is adorned by IIIl1slIlg black and most poisonous snakes and who has unctuous lushkc a hlue lotus. (III

III

__ if'" IMiStll<;{I'IH$Q'I1'I4)1lQ: ~: I I~ ~4fq<94"fNOtiq;q'k'fDf
"

ltm'f.lr;;rQ' ~ ~ Gtl:J0ik,~ ~: 'I'm:

WRRi

llfq~';(jOd 11ffl: ~: II XII II

IIIlltions of the digits from right to left are unit, ten, hundred, thou'Id, len thousand, hundred thousand (/akh), million, ten ' million

tIIr ), hundred million, billion (abja), kharva. nikharva. ,"/r/tpadma. Sanku. jaladhi. antya. madhya. parardha. The value of I h digit on the left is ten times that on the right. This decimal sys-

9

10

LrliivalT of BhaskaracWyl

tem of place values was conceived by ancient scholars to make practical calculations simpler. Comment: Indian mathematicians discovered the decimal system in which place values are assigned to digits wherein values increase in powers of ten. The Greeks and Romans used letters to represent numbers with the result that the progress of Arithmetic was very slow in Eastern Europe. In India the use of the decimal system and the use of ten symbols (for 0, I, .... ,9) to represent any given number, made mathematical operations (addition, subtraction, etc.) easy. What Europeans can "Arabic numerals" were discovered in India and only recently some authors have started calling them "Hindu-Arabic numerals". These numerals were invented some time before 200 B.C. The current DevanligarT numerals have been in use in various parts of

CHAPTER 3

Addition and Subtraction

I,

India since A.D. 400 and the English numerals are their modified 1 \ forms. Although this verse goes up to pariirdha (10 17), there are terms 140 for numbers up to 10 in Sanskrit. I I

- --

r

, I

hull explain the method of addition and subtraction in half a stanza.

II,hliun or Subtraction should be done place-wise from right to left or II .. "I!rsa.

""''''ent: First write down the given numbers one below the other care to see that the digits match the places, that is, for example, U II ".1 ten's places should be in the same vertical line. Then beginlUll' with the unit's place add (or subtract) the digits, then move to I 1011'

In this procedure two things have been laken for grant!!d, " (I) II the sum of the digits in the same place is greater than ten, Ih !lIIIIC has to carry forward the digit in the ten's place of the sum to .' Idl; (ii) The number to be subtracted must be smaller than the "Ih~r .. umber. These things are not explicitly stated. Although addi111111 lind subtraction can be done from left to right also, this mehtod is '1'" III vogue. The method is given in brief because it is clearly ex, II' : .. .' .

I See Datta and Singh. History ofHindu Mathematics; and O.P. Jaggi's Indian Astronomy and Mathematics.

1I

II

12

f'

I

,(

,,'.'. r

LI/wall of Bhaskaracarya

pected of the teacher to fill up the details. If there are more than two numbers to be subtracted from a given nl!ffiber, then the procedure is to subtract the numbers one by one.

\tl ~ ~q*l"flmt ~"14~m"'It'
an) ~ "'tM,q~ ~

O! you smart girl 'Lilavati', if you are skilful in addition and subtraction, tell me the result when the sum of 2,5,32, 193, 18, 10 and 100 is subtracted from 10,000.

CHAPTER 4

Methods

Comment: As shown in the adjoining figure, write the numbers to be added one below the other, keeping their proper positions of digits. Then draw a horizontal line below the (places) 2 last number, viz. 100. Add all the digits in the unit's place 5 get 20, write 0 in the unit's place below the line and to 32 carry 2 to the ten's place. Now add the digits in the ten's 193 place (along with the 2 carried over) and the sum is 16. 18 ten's place below the line and carry over Write 6 in the 10 I to the hundred's place. The sum of the digits on the 100 hundred's place, together with the I carried over, is 3. 360 Write this digit below the line in hundred's place. Thus the required sum is 360. Now as in the next figure, write the sum 360 to be, 10000 subtracted below 10,000 and draw a horizontal line below ...=160 360 and the sign - before 360. In the unit's place, 0 subtracted 9640 from 0 yields O. In the ten's place 6 cannot be subtracted from 0 and so bring I from the hundred's to make it 10. 6 subtracted from 10 gives 4 and we carry i on the hundred's. Next in hundred's place, 4 subtracted from 10 yields 6 with I carried over. Finally, J' subtracted from 10 in the thousand's place, gives 9. The required answer is 9640. This explanation is to be taught by a teacher to the pupils.

, i

l'

of Multiplication

Nnw I tell the method of multiplication in two and half stanze'11f~ ~ E'~rs,",,~~qlqr~ ..r61' '1"~~10f('!JV'8g("q~: ~: ~ ~ lIll I XV II

,"""~: 1J~ ~ ~ ~ ~ ~~: ~ lIll It'" ~~ ~ ~: ~~ ~: ~: II XVI II "j"vR ~~G(~o<J'~M~ lIlll XVII I I

Ir I multiply the digit in the unit's place of the multiplicand by the IIlIiliplier, then the digit in the ten's place and so on up to the last h~11 on the extreme left. I

',m,,,,ent: This method can be used when the multiplier is a small

and the user has memorized the multiplication tables. Of if the product at any stage consists of two or more digits, one 1111 In carry over the part without the digit in the unit's place. There is 1111111

tHIiS ,

13

)., \

101,·-

14

~

no reference to this carrying over in the above rhymes; it is to be understood.

IIIlclct!~'I(iI)cl"lI" cfll'1lqPti ~

'" ~ql"i'(jOIl anm: r i ~ I dwma5fir;t'f1

.-q\tii'1fitot' i IU'8!}O,"

~ if)("Qlfq~

~ ~ ~ " .~ 3i1I;r:

Split Method Split the multiplier into two convenient parts. multiply the multiplicand by each of the two parts and add the results.

Comment: If the multiplier is 45. split it into 40 and 5 as it is easy to multiply by 4 and 5.

I

Factor Method If the multiplier is a composite number. factor it. Then multiply by one factor and then the result by the second factor and so on. (If the I multiplier is 45. first multiply by 9 and then the result by 5.)

Place Method Multiply by each digit ofthe multiplier separately and write the result in each case under its proper place. Then add all the results.

Comment: This product obtained by mUltiplying by the digit in the j unit's place is written down as it is. Then the one by the digit in the ten's place is shifted by one place to the left and written below the I first one. This procedure is carried on until all the digits of the multiplier are used. Finally add all these products. I Method of Adding or Subtracting Add any convenient number to the multiplier and multiply by the result. Then multiply by the added number and subtract this product , from the previous one. Instead of addition of a convenient number one ( can use subtraction too.

l

Comment: To multiply by 45. first multiply by 50 and then by 5. . Subtract the second from the first. The above methods give the fundamentals of multiplication. Toll elucidate them Bhaskariicarya gives the following example.

15

Methods ofMultiplication

LWivatl of Bhaskaracarya

ri

~II XVIII II

O! you auspicious girl with lovable eyes of a young deer. if you 1I:,ve well understood the above mehtods of multiplication. what is the product of 135 and 12? Also tell me what number will you obtain if Ihe product is divided by 12.

Comment: According to the first method we get 135 x 12 = 1620. Since 5 x 12 = 60, we write 0 in the unit's place and carry over' 6. 3)( 12 = 36 to which we add 6 to get 42. We write 2 in the ten's place 'Illd carry over 4. We add 4 to 1 x 12 = 12 to get 16 which is written III Ihe left of 2. (The explanation given by the late Khanapurkar ShasIri concerning this method appears to be incorrect.) To use the second method. we split 12 = 8 + 4. Now 135 x 8 = IOKO and 135 x 4 = 540. So 135 x 12 = 1080 + 540 = 1620. The third method utilizes the factors 4 and 3 of 12. 135 x 3 =405 and 405 x 4 = 1620. Of course. if the multiplier has no factors. then tJlis method cannot be used. If the multiplicand has factors we can inlerchange the multiplicand and the multiplier. The fourth method uses the digits 2 and 1 in unit's and 135 ten's place respectively. We first multiply 135 by 2 and get )( 12 270. Next we multiply by 1 and get 135 which is written by 270 shifting it one place to the left. Then we add the two num135 f(,20 bers to get 1620.

=

=

The fifth method uses 12 = 10 + 2. 135 x 10 1350 and 135 x 2 270. Then 135 x 12 = 1350 + 270 1620. Alternately 12 = 20 - 8. I) x 20 = 2700 and 135 x 8 1080. 135 x 12 2700 - 1080 = 1620.

=

=

=

In stanza XVIII "the number formed by 5. 3 and I" means 135 . ince the digits are written from right to left. I The multiplier is "Divakara" which means the Sun representing 12. In this stanza. Ihere are some vocatives used to address Llliivatl and we conclude that Ohfaskaracarya must have been teaching a female student. I

~ ~ ~:. _

"Digits move in revene 0l'I:l«".

CHAPTER 5

Division

I

Ivision is explained in the following stanzat'iin1rlMOr.J:

~ ~: +t'C ~ ~ I

: " ~"I'4qq(ilj £:1'Nf\i
r

'IfIlm

~

m

9 II

XIX II

It! the largest integer whose product with the divisor can be sub-

t ted from the extreme left hand digit(s) of the dividend. This intei. the first digit of the quotient. If the divisor and the dividend v "common factor, then the common factor can be cancelled and •. division is carried out with the remaining factors.

d

,"ment: To explain the process of division, we consider the examIII the end of stanza XVIII: i.e .• divide 1620 by 12. The first hod is:

•• 17

LllavatI of Bhaskaracarya

18

12) 1620 (1 35

12 can be subtracted from 16 once only and so the first digit of the quotient is I and the remainder is 4. 42 We then have to divide 42 by 12 and 12 x 3 = 36 36 can be subtracted from 42. So the next digit of the 60 quotient is 3. Now the remainder is 60 which is 60 12 x 5. So the last digit of the quotient is 5 and the 00 division is complete. The quotient is 135 and remainderO. In the second method we use the factors 3 and 4 of 12 and 1620. Dividing 1620 by 3 we get 540. And dividing 540 by 4 we get 135.

.!1...-

3)

1620 (540 15

4) 540 (135 _4_ 14

-.2

12 000 ....QL

and

00

~

20

J.L 00

In the above example the remainder is O. If the remainder is not and we follow the second method, we may end up with different remainders. for example:

l!rO

In

427 the quotient is 15 and the remainder 7,

28 and ~ the quotient is 15 and the remainder I. 4

CHAPTER 6

Methods of Finding Squares I I'll

Method

wnfttlrn: 'P~'(iii;j~Sl!I 'f\t/,<4I~ql~ ~!l0lt~~&i': I ..,,)qA'eo,&i ~SlRS;S~"'ffi'<J*'
Th

product of a number with itself is called its square. To square a n\illl\ler use the following procedure: First write the square of the exIle left-hand digit on its top. Then multiply the next [i.e., second] ,II hy the double of the first digit and write the result on the top. N,xl, multiply the third digit by the double of the first digit and write l'Csult on the top. In this way arrive at the unit's place. Next cross ). lirst digit and shift the number so formed one place to the right. '}'t.(JI1 repeat the same procedure. Finally add all the products written III Ihe top and the sum is the required square. leecmd and Third Methods WV'Qq'fq,fia~~~~ a'~o"q"O'Qg(t' ~ I

_19'("'~~: ~: 'flI'~f'Q

1ftlvt

~ 18' I XXI II

19

I

Lillivati of Bhaskaracarya

20 Second Method

Split the given number into two parts. To the sum of the squares of the two parts add twice the product of the two parts. The result is the square.

AI~lh()ds

of Finding Squares

21

Some writers take the sums of numbers between A and B which is 7KII, hclween Band C which is 936 and C and D which is 49 and add ,h .. 'I: 788 Here 2nd number is shifted by one 936 place and 3rd by two places.

----.!2 88209

Third Method Add and subtract a suitable number from the given number. Take the product of the two numbers thus obtained and add the square of the suitable number chosen above. The result is the required square. We QOW use the above-mentioned three methods and give the algebraic theory behind them.

The lale Khanapurkar Shastri and Pandit Hariprasada Bhagiratha not explained this point properly. i'his example can be solved by other methods:

huvc

"""CI/nd Method: 297 = 290 + 7, (290)2 = 84100, (7)2 = 49 and 2 x 290 · 7 . 4060. The sum of these three: 84100 + 49 + 4060 = 88209. I III,.d Method: We choose 3 since 297 + 3

'fld -;m;rt ,. ttgttlf1i 1J..~ ~tt1t'Q 'J('j~(H'4 I q~'dH'4''4S(f'C''Q 1fJ1 ~ ~i=,if.h.jji("'~rP{ II

(297)2 XXII II

O! Friend, if you know the method of finding squares, find the squares of9, 14,297 and 10005. 8

D C

8 1 8

2 2 1

0

9

4 6

9

Comment: We find the squares of 297 and 1000 5 only.

First Method: In row A write 297. We make five vertical columns to B 8 2 facilitate the work. Square of 2 is 4 3 6 which is written above 2. Twice 2 is 4 and its product with 9 is 36 4 which is written above 4, taking , 7 A 9 2 care that 6 is above 9. Then we 7 9 X Z multiply 7 by 4 and write 28 above ;( 11 7 Y 36 with 8 above 7. We then shift 297 to the right and cross out 2 as in the row X. We repeat the procedure: square of 9 is 81 written above the line B and twice 9 multiplied by 7 = 126 written above 81 with 6 directly above 7. Then we shift 297 again, cross out 2 and 9 as in row Y. Square of 7 is 49 wri tten \ above line C. Add all the products above the row A and the sum is 88209.

=

(300)

x

= 300. 297 - 3 =294. Then

(294)

+

(3)2

= 88200 + 9 = 88209. The firsl method uses Ihe formula: 2 2 2 { (I + b + C)2 = a + 2ab + 2ac + b + 2bc + c . In Ihe above examples these are 4, 36, 28, 81, 126 and 49, (and wiUl proper place values 40000,36000, 2800, 8100, 1260 and 49). It I. hl'cresting to note that the square of 97 is 8100 + 1260 + 49 = 9409. 2 2 2 '11111 we get 7 = 49, 97 = 9409 and'297 = 88209. The fonnula used in the second melhod is (a+b)2 = a 2 + 2ab + b 2 2 " lItI,he third method uses a = (a + b) (a - b) + b 2. Of course, the sec" 1111 lind third methods are not so useful when big numbers are invlllvl'd. In the first method care should be taken to write the products III p'!iper places. ny' he second method, (10005)2 = (10000 + 5)2

=

100000000

+ 100000 + 25 = 100100025. WI I 1I1so show the first method which is self -explanatory.

~Hf

22

LWivati oj BhaskaracaryG j

F

1

0

0

- --

1

0

0

0

2

5

2

5

,

\

I

I I

E

0

D

0

0

0

0

I

0

(

0

\

0

I I ,

CHAPTER 7

0

c

0

0

0

0

0

0

0

0

Square Root

I

0 I

0 I

B

I

0

0

0

0

0

0

NWCiI"i'4lfa .... j(t ~ ~OI4,c:f ~ ~

I

r---0

0

~ t1GltI~ ....Iri"lai ~ ~ I I

I

0

(S! 'rting from the unit's place, mark alternately vertical and horizonI

,

I

P

Q R

S

I

~ ~S*'4~ ....R'QCI\'iII'<1q" ~

. t1"fi!!)oj ;;qfl~M ~: ~ fQltc4('{11 XXIII II

0 I

A

......

I I

0

0

0

5

,,(

0

0

0

5

,(

,fJ

0

0

5

X

,fJ

;!J

0

5

X

;!J

,6

,6

5

I

•

hars above the digits so that the given number is divided into ' IOUpS of two digits each with the possible exception of the extreme I h group. The extreme left group will contain either one digit or two dlalta and wiJ) have a vertical bar on its top or on the right digit re'JII.~tively.) From the group on the extreme left, deduct the highest f"U.:tible square of a) (say). Then write 2a) in the neighbouring collilllll; this is called palikli (row). To the right of the number obtained (rrJIIl the above subtraction, write the digit from the next group with a hOlizontal bar. Now divide the number so obtained by 2a,; this quoa2 should not be more than 9. Now write 132 below 2a, after hifting it one place to the right and add. The result is the second /Jmikli. Write the next digit to the right of the remainder so obtained llil

"'"1

23

24

Lilavatl of B hiiskaracarya

'\'I/IIIJrI:

25

nllmber 7 in the root column and 7 )( 2 = 14 to its right with I hcl,)w 8. Add the two to get 594 which is the third parikti. Write the hllli Iii ,it 9 of the given number to the right of 4 to get 49. From this 2 . lItllr"Ct 7 = 49 to get the remainder O. The required square root is the IllIlIIber obtained by writing the digits from the root column in the ..!",In ill which we derived them. Thus it is 297. We can get the same IlIIlIIber as half of the third parikti. III the above method of Bhaskaracarya, we take one digit at a time. lit the current method, we take two digits at a time. low we tabulate the method of finding the square root of 100100025

and from that subtract the square of the second quotient a2. Now to the right of the remainder so obtained. write the next digit and divide this by the second paivai. This gives the third digit of the required square root. Now twice the third digit of the square root should be added on to the second parikti after shifting it by one place to the right. The result is the third parikti. Then write tre next digit of the given number to the right of the remainder and sublract from it the square of the third digit of the square root. Repeat this process. The result is the required square root.

Ihl ll

~~,,~~1{d~"ri~1 ~-'I'II.q.iq~l~ ~ ,;t~~4~ ()~ imfT II XXIV I I

' 001 0002 5(l

J

My friend! If you really have understood the method, then find the square root of 4, 9 and of the squares obtained earlier viz. 81, 196, 88209, 100100025.

(0

JI (K)

root I 0

0

Comment: We'll find the square root of 88209 by Bhaskarliciirya's method. The first thing is to 88209 (2 root parikti make horizontal and verti4_ _ 2 4 1st cal bars 88209. From its 4) 48 (9 18 9 first group, viz. 8 subtract 1L58 2nd 122 the highest possible square 14 7 !L which is 4. We get the first 594 3rd 58) 410 (7 remainder 4 = 8 ' - 4. Now 1 + 2 406 write 8 (from the given 297 297 049 number) to the right of the ..!L... to get 48. , remainder.4 00 2 )( 2 =4 IS the first parikti.

I

I

Divide 48 by 4 and see that the highest one digit quotient does not exceed 9. Here, the quotient is 9. Write this 9 below 2 in the root column. In the same horizontal line write 2 x 9 18 with I below 4. Add ) I the two to get 58 which is the secondpm*ti. Then subtract 36 9 )( 4 ' from 48 to get 12. To its right write the next digit 2 and we get 122. From this subtract the square of9 to get41. To the right of 41 write 0, the next digit from the given ~umber. Divide 410.by the second parikti ~ VIZ. 58 and get 7 as the quotient and 4 as remamder. Next we write f

=

Root

=

I

JU)

o() 1

(0

0

QQ...10 ~ l UU)

100

(0

QQL 1000 QQQQ.. l iIfIO) 10002 10000 25 25 00

(5

5 10005

parikti 2 00 20 000 200 000 2000 10 20010

1st 2nd

3rd 4th

CHAPTER 8

l\fethod to Find the Cube RmIlll7W1 'QO'f: ~: ~

Q.,)"f4't4 'ffit)~: I

.lfitRrlf.1tu'ffit~ 3"lea.'&Q6'?msql~q.,,'q ~ I I

"'iJil"'fl\(~'1 ~ tA: 't41d1"'{iQl

U§4'1""..g~"'lqltiqsM

ttt't!iag.i

~ ~

xxv II

{ffi)~ I

1ISl'd: II

XXVI II

t Method

of a given number is its product with itself thriCe over, If we 3 to find the cube of a number of two digits, say, lOa + b, write a fI' Below this, write 3ib by shifting this result one place to the 2 n v. ll. Below this write 3ab after shifting it one place to the right. 3 I r,!IlW this write b after shifting it one place to the right. Add all the , IIlIs, and the result is the cube. This procedure can be modified by III , IIlg from b but then each time the shifting should be made to the I I f there are more than two digits, then find the cube of the two ,'II

11111

27

LlIlivatI of Bhaskaracarya

28

digits at the extreme left and continue with the procedure given above.

W" XXVII

II

29

~"Ir.t Method

= = = =

(2)3 3 (2)~ 3(2) (7)2 (7)3

~ 'ZIT ~ '(,ftl~61: (lj:etl~qqj"t I

Q4f'lC'ftl.,: ~ ql(.itlflQo(\

II fj fhod to Find the Cube

8 84 294 343

8 84 294 343 19683

Second Method

So (27)3

= 19683.

Or split the given ...umber into two parts. Multipl) times their sum. Add this to the sum of the cubes of the we get the required. cube.

tw~

...<1

Sl~cond

Trike

Another Method The cube of a givem number is the square of the cube of square root of the given number.

Comment: The melbtod given in stanza XXV depends on the algebraic identity: 3 3 (a + b)3 =a + 3a2b + 3ab 2 + b . This further leads tl() 3 (a + b + C)3 = (a + b)3 + 3(a + b)2c + 3(a + b)c 2 + c which can be convoeniently used. The methods given in stanza XXVII depend upon the i d,entities (a + b)3 =a 3 + b3 + 3ab (a + b) and

omA

2 3

(a)

~

= lA

~(20)

Method

27 3 20 73 (7) (27)

=

(9)

~ ~ .hHH+
~ ~ m!l~ "".,ilffiU ~

'f:Asftr

tA

"Q'"11

~ .q I 1tlfffi ~: I I xxvrn

II

Friend. if your comprehension of cubing is deep. tell the cubes of 9. 27 and 125.

Comment: (9)3 = 9 first two methods.

x

9

x

9

= 81

x

9

=729. We'll find (27)3 by the

20 + 7 8000 343 11340

19683 In this method there is no shifting as in the first method. To cube 125 we group (12) 5. Ii = 1738 1728 3(12)25 = 2160 2160 :\(12) (5)2 = 900 900 (5)3 125 125 (125)3 = 1953125 Now we apply the last method to find (9)3. Evidently

3 2

(a) .

= = = =

3

=«3h3 =«3)3)2 = (27)2 =729.

CHAPTER 9

Cube Roots .,Ii q"f'(OQI"'f'i'OIEl=\ t

:J"f'ftt4~I,q.,JI ~ I

"" '1~ ~ ~ ~ ~ ~ q;c.;i ~

_ ....

II XXIX I I

~~ftt'fft''''~At4l ~ ~4"INH"'f41 ~ f.M*lC'f~q qfila if~qlt('I: ~ II

xxx II

rllw a vertical bar above the digit in the tmit's place of the number Whose cube root is wanted. Then put horizontal bars on the two digits if' lta left, vertical bar on the next and repeat until the extreme left~lIlId digit is reached. Prom the extreme left-hand section, deduct the highest cube possi. Ie lind write to the left side the number (a) whose cube was subVliCled. Write to the right of the remainder, the first digit of the next

"diun to get new sub-dividend. Now divide by 3i and write the quotient b next to a. Then wri.te the next digit from the section to the of the remainder obtained above. The next divisor is 3ab2• In the n'xl step lake b3 as the divisor. Continue Ihis procedure till the digits iii Ihe given number are exhausted.

n.ht

31

32

LWivatI of Bhaskaraclirya

Comment: Although a similar method was used in later books in Arithmetic, there are some dissimilarities. Bhaskaracarya takes one digit at a time whereas others take groups of three. Below we find the cube root of 19683. 19683 Palikti Root First we put the bars 2 --119683. First group is like 116 19 and the other 683. 84 3(2)2 (7) = 84 Subtract 23 = 8 from 19. 328 7 The remainder is 11 and 294 3(2) (7)2 = 294 we write 6 next to it to get 343 3 = 343 343 116. The new divisor is 7 000 3(2)2 = 12. We can go up to 9 but further subtractions are not possible and so we take 7 as the quotient. Write this in the root column. 3(2)2 (7) = 84 and 116 - 84 = 32. To the right of 32 we write 8, the next digit. The new sub-divisor is 3(2) (7)2 = 294 which is subtracted from 328 to give 34. We write 3 to the right of 34 and 343 = 73 which when subtracted from 343 gives remainder O. The required cube root is 27. By the same method we now find the cube root of 1953125. The cube root is 125. The current textbooks i9SHB Palikti Root on Arithmetic don't deal 1 1 with square roots and 09 3(1)22 = 6 cube roots. These are now 6 found by the use of loga35 3(1) (2)2 = 12 rithms or by the tables. 12 233 Such means were not 2 23 = 8 8 available in the days of 3(12)2 (5) = 2160 2251 Bhaskaracarya. The above 2160 method of Bhaskaracarya 00912 3 x (12) (5)2 = 900 is quite simple; this sim900 plicity disappeared in later; 53 = 125 125 5 methods. This method 125 may be extended to ob000 taining nth root of a positive in~eger.

CHAPTER 10

Eight Operations on Fractions Reduction or Fractions to Common Denominator: Simple Fraction 3I~"In:t'
mm:

'Ct ..qhfaQ...,~q't I

~ f'
A simple fraction is one whose numerator and denominator are both Integers. To add two simple fractions, first make both the denominators equal. This is done by multiplying the numerator and the denominator of the first fraction by the denominator of the second and vicev' rsa. This method can be extended to more than two fractions. Alternatively, divide the denominators of both the fractions by their common factor and then multiply the numerator and denominator of the fractions by quotients of the other. If a wise man multiplies in this way, there will be a common denominator for both.

33

34

LIlavatT of Bhaskaracarya 35

,'/ hi Operations on Fractions

Eumples ~ qql'4qfft ... ·h q)·"~l)a'=q< 9~t='i('11 ~ .. ~ .. ,·ltq "g~"i":
('omment: This is an example of a fraction which involves 'of'. Fraclion of a fraction is also a fraction.

, ample . 3 I I I I Fnend tell me - + - + - and - - 14 63 , I 5 3

Comment: As per the first method, we multiply the numerator and the denominator of the first by 5 x 3 = 15, second by 3 and the third by 5. Example J

3 I I -+-+I 5 3 45 3 5 45 + 3 + 5 53 =-+-+-= =15 15 15 15 15 We could have added two fractions first and add the sum to the I 15 I 16 3 third: - + - = - + - = - and I 5 5 5 5 16 I 48 5 53 -+-=-+-=-. 5 3 15 15 15 Example 2 I I In - - - we note that 14 and 63 have a common factor 7 and the 14 63 quotients (after dividing by 7) are 2 and 9 respectively. 1 1 9 2 7 1 So- - - = -- = =-. 14 63 126 126 126 18 Note that L.C.M. {3, 5} = 15 and L.C.M. {14, 63} = 126.

Compound Fractions

~~Hq,q~

wm ~ ~

,NlQit,.,,'fl:&t,itl~i(o,:

l)o:j ~: ~

~ ~ ~ lIffi good-minded

.. ,..Pla.,!1iil" I 4Sc::4onrlm«l., "if TJfitm unfi!t

v"""fiMJ{ II XXXIV I I

guy!

A

miser

gave

to

a

beggar

-I x -2 x -3 x -I x - I x -I) th part of a dramma. If you know com( 2 3 4 5 16 4 I und fractions well then, my dear child, tell me how much money the miser gave to the beggar?

('omment: The products of the numerators and the denominators are respectively 6 and 7680. So the result is _ 6_ = _1_ dramma = I kavaclI or Cowrie. 7680 1280 . So the miser gave the smallest coin to the beggar. This shows the nse of humour that Bhaskaliicarya possessed.

Addo and Dedo Fractions <61+i~l ~ l:I'1 ufilqs'('q 1fJ1TI

adil..,).,qs"t\q: I

~: ~ mt ffiI ~ " (1qjqq,~ I ~ ~ ~ ~ 9 a.t 1fTTTr-l11

mvr

xxxv II

Method J An addo fraction is one which consists of an integer plus a simple

The product of several simple fractions is obtained by dividing the product of their numerators by the product of their denominators. Then common factors of the numerator and denominator are cancelled to get the fraction in its lowest terms.

fraction whereas a de do fraction consists of an integer minus a simple fraction. To simplify such fractions, multiply the integer by the denominator of the fractional part and then add or subtract the numerator of the fraction.

36

LIllivatl of Bhaskaracarya

2 Comment: 7 + - is an addo fraction which is nowadays written as 3 2 2 . . 2 7 x 3+ 2 7 -. Similarly 7 - - IS a dedo fractIOn. 7 + - = = -22 3 3 3 3 3 which is a simple fraction. In current textbooks a 'rational fraction' is, defined as

!

b

where a, b are integers and b # O.

I I . . , I I (I) 1~ -.-4,3 - 4' Also simplify (I) 4 +"3 "4;

(Ill)

l

I and add -9 (I v) -1 - - I (-) 2

8

2

To add a part of a given fraction to itself, multiply the denominator of the given fraction by the denominator of the 'part' and write the product so obtained in the denominator. Then add the numerator and the denominator of the 'part' and find its product with the numerator of the given fraction. This is the numerator. To subtract a part of a given fraction from itself, suitable modification is to be made.

(1)

(Iii)

~ ~

35

lZiRl

35

~ ~

I"

I

12-1

4

I

-=12 3

12

:.

~}

. Ix(2+1) = 2=.!.. 6 6 2

%}I ...

3

2 x (8 - I) 3x 8

=.!i =2 . 24

12

:. 7 x (7 - 3) 12 x 7

28 84

=_ . 3

7

.. qliitt'{ I

VlI.,I'f4'"jciQ ~ ",ijqqltc.,'{ II XXXVI I I

it

">Iitft +qlllCiwrtl4t ~ "" ~ ~: ~: II 3!tI +qlllCitl~'; ~ ~ ~: ~: ~'iftijl~ ~ ~ ~ riswlljeiQjq
o friend!

I x (3 + I)

I~}

7

~: 't4sQtlgcm: 1I PtVlC;C4gd: ~: ~

9

If you know how to add a part of a fraction to that fraction and how to subtract a part of a fraction from that fraction, simplify

(IV)

~} I

2 .

:. I x (8 - I) = 2x 8 16

8

;6}.:. 7 x (7 + 9) Ipx7

.2.

11

"4 = 4' 3 - 4 = - 4 - = 4" .

8

I 2 7-2 5 I ---=--= - 35

I

f thoIS result to Itse . If.

2

By way of the present method, it would be 5

~}

0

7

C'"mment: 2 +

IIi)

.. ) I +-I "2 3

(11

~ - i (%) and [% - i (%)]- %[%- i (% )

Method 2

For example, consider the problem of adding .!. th of ..!. to ..!. . The 7 5 5 . 1(7 + I) 8 above method gl ves =- . 35 35 . I I I 7+1 8 The current method IS - + - . - = - - = - . 5 7 5 35 35 .. I 2 I I (7 - 2) 5 I Similarly, "5 - -:; . "5 = 35 = 35 = -:; .

37

Iti" " Operations on Fractions

112 112

=I

CHAPTER 11

Addition and Subtraction of Fractions

reduce them to the common denominator. If one of the terms is I integer, take I as its denominator and proceed. Next add (or subru 'I) all the numerators so fonned to make up the numerator of the 1111. The common denominator is the denominator. Cancel common l I rs, if any, and the final answer is obtained, , 5 7 5 7 4 15+91+156 262 Imment: + - + 4 =- + - + - = = 39 13 3 13 3 I 39 Ich is also the current method. II' I

mpJe

.mr
'fA1t II

XXXIX II

39

40

LfI'iivatI of Bhliskaracaryo

. o friend!

1 1 1 1 1 Tell me the sum of -, -, -, -, - and the result of sub· 5 4 3 2 6 tracting this sum from 3. Comment: L.C.M. of the denominators is 60. 1 1 1 1 1 12+15+20+30+10 8729 -+-+-+-+-= =-=5 4 3 2 6 60 60 20

3- ~ = 60- 29 =~. 20 20 20

CHAPTER 12

Multiplication of Fractions rh III

product of a number of fractions equals the product of their nurators divided by the product of their denominators.

1 7 5 35 ,. 1 \. IImment: - x_x_x_=_ 2

3

2

6

72

I llImple ",q "'CiNflh'l~'" f.!tt;i lft'cN'Iiijtj ~ ~ I

"" ~ ~ '" ~ l:~sftf ~ ~

lht II XLI II

o friend! If you are an expert in multiplication of fractions, tell what 1

1

1

I:he product of 2'3' and 2-=; .as well as of '3 and

•

'21 '!

1 1 7 15 1 I 1 tmment: 2 3' x 2 l' = '3 x l' = 5 and '3 x = '6'

'2

41

CHAPTER 13

Division of Fractions ' ~;:r~~~:1

mSQ ~

jU,."faRlh I I XLII I I

I)ivision of one fraction by another is equivalent to multiplication of Ihe first by the reciprocal of the second which is obtained by interchanging the numerator and the denominator. I!:xample ~'5Ii"",q~t1~" ~ ~

W « .q

~4'1"(J1'tldleu.!~'tlc;~

~I

a ~ ~ I I XLIII I I

o friend, if your intellect is as sharp as the sharp point of a blade of "uSa I grass, then give me the answers to 5 + 2.!. and .!. + .!.. 3

I

6

3

1)00 cynosuroides.

43

44

Comment: and

Li lavati of Bhaskaracarya I 5 7 5 3 15 5 + 2- = - + -= - x -= 3 I 3 I 7 7 I I I 3 I -+-=-x-=-. 636 I 2

CHAPTER 14

Squares, Cubes, Square Roots and Cube Roots of Fractions ~~~~tAr~1 • ~ 'til qc:gRf~ I I XLIV I I

III lind squares, cubes, ... of fractions, find respectively the squares, III)cs, . . . of the numerators and divide them respectively by the Illares, cubes, ... of the denominators.

( '",,,ment:

(2)2 3

= 7 x 7 = 49 3x3 9'

{8l = JP,I =~

"16

Jl6

4'

,~ ~ lf11 ~ ~ '" fit;( I , "" ~ '" tATm1l~ ~ lht ~ ~II XLVII

45

LIlavatI of Bhaskaracarya

46

Find out quickly, if you know how to fmd squares etc. of fractions, the I

square and the cube of 3 - and the square root and the cube root re2 spectively of the numbers obtained.

49

Comment:

4

{49 =J49 =2

'17

..[4

2

3

( 3

3_21)3

CHAPTER 15

= (2)3 ='23=8 7 343 2

f343 _ .J343 _ 7 Vs- 3.{g -2' 3

Readers would have noticed that in Bhaskaradirya's were no symbols for exponents nor for radicals.

tim~~,

......,

Eight Rules Concerning Zero ~u~~li~~:1 : 'fQI('(!j!)VI: li (J~ol~tq ~ II XLVI II

~ ~ 'GIffi li ~ ~ ~:I ~ ~ ~Q~q d'1'1f'ttH'Q ¥: II XLVII II

If :lero is added to a number, the result is the same number; the square

ote. (i.e., square, square-root, cube, cube-root) of zero is zero; any (lion-zero) number divided by zero is khahara, i.e., infinite; the prodII t of a number and zero is zero. (If in some mathematical calculations, multiplication and division by zero are likely to occur frequently then, though a number multiplied by zero is zero,) one should maintain the form of multiplicand (lIId multiplier zero in rest of the operations (until the final operation I~ reached). This is because if a number is multiplied by zero and divided by zero then the result is the (former) number. (For example,

47

48

LillivatI of Bhiska radirya

7 )( 0 should not be written as 0 because 7 )( 0 + 0 is 7; and if one writes 7 )( 0 0 and if this result is divided by 0 then the result (0/0) will become indeterminate, while 7 )( 0 + 0 7.) Similarly, if zero is added to or subtracted from a number then the number remain s immutated.

=

49

/':ight Rules Concerning Zero

Visnu (Almighty) due to dissolution or creation of abound ing living beings.)

=

= JO = = =

Comment: a + 0 = a for all numbers a. 0 0 = 3.JO = O. For any non-zer o number a, aIO 00. a )( 0 0 for all a. For non-zer o a, axO - 0- = a. a ± 0 O. Especially for the rule alO 00, Bhaska ra em2

=

=

3

phasizes in his Algebra that whenever such a situation occurs, it remains immuta ble in form and concept both, that is, it must be written as alO, and that any (finite) number added to it or subtracted from it will not alter its value. He points out that this type of mathem atics is used in astronomical calculations. The exampl e given by Bhaskara in L'illivatI (see Stanza XL VIII) and mathematical situations occurri ng in his astrono my suggest that by "110 is infinity" he means " lIh ~ 00 when h ~ 0 through positive values". However, the symbol 00 was not in vogue in ancient Indian mathematical works. Not only this, but several astronomers and mathematicians did not agree to Bhaska ra's concept of infinity . For example, Jnanaraja (c. A.D. 1503) says that infinity does not remain immutable when something is added to it or subtracted from it. However they used khahara for infinity. Khahar a means a fraction having a non-zero number in its numerator and zero in its denominator. It seems that the purpose of this usage of this symbol was two-fold: It was a notation for infinity. The khahar a-form (such as 5/0) was used quite safely for further calculations or reversing the process of mathematical operations as and when needed. The following stanza from Bhaskara's Algebra describes the nature of khahara (infinity).

31~~ili''(: ~

"'

~ Q~~'4~ f.!t:~ I

~ ~I~q~ilil~~~ ~ ~II 'There is no change in infinite (khahara) figure if some thing is added

~o or subtracted from the same. It is like: there is no change in infinite

Examp le

.,qtqg",q ftt Q; 1I'f v;;i ~ ~ 6:]oliWoq q,.1 ,............... ~ ~~ -I." ~. d~ 11 l{l11( ,......,m f.!tGmlgcmfftfil'" ~: Uldfft.. .

-=

!'lin

~:r"

II XLVII lII

~q~~ ~iiitqsqo" 1ft ~ ~: II XLVllI A II

JO.

Find (i) 5 + 0, (ii) 0 2,03, 3.JO • (iii) 0 )( 5, (iv) 10 + O. (v) A certain number is multipl ied by 0 and added to half of result. If the s~ so obtained is first multipl ied by 3 and then divided by 0, the res~lt IS 63. Find the origina l number : It is obtained by 'reve~se process (cf. next chapters). (i) 5 + 0 = 5, (ii) 0 2 = 0 3 = = 0, (iv) 10 + 0 = 10/0. (v) If the number is x,

JO

Comment:

3.JO

= 0, (iii) 0 )(

5=

1 (x x 0 + - x x 0) x 3

2

o

=63

9x x ~ =63_ :. x =14. 2 0 . I This is the answer Bhaska radlrya would expect. In modem tennmo 1 (x x h + - x x h) x 3 , I' 2 =63. ogy this is eqUIvalent to h ~O h that is

=

This gives x 14. It' unfortunate that this modem concept of limit in traditional I~nIS Id not be understood properly even by traditional IndIan guage cou ' 1 I savants. As a consequenc e, the theory of modem differen ual ca cu u!t was delayed by about 4-5 hundred years.

CHAPTER 16

Reverse Process u~ ~ ~

tR 1Pf

wi

~ ~I

~ ~ ~ ~ ,(1i=tlwRt,,1l11 XLIX I I ~ ~ ~ (1qrc(l~'~ ~ ~:

'I£1Ii

~t1@t~f('j'('('f~ ~ t)"jiffi4q, II L II

In this 'reverse process' of obtaining an unknown quantity from the known ones, the divisor should be taken as multiplier and vice-versa, the square as square root, additiqn as subtraction and vice-versa. Again if by adding to or subtracting from a given number its part, we should find the indicating fraction, viz. _N_ or _N_ as the case

D+N

D-N

lIlay be and then apply the reverse process. Of course, the given numher is to be multiplied by the indicated fraction. Next, subtract or add this result to get the unknown. 'omment: This will be made clear by examples: 0) Add 10 to a certain number. If 4 is subtracted from 215th of the sum so obtained, the result is 12. Find the original number.

51

LUavatI of Bhaskaracarya

52 Solution:

Solution: We have to add

~th

~th 8

of itself yields 33.

of the number. So multiply 33 by

3 3 3 . - - = - to get 33 x - = 9. So the answer \s 33 - 9 = 24. 3+8 II II

~ th part

=

14

14 + 14 (_1_) = 3-1

21

21

8

(iii) Find the number which when

53

.J196

12+4= 16. 16 + 215 = 40 40-10 = 30 Ans.

(ii) Find the number which when added to

Reverse Process

x

7 = 147

147 - 147 (_3_) = 3+4 84 + 3 =

of it is subtracted yields

84 28 Ans.

By Algebra, in the equation last but one above, x - x x

8

7

im-

= 84 .

15.

plies x x (7 ; 3)

3 ' Sol utlon: -3 -:_ 8- 3 5'

Hence, x = 84 x - = 147.

7 4

3 15 x -=9 5 .

Bhaskaradirya takes

15 + 9 = 25 Ans.

z + z x ~ = 147. Here also z x ~ = 147 or

4 z=147x-=84 7 .

Esample ~: 'Mq'(~4*,;fih':

~ = 84

io the required number is 84/3 = 28.

lfI
~~: ~ ~ iqqIW,
1. 
   ~~~~~Fi~~1
   
   -nm ~ it ~ ~ ~ ~~qf<JiQI'(11 LJII A certain number is multiplied by 3. To this product add its Divide the sum so obtained by 7 and then subtract
   
   .!. 3
   
   ~ th 4
   
   part.
   
   of the quotient
   
   from the quotient so obtained. Subtract 52 from the square of this remainder. Add 8 to the square root of this result. Lastly, by dividing this sum by 10 We get 2~ Then 0 you fickle-eyed girl, if you know the 'reverse process', tell me the original number.
   
   Comment: Using the reverse process and retracing the steps we get: 10 x 2 = 20 20 - 8 = 12 Ii = 144 144+52 = 196
   
   CHAPTER 17
   
   To Find an Unknown Quantity (Subject to certain conditions) \jt"q;t(iN
   To discover the unknown number, begin with any convenient number . Then according to the conditions given in the problem, carry on the operations such as multiplication, division etc. Next multiply x by the I ult given in the problem and divide this product by the number ohtained above. Thus we get the unknown by the method of supposiIon. ~
   
   ample 1
   
   ~:~~:~:I ,~,,~: ~ '<1~1<"'l'1f1'
   55
   
   56
   
   Lflavatf of Bhaskaracarya
   
   A certaill number is multiplied by 5, and
   
   ~
   
   ~xample
   
   Comme"t: Suppose the original number is 3. Then 3 x 5 = 15. x
   
   .!. = 10 .
   
   ~ 3(~)+3(~)
   
   17 12 + 18 + 12 + 9 =51 -= +3(±)= 12 4 12
   
   Siace the final result is 68, the unknown number :::
   
   ~= 68 x g
   
   =48Ans. 17 H 17 1 In this method of supposition, we have to add or subtract only parts of or multiples of the original number. Otherwise the answer may no
   
   be correct as the following example shows: Whtn 15 is subtracted from five times a certain number and the remainder is divided by 5 we get the quotient 29. What is the number?
   
   Suppose the original number is 10. Then 10 x 5 =50.
   
   IJ.~ ~ q~'('ld ~ l!I ~ ~ omi ~ l!I ~ tNt'liit'1 fltlJJ: I ~ ,utC'Iitl: tq'1q'ltt~: ~ ~ ~~: ~ .. ~~;ft&llil LlV II
   
   cave. One-sixth and
   
   ! th of
   
   river. One-eighth and
   
   !9 th of one-eighth were sporting in a pond fulI
   
   Sinte the result is 29, the unknown number =
   
   one-sixth were drinking water from a
   
   Comment: Suppose the number is 1. Then the last group of elephants
   
   11-
   
   (f f j) -(i i +
   
   x
   
   +
   
   x
   
   is
   
   ~ ) - G+ i x i) Jth
   
   Now the number in the brackets is 1 _
   
   4
   
   which
   
   is
   
   part of the group.
   
   ~ _ .±. _~ = _1_ 3
   
   21
   
   36
   
   252 '
   
   number of elephants = 4 x 1 = 4 x 252 = 1008 AIlS. 1/252 xampleJ
   
   1)0 the
   
   29 x 10 hi h' t 7 w C IS no
   
   correct. The correct answer is 32 and since 15 is not expressed as a part or multiple of 32, the method of supposition does not work. However. in this case. the correct number may be obtained by 'reverse process' (cf. the preceding chapter). In fact,
   
   5
   
   7
   
   )f lotuses. The lover king of elephants was leading three female elephants. If this was the situation, how many elephants were there in the flock?
   
   50 - 15 = 35. 35 = 7. 5
   
   29x5+15=32.
   
   2
   
   Of a group of elephants, half and one-third of the half went into a
   
   3 10+ 10 = 1. 1
   
   57
   
   of the product is sub-
   
   tracted fran the result. This is divided by 10 and to this quotient, half, one-thini and one-fourth of the original number are added. If this is 68, find :he original number.
   
   15-15
   
   To Find an Unknown Quantify
   
   aPiMi6Qi'tt,ftecit,G'Etjt''4e8: '~"4.,~R1:4f lR g4vr WI !JIfjjQC;qq ~: ~ ~: I
   
   ,..,i"tOl'fi"ttf&li ~Q'I'&J'~ ~ II LV II
   
   'rom a bunch of lotuses,
   
   ! rd are offered to Lord Siva .!. th to Lord
   
   V
   
   1
   
   1~J:lU.
   
   1 h
   
   6' t
   
   to the Sun,
   
   3
   
   '4 th
   
   ' 5
   
   to the goddess. The remaining 6 were
   
   I.IITered to the guru. Find quickly the number of lotuses in the bunch.
   
   LWivat1 of Bhiiskaracarya
   
   58
   
   (~+! +! +!) = 1 _ 20 +12 +10 +15 3 5
   
   6
   
   1 _ 57 = 1_
   
   4
   
   60 . 6x 1 So the total number of lotuses IS - - = 120. 1120
   
   60
   
   .!2. = .J..- . 20
   
   20
   
   1- (..!. +..!. +.!. +! +~) 634530
   
   ~ 41~l"ri ~
   
   ~ ~: tllJ.,(i('t.f(t: ~~ ~: I 1IJI
   ~~
   
   ~ 1li~ >.flfiN~
   '"
   
   In a coital sport of a couple, the lady's pearl necklace was broken. One-third of the pearls fell on the ground, one-fifth went under the bed. The lady collected one-sixth and her lover collected one-tenth. Six pearls remained on the original thread. Find the total number of
   
   Comment: Suppose the total number of pearls is 1.
   
   The final number i s 6. Remaining pearls
   
   1- 10 +20 +15 + 12 60
   
   =
   
   1- 59 =~ 60
   
   +2
   
   60'
   
   So the number of bees in the colony
   
   =
   
   =
   
   1-
   
   (.!.3 +.!.5 +!6 +.J..-) 10
   
   1 _ ( 20 + 126~ 10 + 6 ) 48
   
   4
   
   1
   
   1--=1--=60 5 5 6x1 :. the total number of pearls 30. 1/5 From this example it appears that amourous problems were not taboo.
   
   =-- =
   
   Example 6
   
   "'4d4g'lci
   
   liIS4fj"l~ ~
   
   ~ ~lMi~<11: ~
   
   1rlW G,IQl"l41",,< ~ $'1'*,'llJI41,(1 A~1t'<:Q"''lf ..... ql c1lld4100Q: Rlm'J:
   
   A pilgrim carried a certain amount of money. He gave away half the nmount (to Brahmins) at Prayaga. He spent two-ninths of the remaining amount in Kashi. One-fourth of the remainder was paid as duty. lie then spent 6IlOth part of the remainder in Gaya. Finally, he returned home with 63 ni$kas. If you know the fractional residues, find the amount he carried.
   
   Comment: Suppose he had
   
   ni$ka to begin with. He spent 1 in
   
   1 I'rayaga and so - is left.
   
   Example 5
   
   2
   
   ~: ~ 1"1'<~q;,(1: ~: ~
   
   Ql;Pci(i~
   1Jlffi!I
   
   "Ill:"
   
   = _1_ =60. 1/60
   
   ~ c;iZlQifloj ~ ~ ~~: ~II LVIII II
   
   =
   
   ~ ~
   
   =
   
   ~ glCldl
   perals in the necklace.
   
   ql
   friend! One-sixth of the bees in a colony entered a parali flower, one-third went to kadamba tree, one-fourth flew to a mango tree and one-fifth went to a tree blooming with campaka flowers. One-thirtieth went to a beautiful bed of lotuses bloomed by the Sun's rays. If only one bee was roving about, how many bees were there in the colony?
   
   Comment: Suppose the total number of bees is 1. Then the fraction which represents the remaining one bee is
   
   Example 4 i!f'HdlH<'f'60lJ1
   
   59
   
   o
   
   Comment: Suppose the total number of lotuses is 1. Then the number of lotuses left is 1_
   
   To Find an Unknown Quantity
   
   lfd
   
   iIS1 ~ ;'I't16IJ111 LVII II
   
   In Kashi he spent
   
   ..!.. ~ =..!. 299
   
   nnd he was left with
   
   ..!.2 - ..!.9 = .2. 18'
   
   2
   
   LIlavati of Bhaskaracarya
   
   60
   
   To Find an Unknown Quantity
   
   61
   
   Belt needed ..!.. x ~ = 4 4 16 1 3 1 Balance was - - - =-. 4 16 16
   
   2..
   
   . 7 1 7 Duly paJd = x "4 = 72 ;
   
   Is
   
   7 7 21 7 and so he has - - - = - =-. 18 72 72 24 7 6 7 Amount spent in Gaya = 24 x 10 = 40' 7 7 When he left Gaya he had 24 - 40
   
   ~th
   
   of the total jewels = 16jewels; 16 and so the total number of jewels were 16 x 16 = 256.
   
   7
   
   = 60' 63
   
   :. By the method of fractional residues, he had 7 160 = 9 x 60 = 540
   
   Example 7 ~s1\4ia'(¥:t'bi,qazitj ~
   
   tm:
   
   ~,~ .. fflI!Ju~ ~ ~ dC'IIqitl""StR: I
   
   ni$kas.
   
   ~ ~~O«iIl'~4>Mfi1q~
   
   Example 6A
   
   ~ ~ ~
   
   ~~ IJIq~11RIi ~ 1J.1'1 ~sfifm ~+et1#,v"fi;"'''S:m<'T ~ ~: ~ I ~ j"''''(1q~4~1''u,: " .. ~qst5Q"d:
   
   o deer-eyed
   
   ~ q~Ii(lfi,q,~ ~ ~
   
   1lvtri
   
   ~ ~ II LIX II
   
   A lover gave his fiancee some jewels for making ornaments. She used one.eighth of them for an ornament for the forehead. She used 3mh' of the remaining for a necklace. Half of the remainder were used to make annlets. Three quarters of the remaining jewels; along with little tinkling bells, were used to make a belt. Finally, she put 16 jewels in her wreathed hair. Find quickly the total number of jewels.
   
   Comment: .!.th jewels were used for an ornament for the forehead. 8
   
   So
   
   ~ 8
   
   remained.
   
   Those used for the necklace 3 4 I 8 8 2 1 1 1 Armlets used - x - =- ; 224
   
   Balance
   
   7 8
   
   ---= -=-.
   
   1
   
   balance was -. 4
   
   733
   
   =-8x7 - =-. 8
   
   d
   
   ~~:66Qi ~II LX II
   
   one! there was a colony of bees. One-fifth went to the
   
   kadamba tree, one-third to Silindhra tree.
   
   (i -i)
   
   x 3 were roving
   
   around kuraja tree. One bee was attracted by the fragrance of ketakI lind malati creepers and was moving here and there. Find the total number of bees.
   
   Comment: In the third line there is an exquisite figure of speech. It was as if the fragrances of ketakI and malati were indistinguishable lind the bee was unable to choose. Consequently the bee went once towards ketakI and then towards malati and vice-versa like a lover ICing called simultaneously by two messengers of his two beloveds. Suppose the number to be found is 1. Then
   
   [1-G+ ~) - (~. - i) x 3Jth I. SO the number of bees = 15.
   
   part is 1, i.e. 1- 18 - 1(j5 5
   
   = 1~ th
   
   part is
   
   CHAPTER 18
   
   Method of Transition
   
   Suppose we have to find two numbers whose sum and difference are given. Then add and subtract the given numbers and divide them by 2 to get the two numbers. This is called the method of transition.
   
   Comment: Let the two numbers be a and b. We are given c = a + b, and d = a-b. c+d , c-d Clearly a = - - • b = - - .
   
   2
   
   2
   
   ~: ~ ~ ~: qqta,,~: I
   
   ~
   
   mit
   
   ~.q lIffi ~ ~ ~II LXIII I
   
   o my dear child! if you know the method of transition. find two numbers whose sum is 101 and whose difference is 25. Comment: By the above method, a=101+25=63' b = 101 - 2S =38. 2' 2
   
   63
   
   CHAPTER 19
   
   Square Transition CASE -1
   
   If Ihe difference of Ihe squares of Iwo numbers is divided by Iheir difference, we get Iheir sum. Then using Ihe method of Iransilion, we gel Ihe numbers.
   
   Comment: In Algebra we have i
   
   a2
   
   - b2 = (a + b) (a -
   
   b).
   
   b2 .. ---=a+b. a-b _
   
   [sample '("q)4q)ffq)·~~ tfcop('l}1t" ~:wat I ~ dI mft tftli .rfiJltfOll~ II LXIV II
   
   fitIri
   
   a malhematician! len me two numbers whose difference is 8 and Ihe difference of whose squares is 400.
   
   65
   
   66
   
   LIliivatI of Bhaskaracarya 400
   
   67
   
   Square Transition
   
   Comment: - = 50 8 a + b = 50, a - b = 8. 50+8 50-8 a=-2-=29, b=-2-=21.
   
   Similarly,
   
   CASE-l/
   
   2nd Mehtod: If x is any number, then
   
   2 2 a4 2 2 b-a-I=4+ a + l - a - l =
   
   a = I + 2x 2x
   
   $'Ic:;1'~,<1IC1~'(\' ~ ~ ~~~., I
   
   ~: ~ ~<~, ~ tM mtr:11 LXVII
   
   ~ ~u)IIC&d ~ ~!lfIit~ ~ I '1if"~f?lqM Q 11IiT llmIi 'tfQf ~:
   
   The second line of stanza (LXVI) poses the problem: Find two numbers 2 2 a, b such that a + b ...:.1 and b2 _ a 2_ I are both squares (a < b). First Method (lXV): Take any nwnber x (say). Compute a = (8x 2 - I) 12x.
   
   8x - I a2 a = - - - , b=-+l. 2x 2
   
   a2
   
   = -4
   
   4
   
   a 4"
   
   + 2a2
   
   2
   
   (a + 8) 4
   
   -
   
   [
   
   = (8X2 -
   
   1)2 (8X2 + 1)2 64x
   
   4
   
   2 I6x + I + 32X2 ]
   
   4X2
   
   =[64X2 ~ 1]2
   
   J
   
   8
   
   (49)2 R d 63 2 2 2 ("8 ) ,b -a -1= 4 . eaerscan
   
   ..
   
   (II) X
   
   = I, a =
   
   3 b -, = I, 2 2 2 a +b - I
   
   = -94' a2 -
   
   2 I b - I = -.
   
   4
   
   NOTE: We can extend the method to get
   
   + a2 + I - I =
   
   64X
   
   2x
   
   check with x = 2, 3, 4.
   
   2
   
   4
   
   4x
   
   The second method is easier than the first. 7 57 (i) If we take x = I, we get a = -, b = - , a2 +b2 -1=
   
   Comment: Jst Method: Of the two methods, the first one i s rather: tedious. Its algebric explanation is as follows:
   
   4"
   
   (I + 2X22 )2 +I - = I (I + 2X2 )2
   
   .' _b' - I = 1+ 4x' ::,x' - 8x' = (' -2~X'
   
   Then compute b = i 12 + I.
   
   Now a 2 + b2 - I = a2 + a
   
   b= I
   
   ,
   
   2
   
   Second Method (lst line of LXVI): Take a number x . Then the two numbers are 1I2x + x and l.
   
   2a 2)2
   
   2
   
   2 2 a+b-I=
   
   II LXVIII
   
   (
   
   2 - C
   
   i
   
   + b2 - c2 = d2 and
   
   i -
   
   b=
   
   2
   
   = e.
   
   Another Derivation of the First Method: Let the two numbers be band 2 2 2 II + I, (a + 1)2 - b - I = a + 2a - b and so it is a square if we choose 2a = b2. Now (a + 1)2 + b2 - I = i + 2a + b2 must also be a square, 2
   
   4
   
   b b I. e., by choosing a = - , - + 2b 2 2 4 2
   
   b
   
   =-
   
   2
   
   4
   
   (b2
   
   + 8) must be a square.
   
   II is sufficient that b + 8 is a square. So b = 2t -
   
   8x2 2
   
   h + 8 = 4t
   
   2
   
   -
   
   .!. t
   
   for some t, since
   
   I + 8 = ( 2 t + t1)2 . Now put t = 2x. 4 + ""t2
   
   . LWivati of Bhaskaracarya
   
   68 I Then b = 4x - 2x
   
   8x 2 - I b2 and a + 1= - + I are the two numbers. 2x 2
   
   =~
   
   Esample
   
   m4)44l: 1'~fa4)1'1~ ~ ~ 1fIri': (f\
   
   lf1{
   
   Square Transition A derivation of this method is given below:
   
   Suppose the two numbers are a + I and b. Then 2 2 2 2 (a + I) + b - I = a + 2a + band . 2 2 2 (a + 1)2 - b - I = a + 2a - b must both be perfect squares. So we put 2 2 2a = n and b = 2an which makes them squares. Thus b2 = 2an = n3
   
   ~ 81
   
   ~ "\tIII~I~ ~~ ~:
   
   4
   
   and :. a + I = 8x + I.
   
   ~cUcm'ICI'lfittd ~: I I LXVIII I
   
   find two numbers such that whether I is subtracted from the sum of their squares or difference of their squares. the result is a perfect square. 0 friend! even intelligent mathematicians who know the six algebraic methods fumble in trying to solve this problem.
   
   Comment: By the first method, choosing x =
   
   ..!.x8-1
   
   a=..:!4~--
   
   ..!. 2
   
   mm I
   
   ~ ",*t'Q~qsQf II LXIX I I
   
   Some find algebraic rules as difficult as the arithmetic ones. However. many would not find any difficulty in the above six or other rules. ~~~~,.~~:
   
   a 3 + I= 2 2
   
   I and b = -
   
   In arithmetic. "the rule of three" is the essence whereas in algebra. a clear intellect is necessary. Many mathematicians may grasp easily but others would require explanations.
   
   I Ix 2
   
   3 2
   
   a=--+I=- andb= 1.
   
   Esample ~ ~ ~ <'I,qliC'fi!j0ft glA: I ~ ~~~II LXVIII I I 3
   
   Choose any number x. Then a = 8x4 + I and b = 8x are the two num- . bers.
   
   Comment:
   
   1lui '.l4fTfiJJq
   
   'lCl"'lClI.,i
   
   ~ ..,!;ift."qffi q"<'4jq;j~ II LXX I I
   
   2 By the second method. choosing x = I,
   
   mft
   
   qJC!hl~
   
   ~
   
   we have
   
   2
   
   2 x..!.
   
   ~
   
   69
   
   a 2 + b2 _ I = (8x 4 + 1)2 + 64x 6 - I 8
   
   4
   
   6
   
   = 64x + 16x + 64x = [4x 2 (2x 2 + 1))2; a 2 _b 2 _1 =64x 8 + 16x4 + 1-64x6 -1 = [4x 2 (2x 2 - I)f
   
   CHAPTER 20
   
   Quadratic Equation 9vtt"l'l('f)'1j?lf(l ;m){"C:f(I ~ ~I
   
   wt ~"' ~ ~ ~ U"S'
   the above -
   
   4
   
   and find the square root of the result. Next subtract or
   
   add (as the case may be)
   
   ~ 2
   
   and square the result. This is x2•
   
   ~ (ijh'*''''j?t: ~ '
   If we are given x ±
   
   ..!. n
   
   2
   
   x ± bx. then multiply by
   
   _1-. and proceed I ±_ n
   
   as before.
   
   71
   
   LT/avatT of Bhaskaracarya
   
   72
   
   Comment: (LXXI): This is the standard method in quadratic equab2 2 . . h 2b b =c + - . lions. x2b ± x=C. IS gIVen. T en x ± x + -
   
   4
   
   C
   
   Then x ±
   
   ~
   
   +
   
   b2
   
   b
   
   2
   
   4
   
   2
   
   4"' Suppose c + 4" =p .
   
   r
   
   =p and x2 = (p +
   
   %
   
   In current practice we take ax 2 + bx + C = O. Bhaskaracarya takes a = 1 and --(; > O. He also takes only positive square roots, whereas we
   
   Quadratic Equation
   
   ~ ~I"'N'tnl('j~1
   
   t1MI
   o learned one, find the square number x2 such that x 2 + 9x = 1240. Comment: We have to solve x2 + 9x = 1240. x 2 + 9x + !.!. = 1240 + !.!. = 5041 4 4 4
   
   :.(x + ~r =(:lY
   
   !
   
   n
   
   49 16
   
   1('1<:"q;
   
   r i 4lM'rrrcn,r#tA
   
   -i
   
   +-
   
   = 961.
   
   ~ ~
   
   ~ ¥~fJ;4j(:ijr(iffl
   
   ~ ~flg'I'lf4 ,. ~
   
   There was a flock of swans on a lakeside. Seven times half the square root of the number of swans were moving about near the lake. One amourous pair of swans was playing in water. How many swans were there? 2 Comment: The quadratic equation is x x = 2. X
   
   2
   
   ~ ..., .. q~;ftq""D"I<'lCi""'h""'k'Iej(t I
   
   ~ ~~«'''~'4'<J' I ~ ~fAq)('1~ ItH"'~'C11·1f't ~ ~ ~ '«I('f'!F",gQIVI't II LXXIll I I
   
   2
   
   :. x
   
   ~ ~fl~('ff'Q
   
   1f'(f('f~('f'F"<:c"R ~
   
   7 :. x2 - -
   
   2
   
   2
   
   Example 3
   
   Example 1
   
   ri
   
   71 9 :. x= - - - = 31
   
   1±!
   
   n
   
   2
   
   Example 2
   
   ..[l6 = ± 4 . So in
   
   1±!
   
   -~) 4 =2 4' and x =4 , x = 16.
   
   :. (x
   
   the above working x > 0, p - ~ > O. Also in2 stead of solving for x, Bhaskariicarya solves it for x . 2 (LXXII): Here the equation is x2 ± x ± bx = c. When simplified n this becomes x2 ±~ = _c_ and then (LXXI) can be used.
   
   take
   
   73
   
   81 =2 + -49 =16 16
   
   litml
   
   ~ 1fi: II LXXV II
   
   A flock of swans contained x2 members. As the clouds gathered, lOx went Miinasa Lake, and
   
   ~
   
   2 x flew away to a garden of Hibiscus Mu-
   
   ta/is. The remaining three amourous couples played about in the water. 0 young woman, how many swans were there in that lake full of beautiful little lotuses? Comment: Here the equation is x2 - lOx x2 = 6. 8
   
   .!.
   
   7
   
   2
   
   :. - x - lOx = 6
   
   8
   
   •
   
   2
   
   80x
   
   48
   
   .. x ---=7 7
   
   Lilavati of Bhaskaracarya
   
   74
   
   Quadratic Equation 1936 48 (40)2 (40)2 2 80x -=-. x - - + - - = -7+ -49 49 49 7 ., 2 40 44 + - 7 = 12 and x =144 . ... X = - 7
   
   (2X2) = 2 i e , Comment: Here the equatio n is 2X2. - X-! ' .. 9 2x2 --x=2 9
   
   Example 4
   
   9
   
   2 :. x - - x =9
   
   m
   
   qpj: ~ -qJ.lo,"cri ~ ~ ~ ~ (f1Ui(.,oj 'l~'Ifig11l'''''1'
   
   :. x2
   
   ~ ~ 1U'Ji tJf\ii ~ ~uhil~ ~:
   
   ~
   
   tR1II
   
   a ~: m
   
   I, LXXVI, ,
   
   2
   
   X
   
   Conr",ent: Here the equation is x - i.e., x
   
   :. x
   
   2
   
   -
   
   :. x =
   
   2
   
   -
   
   2
   
   - 4x = 6 + 3 + I
   
   •.
   
   = 10
   
   r (~r
   
   f
   
   16
   
   16
   
   225 16
   
   =
   
   "4
   
   Examp le 6
   
   m 't'~ji(":'C::"~:
   
   ~:
   
   ~'f1~,:q,
   
   urn; tlt1C,c::t,4S
   8x = 20
   
   8x + 16 = 20 10, x2 = 100.
   
   9
   
   2
   
   15 9 2 :. x = 4 + = 6 :. 2 x = 72 .
   
   x\
   
   x
   
   ! x + !!. = + !!. =
   
   -
   
   :.( x -
   
   Arjuna became furious in the war and, in order to kill Ka~, picked 2 he destroyed up some arrows (say ). With half of the arrows (Yl all of Kart:la's arrows. He killed all of K~a's horses with four times the square root of the arrows (4x). He destroyed the spear with 6 arrows. He used one arrow each to destroy the top of the chariot, the flag, and the bow of Kart:la. Finally he cut off Kart:la's head with another arrow. How many arrows did Arjuna discharge? 2
   
   75
   
   LXXV DIII
   
   2 . o a certam number (x ), the 18 times of the square root of the num-
   
   her (viz., 18x) is added. When one-third of the number (ViZ.
   
   + 16 = 36 = 62
   
   2 Comment: Here the equation is x + 18x +
   
   Jln.o.F"'C("f'ic4 "Iffffifl ~
   
   M
   
   II
   
   From a group of black bees (2x\ square root of the half (x) went
   
   . 8
   
   the rna/ali tree. Agam
   
   "9
   
   !
   
   3
   
   2 :.4x + 54x = 3600
   
   ~ftJi'f ... qq"ji"t
   LXXVII
   
   is
   
   tell lidded to this sum, the result is 1200. If you know arithme tic well, the number.
   
   Example 5
   
   qR'tNtj aj ~ f.w.I1{ ~ ~ ~ ~~:cf&l,'t II
   
   ~ x2 )
   
   to
   
   27 2 • .. x +-x= 900 2 2 :. x + 27 x + 729 = 900 + 729 16
   
   2
   
   th of the bees went to the ma/ati tree. Of the
   
   remaining two, one got caught in a lotus whose fragrance captivated , him; he started wailing and his beloved responded. Then, 0 beloved how many bees were there?
   
   :. (x + :7r =
   
   151~9 =c~3r
   
   123 27 :.x= ---= 24 4
   
   :. x 2 = 576.
   
   4
   
   16
   
   x2 = 1200
   
   CHAPTER 21
   
   The Rule of Three S4'1IUI~tOl ,. ~'1I;1\JII~:
   
   ;jf,fJ""d41+C'
   There are three quantities involved herein. The first one on the left (a) is called pramliI:Ja (scale), the second (b) phala (result or fruit) and the third (c) icchii (desire or requisition). The answer to be found (d) is called iccha-phala (desired result). Here a and c must be of the same . . . d bxc kind and b should be dIfferent from a and c. The formula IS = - -
   
   a
   
   which is of the same kind as b. In a reverse case, the inverted rule of three is applied.
   
   77
   
   LTlavatl of Bhaskaracarya
   
   78 Comment: (i) ~ b
   
   =..:.d
   
   in case of direct variation. and (ii) ab
   
   =cd in
   
   case of inverse variation. So in the first case d = b x C and in the
   
   a
   
   79
   
   The Rule of Three Example 2 gcpeqs1'N~~"\lCiJ1
   
   -.C't"ti<4d
   
   ~....,q9""'9
   m ~ ~: ~: ~: fit;lrrq~ m! ~ II LXXXII I I
   
   axb
   
   second case, d =- - .
   
   63 palas of purified camphor is sold in the market for 104 ni!jkas. o friend! tell me, after giving a thought, what will 124"1 palas fetch.
   
   C
   
   1'111,,",, ~ ~ ~
   
   umrn I
   
   'ftCfl"ffIt; "ffiiI jlj "~(tll>)f,,~: II LXXX I I
   
   Comment: This is an example of direct proportion. I
   
   ,63 P : 104 N : : 12'4 P : d.
   
   If, in a given situation, the desired result (d) increases (respectively decreases) as the requisition (c) increases (resp. decreases) then the simple rule of three is applied by an adept mathematician.
   
   So the answer is 20 ni!jkas, 3 drammas, 8 paI)as, 3 kakiI)1s and II ~
   
   Example 1
   
   varaJakas.
   
   ~ ~ ~ ~""'t1'<'1 .. ~a~fi\li~ I ~ ~ ~ ~ ~ ~....,.,q~., ~II LXXXI II
   
   If 2
   
   f
   
   (pala) of saffron costs
   
   ~N
   
   (ni!jkas) 0 you expert business-
   
   man, tell me quickly what quantity of saffron can be bought for 9 ni$kas. m~
   
   COMment: 5 guifjas = I m8# and 64
   
   guiIjas.
   
   2
   
   f
   
   pala = 800 guifjas.
   
   = I pala. So I pala = 320
   
   The rule of three gives
   
   5
   
   3
   
   5
   
   '7 : 2"
   
   So d = 49 x 104 = 182 = 29~ N. 4 63' 9 9
   
   9 x '2 105 : : 9 : d and thus d = 3 17 = T palas. This is direct pro-
   
   portion, since more money buys more saffron.
   
   NOl'E: Here kumkuma means saffron and not the red powder that is applied to a forehead, as the late Khanapurkar Shastri takes it to I
   
   mean. In 1937, two rupees bought 82" lolils of saffron. The red powder is cheap even now.
   
   Example J c:wuii~'" ~ t"R4t'ro:§a~1 "MI
   
   lht
   
   ~O''ft'<'lt'(ll ~ ~ ~ I I LXXXIII I I
   
   If I.!. kharikas of rice can be bought for 2 drammas, then how much
   
   8 rice can be bought for 70 paI)Os?
   
   Comment: This is also an example of direct proportion. 9 32 P: - K: : 70 P : d 8 :. d = 70 x 2. x_I = 315 = 2 59 K. I 8 32 128 128
   
   (Here P is for paI)a and K for kharika.) The rule of three perfected in India, and radiated in the eighth century. In Europe, it was held in very high esteem and called the Golden Rule.
   
   CHAPTER 22
   
   Inverse Proportion $'llI'''~ ~ ~ ~ ~ ~, Uffii ~ ffiI 1fti "~M",)f~t: II LXXXIV II If, in a given situation, the desired result (d) decreases (resp. increases) .as the requisitjon (c) increases (resp. decreases) then the inverted rule of three is applied by an adept mathematician. Comment: Common sense (which is uncommon!) should be used to determine whether the fourth term is going to be smaller or larger than the second term. Then use the rule of three. At present unitary method has replaced the rule of three and in the West, the rule of three has become obsolete. Naturally rules of five and higher numbers have also disappeared. The authors are of the firm opinion that the rule of three is simpler than the unitary method. ~ ~ ~ ~ ~ ~ ~
   
   Uffii
   
   Ptf.tl
   
   ~ ~:II LXXXV II
   
   81
   
   Lilavatf of Bhaskaracarya
   
   82
   
   Vigorous age of a creature is inversely proportional to its price. Same is true of the relations between the desired fineness of gold and its weight, and given unit of measure and the quantity of grain.
   
   Example 1 ~ l\ct "':e¥llfffi;qg""'jHl ~ ~1.:'.~qsq~'RIC:l ~ I I LXXXVI I I
   
   If the price of a sixteen-year-old female slave is 32 ni!jkas, find the price ci a twenty-year-old one. An ox which has been yoked for two years cost's 4 ni!jkas: what will an ox which has been yoked for six
   
   CHAPTER 23
   
   years cost?
   
   The Rule of Five
   
   Comment: (i) Price decreases as the age increases. Sod=
   
   ~= 128=25~N. 20
   
   5 5 . 4x2 4
   
   (ii) Here the price IS - 6 -
   
   1
   
   ='3 = 1 '3 N .
   
   Elample 2 ~ ~ ~ fmlvl ... q~IQJ~ I cro « ~~~d't I I LXXXVII I I
   
   ~ ~Iilqoi~
   
   One 'gady8r)aka' (= 48 'guifjas') of 10 carat gold can be bought for 1 N. How much 15 carat gold can be bought for the same price?
   
   Comment: Finer the gold, less can be bought for the same price. So quantity of 15 carat gold =
   
   10
   
   48
   
   15 x 1=32 gUifjas.
   
   .j.,tfIC'I ... q'lJftlIlS'~¥~~ 'M'~C:l' I ~
   
   ..!(,~~ ri
   
   ~ ~II LXXXIX I I
   
   In the case of examples on the rules of five, seven, nine, etc. keep the antecedents of all proportions in the numerator. All the other terms, except the desired result. should be kept in the denominator. The product of the numerators divided by the product of the denominators is the required result.
   
   Example 1 ~ ~ ~ q3iqctii"lR'l~ ~ ~ ~ ~ ~ ..).nli.", I
   
   1f11 Example 3 ~~mfI~~1 ~ ~ Uffii cro q~ICi¥" ~ I I LXXXVIII II
   
   m
   
   {tID
   
   lP.Rl
   
   '1i'f lIS i"Il "'d'lf'Ql *(
   
   witA~~~IIXCII
   
   There are three problems in this.
   
   With a measure of 7A (acJhakas), a certain quantity of grain measures 100 units. How many units will there be if the measure is 5A?
   
   Comment: This is also an example of inverse proportion. The answer . 100 x 7 -140 uru'ts.
   
   IS - - -
   
   S
   
   83
   
   LTlavati ofBhaskaracarya
   
   84
   
   The Rule ofFive
   
   (1) If 100 ni!jkas (= N) fetch 5 N interest per month, find the interest on 16 N for one year.
   
   Interest
   
   85
   
   , , 26.
   
   , . "'5 . x
   
   16 x 12 x 5 __ 48 __ 9 _3 N =---100 x 1 5 5
   
   direct} direct
   
   In a tabular form, the rule of five is: 100 N Principal: 16 N princiPal} I month: 12 months : : 5 N interest: x
   
   Example 3
   
   (2) The above problem is altered: at the same rate as in (1), find the
   
   period to fetch 9 ~ interest on 16 N.
   
   In the above, the first proportion is inverse since if the principal is less, the period is longer to gain the same interest. 100 xl 48 So x = - - x - = 4 x3=12 months. 16x5 5 (3) Suppose we are given the period and interest and we have to
   
   1 M : 12 M x=
   
   For 100 N (ni!jkas) 8 pieces of multi-coloured embroidered (clothing) material, each measuring 3 cubits x 8 cubits, are available. 0 businessman, if you are proficient in business, tell me quickly the price of
   
   . 3 -1 cu b'Its x -1 cubit, a pIece 2 2' Comment: This is an example of the rule of seven.
   
   3 cubits: .!. cubit 2
   
   find the principal (x). Here 100 N : x
   
   ~ ~~f?q ~
   
   tul~~~ ~~ ~ ~ CJfOjiq,fO,"'Q* ~ ~'I XCII Ii
   
   100 N : 16 N ) ' mverse } 48 : : 1M : x 5 N:"'5 N direct
   
   N) : :
   
   (fi(jliCtilfi4n'
   
   ~~,<,«f~~~'lQflq;, ~ ~ tml{1
   
   Lettbe period be x months.
   
   5N:48 5
   
   ~ ftntm:
   
   d'} nect
   
   8 cubits: '2 cubit" 2 8 :1
   
   inverse
   
   100 x 48 1 x-= 16 N, 5 x5 12
   
   direct} lOON : x direct direct
   
   100 7 1 .. x = - x - x - x - - 1 2 2 3x8x8
   
   = 175 N .
   
   Example 2
   
   192 ~'SIiM'~ ~ ~ffqltqiC1I"'"
   
   q3iQ
   
   ~: 'rllli(iiqrtQAk""'~:
   
   If the interest on 1()() for 1
   
   ~qsqqitll: I
   
   qSC:1:Jf4Jt1i f$l{ I I XCI
   
   .!3 months is
   
   1
   
   62 - for 3 - months? 2 5
   
   Comment: The rule of five is:
   
   II
   
   5.!. what will be the interest on
   
   5'
   
   Example 4 ~ l)S4ffil(lj"'~!lC'fI: ~
   
   "g4.d!JC'ft ~ tTt11I I
   
   ~ ~ q'l(t1~i('f~.n,C't'f~
   
   ~ ~ l)1ri "g4~t1I: ~lI'l:''''~'Wd~~~11
   
   xcm
   
   II
   
   86 30 planks of wood, each measuring 14 cubits x 16 fingers x 12 finget~ cost 100 N (n;'$kas). Find 0 friend. the cost of 14 such planks WhOM dimensions are each 4 measures less than the former.
   
   I I
   
   Comment: This is an example on the rule of nine: 14 cubits : 10 cubits 16 fingers: 12 fingers 12 fiIOgers : 8 fimgers
   
   : : 100 N : x
   
   All are direct proportions.
   
   : 14
   
   30
   
   :. x=
   
   lOx12x8xl4xlOO 50 2 =-=16-N 14 x 16 x 12 x 30x 1 3 3·
   
   CHAPTER 24
   
   Example 5
   
   Rules for Barter
   
   ~ l} "Aq ....H~(1g~(14l 'I~id ~:
   
   ~"lql.,q"lq ~'U4I~.,i ~ ~ I ~ 1) ~ ~ ~ qg4lJidi:
   
   a-.ri 1* ~ "lcqsfil~lfuffi'''~~
   
   «I I XCIV I I
   
   If 8 drammas are transport charges for carrying the planks, of the previous example, through a distance of 1 kosa. find out the charges for transporting the second set of planks (measuring 4 units less III each dimension) through a distance of 6 Iwsas.
   
   .~~~~~IIXCVII
   
   In different kinds of goods are to be exchanged, use the rules of three, five etc. as explained earlier. However, the price and the quantity of goods are in inverse proportion. Example
   
   Comment: This is to be solved by the rule of eleven. 14 cubits : 10 cubits 16 fingers: 12 fingers 12 fingers: 8 fingers :: 8 D : x All are direct proportions. 30 : 14 1K :6K lOx12x8xl4x6x8 .. x= 14x16x12x30xl = 8drammas.
   
   mt
   
   ~ ~ WSSWtld':i
   l"MJf.'t
   
   crnfll: r i
   
   (lf3qIA
   
   (1~~q~., ~
   
   fi!t?r II
   
   XCVI II
   
   In a market, 300 mangoes can be purchased for 1 dramma = 16 paise . However, 30 pomegranates of good quality are available for 1 paisa Find out quickly how many pomegranates can be exchanged for 10 margoes.
   
   Comment: First Method: As per Bhliskarliclirya's method.
   
   87
   
   LUavati of Bhaskaracarya
   
   88 300 mangoes: 10 mangoes} . . 30 : x 16 Paise : I Paisa . .
   
   lOx 30x 16 :.x= - - - 300 x I
   
   ~rect
   
   mve~
   
   }
   
   16 Pomegranates.
   
   Second Method: First we determine the number of pomegranates equivalent to 300 mangoes = 16 x 30 =480 300 mangoes =480 pomegranates 480 10 . . 10 mangoes = x - = 16 pomegranates. 300 I
   
   CHAPTER 25
   
   Simple Interest W'UUIIII,d.,
   
   ~ ~ fltfit~ ~ ~ "I
   
   '('qq)·r~ " ~ ~ " ~~ WI ~ ~: II XCVII II (To compute simple interest and principal.) Multiply the standard principal (100) by the standard period (I month or I year). Next multiply the given period by the given rate of interest. Keep the two products a, b separately. Multiply a by the amount and divide it by (a + b) to get the principal. Similarly, the amount multiplied by b and divided by (a + b) yields the interest.
   
   Comment:
   
   A R
   
   = amount, P =principal, I =interest, = rate of interest, Y =Period.
   
   Po = standard principal (usually 100) Yo = standard period (I year or I month).
   
   P
   
   =
   
   A x Po x Yo Po Yo + RY
   
   89
   
   LIiiivati of Bhiiskaracarya
   
   90
   
   1=
   
   AxRxY Po Yo + RY' According to the Jresent method. first calculate the interest on Rupees (Rs.) 100 foc the given period and add it to Rs. 100 to get the amount for Rs. 100. By the rule of three then find the principal and then inter-
   
   P _
   
   100
   
   I -
   
   100
   
   - - + --- + - RI Y I R 2 Y 2 R3 Y 3 Interest in each case is: 100
   
   P
   
   P2 R2 Y 2 100
   
   P2
   
   :. Ito- = -100-=--
   
   Example
   
   ~=
   
   100
   
   PI
   
   100
   
   + P 2 + P3
   
   100
   
   100
   
   - - + ---+-RI Y I R2 Y 2 R3 Y 3 RI Y I R2 Y2 R 3 Y 3 from which we get the formulae for Ph P2, P3.
   
   ~ ~ ~ ~ (iqh'll=t'l'<'tl
   
   ~ ~ ffiI ~ ~ ~II XCVIII II
   
   When the interest rate is 5% per month, the amount after one year is 1000 N (ni~as). Find the principal and the interest.
   
   Comment: The above stanza does not mention 'per month' but it seems that in those times the interest was calculated on monthly basis.
   
   P= 1000xl00x I =625N 100xl+12x5 375 N . 100 x I + 12 x 5 We could have got I =A - P = 1000 - ~25
   
   100 ' R, Y I 100
   
   (PI +P 2 +Pl )x - -
   
   PI RI Y I
   
   est.
   
   91
   
   Simple Interest
   
   qcqQq;Aq;qg"O>tlth ~ ~~~I ~ tI'<1<:tI
   94 N (ni~as) were divided into three parts and were lent at 5 per cent (per month) for 7 months, at 3 per cent for 10 months, at 4 per cent for 5 months. If the three parts yield equal interest, find them.
   
   I = 1000 x 12 x 5
   
   =375 N. The interest
   
   375 Non 625 N for one year appears high. ~ ~f~ffi'l: ~ t1Jafflq;f('I .... "I'><11\~ I ~~'t .. *'Hq fitflI~: ggqM;:S~ ~~ ~ II XCIX II (If several parts of a certain principal bear different rates of interest
   
   for different periods and yet yield the same interest, to find these parts.) Take the product of the standard principal and the standard period. divide this product by the product of respective periods and ' rates of interest, and write these quotients separately. These quotients multiplied by the given prin<;ipaJ and divided by the swn of the quotients written separately are the desired parts of the given principal.
   
   Comment: Using the notation of the previous example with suffixel( to denote the parts (we consider three parts):
   
   Comment: Here PI + P 2 + P3 = 94
   
   I I I I I RIY I =35'R 2 Y 2 =30'R 3 Y 3 =20 I I I 1 1 1 47 --+---+---=-+-+-=RI Y I R2 Y 2 R3 Y 3 35 30 20 420 94 420 1 94 420 ' I PI =-1 x 47 x 35 =24,P2 =T x 47 x 30 =28
   
   94 420 1 P3 = - x - x - = 4 2 . I 47 20 Now Bhaskaracarya gives a formula to compute the shares of profit when total profit and individual investments are given.
   
   An individual's share (after business) is the individual's investment llIultiplied by the total output and divided by the total investment.
   
   92
   
   LWivali of Bhliskaraciirya
   
   Comment: If a, b, c are investments and x is the output, then the shares are ax/(a + b + c), bx/(a + b + c), cx/(a + b + c) respectively. q"ml~f)ti~l 'IOIif)I<e"~:
   
   ~~~I mGll ~~~fatJa\ ~:
   
   ~~~~~IICIIII
   
   Three grocers invested 51, 68, 85 N (ni~as) respectively. Skillfully they increased their total assets to 300 N. Find the share of each.
   
   Co"'ment: Total investment = 51 + 68 + 85 = 204 N. Their shares are: 51 x 300 = 75 N 68 x 300 = 100 N 85 x 300 = 125 N. 204 '204 ' 204 Profits are 24 N, 32 N, 40 N. We have now a formula concerning filling up of reservoirs (pools, Jakes, tanks).
   
   One divided by the sum of tho reciprocals (of the times taken by the sowces to fill up a pool) is the time of'filling (the pool when the sowces are used simultaneously). Comment: Suppose the sources take times tit t2, ... to fill up a reservoir. If they are simultaneously used then the time taken I
   
   = -:----:,--- We use the same method at present. Of course at I I -+-+ ... tI t2 Bhaskariidirya's time there were no taps.
   
   93
   
   Simple Interest
   
   Four streams flow into a pool and separately they take 1 .!. ..!. ..!. , 2 ' 3 '6 days respectively. If all the four are used simultaneously, find the time required to fill up the pool.
   
   Comment: As explained in the previous stanza. Time =
   
   I
   
   1+ 2 + 3 + 6
   
   tf'l
   ~thday 12
   
   .
   
   The four streams can fi II up I, 2, 3, 6 pools in one day and so together they can fill up 12 pools in one day. So time for one pool =
   
   J..day . I2
   
   Formula concerning parts of grain qvfJ: 'f4'l~I"'~: ~~~~I ~~.q"ym~
   
   ~~~~:IICVII
   
   Form products of each cost, the reciprocal of the respective measure of grain (available for the cost) and the corresponding quantity (or proportion of the grain to be purchased). Each product (thus obtained) multiplied by the total price to be paid and divided by the sum of the products will be the price of each quantity of the (corresponding) grain to be purchased. (Sum of these quotients is naturally the total price to be paid.) Quantity (or weight) of a grain (when proportions of grains to be purchased are given) is the (proportion of the) quantity multiplied by the total price to be paid and divided by the sum of the products (obtained earlier).
   
   Comment: For the i th grain, suppose, gj measures cost dj and we need quantity qj. Then for money worth d units,
   
   d· ---.!.x 8' xd
   
   Example
   
   Q ~ ~"~~a\<J"~:
   
   =
   
   cost 0 f
   
   ·th 1
   
   .
   
   gram =
   
   8.'
   
   •
   
   -;----'~-:-----
   
   ~
   
   d2 q\+-xQ2+'" 8\ 82
   
   94
   
   Lfiavatl of Bhiiskaracarya
   
   Now to find the quantity of grain gl in the mixture substitute the proportion of the grain gl f or
   
   ~ qj gj
   
   in the above formula.
   
   Example 1
   
   ~ ~ ~ ~q't<: tI~, ~ qfVI4'itl,~v": I ~s.Wl t'w§i1itlg"t'f ~ f$j ~ ~lm: ~II eVI II
   
   _SJr('!)
   
   .
   
   f' .. I nee and 8 units of mi;Jg beans can be bought lor one
   
   dramma.
   
   0grocer, Ihave 13 kakirjis and Iwant rice and miing beans '
   
   2"
   
   WIltS 0
   
   (a kind of kidney beans) in the Jroportion 2: 1. Quickly give me the grains so that we can cook, eat and start on our next journey by cara-. van.
   
   Comment:
   
   1 2 13
   
   Cost of rice = -7/2
   
   1
   
   lIi11~ ~ ~\5Cfi11,a~", ~ ~ hql"<:"1'Q<'1t'Q ~ ~ ~ 1\q I
   
   ~ "OmS'ffl: ~ ~ -q ~ CU"l "'lth)4i,,,':at"\IC~{q ~4fI.n..,.,~'t II eVIl II
   
   Good quality camphor costs 2 N (ni!jkas) for 1 P (pala), sandalewood 1 1 1 costs - 0 (drammas) for 1 P and aloe costs - 0 for -2 P. 0 you
   
   8
   
   8
   
   grocer's son (who pleases his mother) what quantity of each is available in the proportion 1: 16:8 for 1 N?
   
   Comment: Noting that 1 N = 16 0 we get
   
   (4 13) (4 1) 1
   
   x -1 x -64 = _ x _ + _ + - = - dramma _l-x~+!Xl 7 64 7 8 6 7/2
   
   ·h ' whIC is gIven as -13 dramma. 64 So 39 x =Q. . x = - 7 measures. .. , 24 56 64 Example 2
   
   md (fI/!(iijQ''14iiqCjf) ~ ~
   
   31
   
   95
   
   Simple interest
   
   32 cost of camphor
   
   =
   
   8
   
   Q) (i
   
   cost of mUng beans = (! x! x + +!) = ~ dramma 8 1 64 7 8 192
   
   =
   
   13) (4 1) 7
   
   . 0 f nce ' = (2 x quanLIty + - + - = - measures 64 7 8 12
   
   cost of sandalewood
   
   . of m -rng beans = (13) (47 + -81) = -247 measures. quanLIty 1x 64
   
   =
   
   + -
   
   We can prove this formula by Algebra. Let x be the measure of mUng beans. Then rice = 2x measures. x 8 4x
   
   Cost of mUng beans =
   
   - drammas
   
   cost of rice
   
   =
   
   "'"1
   
   So the total cost
   
   =
   
   x 4x 39 - + - = - x drammas 8 7 56
   
   ost of aloe
   
   lluantity of camphor
   
   =
   
   32 x l I 1 - - + - xI6+ - x8x2 1 8 8
   
   x 16
   
   32 x 16 = 128 = 14 ~ 0 36 9 9 1 - x 16 x 16 8 8 =-0 36 9 1 - x 8 x 2 x 16 8 8 = -0 36 9 2 4 14-+ 32 =- P
   
   9
   
   9
   
   !+.!.=64p 9 8 9 8 1 32 -+-=-P. quantity of aloe = 9 4 9 We can work out this problem by algebra. Suppose camphor is x P. Then sandalew~ = 16x P, aloe 8x P.
   
   lluantity of sandalewood
   
   drammas.
   
   =
   
   x1
   
   J
   
   =
   
   96
   
   LlIlivati of Bhaskaracarya
   
   Total COSI
   
   to each of the others one jewel from his lot. This made their jewels worth the same. 0 friend! find out the prices of-these jewels.
   
   =32x + 16x x.!. + 8x.!. = 36x = 16 D 8
   
   97
   
   Simple Interest
   
   4
   
   4 :. x=-P . 9
   
   Comment: We give three methods. (l) Bhaskaracarya 's: Jewels 8, 10, 100, 5. Number of gifts q=l,
   
   =
   
   =
   
   Formula for Exchange of Jewels
   
   number of persons n 4. So the remainders are 8 - 1 x 4 4, 10 - 4 x 1 6, 100 - 4 x 1 96, 5 - 4 x 1 1. If we suppose x 96, the prices
   
   "'
   of jewels are ruby =
   
   =
   
   96 = 24, sapphire 4
   
   =
   
   9:
   
   = 16,
   
   diamond =
   
   1
   
   =
   
   =
   
   th
   
   X
   
   - - - where x is the supposed worth of each. It can also be found ai - nq oJ
   
   =
   
   second method we get the product of the remainders 4 x 6 x 96 x 1 2304 and we get the prices of the jewels in the proportion 576, 384, 24, 2304 and if we choose 1 as the price of a pearl, the proportion is 24, 16, 1, 96 as before. (2) After the transfer of the jewels, the 1st merchant had 5 R and one each of the other three or 4 R and one each of the four type of jewels. 2nd merchant had 6 S and one each of the four type of jewels. 3rd merchant had 96 P and one each of the four type of jewels. 4th merchant had 1 D and one each of the four type of jewels. Since their worth is the same, we get 4 R = 6 S 96 P 1 D. We choose Ihe price of 1 pearl to be 1 Rupee and we get the answer as before. (3) A Igebraic: If the prices of the jewels are a, b, c, d respectively, Ihe equations are 5a + b + c + d a + 7b + c + d = a + b + 97c + d = a + b + c + 2d t', 4a = 6b = 96c d. If we choose c = 1 Rs., we get the same answer I. before.
   
   Comment: Compute llj _ nq ; the price of a jewel with the i person is
   
   IT..
   
   =
   
   96 = 96, Here 96 is the L .C.M. of the remainders. If we follow the
   
   (Suppose n perSOns possess aJ, a2, ... an jewels respectively. Suppose each one gives q jewels to each of the others and then their jewels are worlh the same.) From the numbers of jewels, subtract the product of number of persons and the number of jewels donated (uniformly) to each of Ihem. An assumed' number (which could be the L.C.M. of the remainders or any multiple of the L.C.M.) divided by the remainders will give Ihe prices of the (corresponding) jewels.
   
   by
   
   =
   
   =
   
   (a - nq) I
   
   ai -nq The price of the jewels will naturally depend upon the supposed number x. Perhaps the answers will be fractions. If the L.C.M. of the
   
   =
   
   remainders {(ai - nq) : 1 ~ i ~ n} is divided by a; - nq, the prices will be integral.
   
   =
   
   =
   
   Formula to find the Weight of Pure Gold Example (Exchange of Jewels)
   
   "l,oiq uili5Rl4)<,'
   "IFaICfQloeqs~"i: oi\i'i'=uqi !lCfiilcrtCilI.,i ~ ~ 'I( ~ ~MqfitlGtlq ..i ~ tFf1l1 t(f
   lid
   
   ?lC:M'I?J'~
   
   ~~ Oi"'~4'QqC1f: I
   
   ~ ~~ ~d~if~&l111 CX II
   
   ( n, lind the weight and fineness of gold if two or more kinds of gold .q I I
   
   CIX I I
   
   1\
   
   •
   
   Four merchants had 8 rubies, 10 sapphires, 100 pearls and 5 diamonds respectively. They became friends on their journey and each one 'gave
   
   melted together.) The fineness of the mixture equals the sum of
   
   II. products of weight and fineness of the constituents, divided by the h,lal weight. If the same sum is divided by the weight of pure gold,
   
   Wavati of Bhaskaracarya ale gets the fineness of pure gold and if it is divided by the fineness. ~e result will be the weight of pure gOld.
   
   Comment: If the fmeness of the mixture is x and weight y then xy
   
   =
   
   99
   
   . imp/e Interest
   
   the the given compon ents. The remaind er divided by the weight of . compon ent (whose finenes s is not known) gives the desired fineness
   
   Comme nt: (Using the notation of the previou s stanza):
   
   n
   
   I
   
   1= I
   
   xy-
   
   xi Yi and the result follows.
   
   XiYi
   
   i .. j
   
   y"":'j- x J =- - -
   
   =
   
   tiote that y Yl + .. , + Yn'
   
   Examp le
   
   bample
   
   ~",I6N
   m
   
   I
   
   ~ ~ ~ ~: ~ 1qu f<,fVliq ~ ?fi: II eX111
   
   J golden mathematician, four types of gold 10 M (m4ir) of 13 C :.arats), 4 M of 12 C, 2 M of II C and 4 M of 10 C are melted to!tther to fOIll\ a new one. Find its fineness. If this is purified and 16 ~ gold is obtained, what is its fmeness? If the mixed gold when pun::ed has 16 C fineness, what is its weight? ClJmment: Sum of the products = 13 x 10 + 12 x 4 + II x 2 + 10 x 4 = 130 + 48 + 22 + 40 = 240. 240 240 Fin f enesso the mixture = 10+4+ 2+4 =20= I2C "hich naturally lies between 10 C and 13 C. Fi1eness of the purified gold
   
   L
   
   = 240 = 15 C 16
   
   '
   
   Iflhe fineness is 16 C, weight = 240 = 15 M. 16 Here 5 M of impurities were burnt out.
   
   ({hlqv, f q~",'lI"'''' 3(S'ltdqvi'Rl .... ahill ~ ~ 3($1".,qvi'Rl « ~ II eXIII II
   
   vm; lfd
   
   8 M (m~o;) of 10 C (carats) , 2 M of II C, 6 M of unknow n Care mixed together to fonn 16 M of 12 C. What is the fineness of the unknow n?
   
   = 192-8 0-22 = I5C .
   
   6 ~~ 1~,"rMOf: ",\*",qcJf4"1ftq)~tft~ I aI ~"qvlff~;j""4).,qu(fC"~ .... cm)stafaalft.:tui ~ II exlV I I
   
   Multiply the finenes s of the mixture s by the sum of the known of weights of the compon ents of the mixture. Obtain the differen ce onding corresp the this and the sum of the product s of fineness and ce of weights of the compon ents. The outcom e divided by the differen of that and known) not is weight fineness of the compon ent (whose the mixture gives the desired weight.
   
   .
   
   lin the above notation , Y3 =
   
   r~rmula: Frorn Mixture to Component
   
   ~m"",q~d"O'~tt ~qVfd,ol~iNtf.!n",
   
   ~fHqQtl~4tzQ'~ ~1h\1ijoff4 ~ 1l1m1Tt{,11
   
   16 x 12 - 8 x 10 - 2 x II 6
   
   Comme nt: Here x =
   
   exn II
   
   M\iltiply the weight of the mixture and its fineness . Subtract from this, Lx sum of the products of fineness and the corresponding weights of
   
   x (y I + y 2)
   
   -
   
   x I YI _
   
   X
   
   x
   
   -
   
   x2 Y2
   
   l
   
   Examp le
   
   u'1,
   J
   
   LrliivatT of BhaskaracBry
   
   100
   
   3 M (m~) of 10 C (carals), I M of 14 C are mixed wilh SOIl)~' quantity of 16 C. If lhe fineness of Ihe mixlure is 12 C, find Ih~ weighl of 16 C gold. By Bhaskaracarya's melhod = 12 (3 + I) - 10 x 3- 14 xl =48 - 44 = I M Y3 16-12 4 .
   
   Comment:
   
   Weights of Components ~~ ~: ~ ~: t
   ~~ ~ ~ ~ ~tq,.,('q<J)4oltit*' I I CXVI II
   
   (If fineness of a mixture and its two components are given, to fin 'J weights (indeed the ratio of the weights) of the components.) Obtain' , the difference 0 f the fineness 0 f the mixture an d that of the componen with greater fineness; and the difference of the fineness of the mixt ,I and that of Ihe component wilh smaller fineness. The two difference~ multiplied by an assumed number will give weights of the compo , I nents.
   
   CHAPTER 26
   
   Combinations qctSlGcM'diCl ai"ctSJ t.lffifI 1fJ\Rn: ~: I tR: ~ :d:1 0
   ~~~~I ~~~~lleXXII
   
   Example ~'C!crt1ft~ ")UI
   q
   
   &lCtlquj'CjqVf ~ ~: ~
   
   \ilRll{1
   
   q II
   
   exvn II
   
   o friend, two small balls of gold of 16 C and
   
   10 C were melted tot gether to form a mixture of 12 C. Find out the weights of the balls.
   
   Slarting with the nwnber n write down n, (n - I), (n - 2), ... , Divide n n-I n-2 Ihem by 1,2, 3, ... to get -, - - , - - , .... Then the number of I 2 3 combinations of n things taken I, 2, 3, . . . at a time are 1\ n(n-I) n(n-l)(n-2) . - " , ... respecllvely. Or the number of I Ix2 Ix2x3 combinations of n things taken r at a time are [n (n - I) (n - 2) ... (n - r + I)] + [I x 2 x 3 x ••• x r). This can be used to solve the problem when r = I, 2, 3, ....
   
   If the mixture weighs y M (m~), then 12 - 10 Y 16 - 12 2y Yt =y x 16-10 ="3'Y2 =yx 16-10 = 3"
   
   Comment:
   
   So the proportion is 1:2,
   
   lOl
   
   '103
   
   Combinations
   
   102
   
   LIlavatI of Bhiiskaracarya I
   
   This is useful in prosody to discover all possible meters, in archi-
   
   Also
   
   (g) + ( ~) + ... + (~) =2
   
   n
   
   tecture, medical sciences, Kham!meru·, chemical compositions etc. I \
   
   If we take the combinations of all the four lines as.. above, the total
   
   am omitting these (applications) for the sake of brevity.
   
   number of combinations will be 64 x 64 x 64 x 64 = 16777216.
   
   Comment: Modem notation is (n) = n(n-I) .. ,(n-r+1) . The
   
   Example 2
   
   r r! first line of the stanza is not clear, as no mention is made regarding the starting number in the product.
   
   (q;'jliJlI~fi'Sf"'~a-Q~) ~ 1) ~:
   
   R
   
   ~~ q~l'(~'(~~f(1~ltIf('i1'~,"~ I
   
   \!"'ft''iD1~1lfi'Q1 ~'("'S"'''''41~'''~: ~~:~~~~~:IICXXIIII
   
   Example I
   
   lffifR
   
   A king had a beautiful palace with eight doors. Skilled engineers had constructed four open squares which were highly polished and huge. In order to get fresh air, I door, 2 doors, 3 doors, ... are opened, how
   
   ~ llT1OZn:~: ~~: ~I
   
   ~""~1,«W4'~ ~ ~~ 'l~ II
   
   CXXI II
   
   o friend, there are six letters in the fourth line of gayatrI meter. If we choose only 1 g ( 'guru' means a long vowel in Prosody), 2 g or 3 g, how many meters are possible.
   
   Comment: If the fourth line contains six l ('Iaghu' == short vowel), then the only combination is UUU . If we have one g and five t, < then (i) gUUl (ii) 19UU (iii) UgUl (iv) UlgU (v) UUgl (vi)
   
   many different types of breeze arrangements are possible? How many kinds of relishes can be made by using I, 2, 3, 4, 5 or 6 types from a sweet, biller, astringent, sour, salty, hot substances?
   
   Comment: As per the above formula,
   
   (~), (~), (~), ... , (:).
   
   the possibilities
   
   are
   
   These are I, 8, 28, 56, 70, 56, 28, 8, I re-
   
   tUUg are possible. If there are 2 g and 4 i, combinations will be'
   
   spectively.
   
   15.
   
   Total number = 2 = 256. The number of different relishes (chutnevs) = 2 + 12 + 30 + 20 = 64=~ .
   
   As seen above if there is 1 g, number of meters =
   
   For 2 g , (6) = 2
   
   6 x 5 = 15 Ix2
   
   3 g,
   
   (~) =
   
   6x5x4
   
   4g ,
   
   (:) =
   
   (~)=15
   
   5
   
   (~)
   
   =
   
   (~)=6
   
   6 g, (:)
   
   =
   
   l.
   
   g,
   
   Ix2x3
   
   •Khanc!meru is Pascal's triangle.
   
   (~)
   
   = 6.
   
   8
   
   Pascal's Triangle
   
   20
   
   Bhaskaradirya has referred to "khancJameru", which is known as Pascal's tria'ngle. This was known to ancient Indian mathematicians. Pirigalacarya had used it in the construction of meters (in Prosody). In this triangle, every line is connected to the next one. Each line consists of the binomial coefficients as in Newton's binomial theorem. See the following scheme.
   
   104
   
   Lrlavali of Bhiiskaracarya
   
   Pascal's triangle (x + y)o
   
   0
   
   (x + y)1
   
   1
   
   1
   
   (x + y)2
   
   1
   
   2
   
   1
   
   (x
   
   + y)3
   
   1
   
   3
   
   3
   
   I
   
   (x
   
   + y)4
   
   1
   
   4
   
   6
   
   4
   
   1
   
   (x + y)5
   
   1
   
   5
   
   10
   
   10
   
   5
   
   1
   
   + y)6
   
   1
   
   6
   
   15
   
   20
   
   15
   
   6
   
   (x
   
   4
   
   4
   
   For eXaI{lple, (x + y) = x4 + 4x 3.Y + 6x.2.y2 + 4xy.3 + Y .
   
   ~
   
   1
   
   I CHAPTER 27
   
   Also note 5 = 1 + 4, 10 = 4 + 6, 10 = 6 + 4, 5 = 4 + 1 in the 6th line.
   
   (~) + (r ~ 1) = ( n ;
   
   This is based on the fonn'ula
   
   Progressions (Series)
   
   1). In Bhaskara-
   
   dirya's times, the above was written like a 'meru' (mountain).
   
   o ~~<6fq<~<Mtiq;gl>: ~ :ct"~d'&J' I
   
   I I
   
   3 4
   
   5
   
   -m ~ ~ fitf.!tt;ft ~ ~
   
   2
   
   I I CXXIII I I
   
   3 6
   
   10
   
   -m:t :ct ¥f(twactQ't
   
   1+2+3+ ... +n=
   
   4
   
   10
   
   n (n + I)
   
   .
   
   2
   
   5
   
   1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + . .. + n) n (n + I) (n + 2) 6
   
   Example f(Cfi1~.,i '1C1I~I'1i ~ ~lflf<"t(iJrf..,
   
   .q I
   
   . ~~lflf<"t~<s4'f'" ~~ ~ ~II
   
   CXXIV II
   
   Find I + 2 + ... + 9 d 1 + (1 + 2) + ... + (I + 2 + ... + 9).
   
   'ommen': As per (CXXIll) the answers arc 9 x 10 = 45 and 2
   
   I)xIQxll
   
   . - 6-- = 165 respeclively.
   
   105
   
   106
   
   LTIwatT of Bhaskaracarya Progressions (Series)
   
   f
   
   i(i+l) Incurrent notation we write ~ ~) , LJ 2 i=1
   
   Example 2
   
   ;=1
   
   The first one is an Arithmetic Progression, and is a special case of 3
   
   + (a + d) + (3 + 2d) + ... + (3 + (n _ I )d) =.!!.. (2a + (n - I) d).
   
   2 3 Fonnulae for ul, rn can be established by Mathematical induction and are found in current books. ~ ~ ~ "q;~~., ~ ~:I "~dfQ ~: .. ,,~q;,riO>"'~6Vj"i<'l'jjt): II CXXV II
   
   12 + 22 + '" + n2 3
   
   3
   
   1 + 2 +." + n
   
   =[ n (n + 1) 2
   
   ~: ~
   
   J
   
   ~ 1"(i9)S1tIlf5I
   
   qlI:
   
   ~-~ ~ lI"l:' ~
   
   1(
   
   mr 1) I
   
   ftil{
   
   II CXXVIII II
   
   If the first term of an A.P. is 7 the common difference is 5 and the number of terms is 8, find the middle term, the last term and the sum.
   
   Comment: f.
   
   = a + (n -
   
   7 +-42=49 -
   
   W
   
   2
   
   Sum
   
   = n (n + I) (2n + 1) 6
   
   3
   
   107
   
   2
   
   =8 x
   
   49
   
   -
   
   2
   
   I)d
   
   =7 + 7
   
   5
   
   x
   
   = 42 and
   
   middle term
   
   =
   
   h'IC h'Isnotatermmt . heA.P.
   
   = 196.
   
   ExampleJ
   
   and ~
   
   4Af"giICci ~ ~ ~~
   
   ll¥: I
   
   Arithmetic Progression
   
   ~W~q~pnll"l:'~~~: II CXXIXII
   
   &l41Q<&1"'J41 ~ ~ 'f9T<~ ~I ~ q<~.:tfU'~ ~ 'I'l'fO'n:t ;;r ~" CXXVIII
   
   A gentleman gave 4 D (drammas) as charity (to a Brahmin) on the first day. For fifteen days he continued his charity, everyday increasing his contribution by 5 D over the previous day. Find out the total charity.
   
   If the first term is a and the common difference (C.D.) is d, the nih term of \he A.P. is given by I. =a + (n - I )d. The sum of the n terms .. n a+ I. IS gIVen by - (a + I.) . Here - - is the middle term. 2 2 Example 1
   
   =4, C.D. =d =5, n = 15. Last term l =4 + 14(5) = 74. . a + l 4 + 74 MIddle term m = - - =- - =39. 2 2 Total charity = 15 x 39 = 585 D.
   
   Comment: Here a
   
   Fonnula to find the first tenn of an A.P. 3
   
   Tell me the sums of 12 + ... + 92 and 13 + ... + 9 •
   
   = 9 (10)(19) = 285
   
   Comment:
   
   12 + ... + 9 2
   
   and
   
   X 13 + ... + 9 3 = -2-
   
   6
   
   [9 10]2 = 2025.
   
   'ro find the first term of an A.P., divide the given sum by the number lit" terms. Multiply the number of terms minus one by half the comIlIOII difference, and subtract this result from the quotient already ob. S (n-l)d lilllled. [Thus a = - .] n
   
   2
   
   ,. ,. 108
   
   109
   
   Progressions (Series)
   
   LTIlivatT of Bhllskariiciirya
   
   Formula to find the number of terms of an A.P. Example ~afQi(4js#h4fiQ4C"'tt ~~q
   ~mi~~~1 ~ -:;pj
   
   lI'li ~
   
   m
   
   ~ ~
   
   ~ ~ II CXXXIII
   
   =7, d= 3.
   
   105 6x 3 Comment: a=----15-9=6
   
   7
   
   2
   
   .
   
   {2Sd + (a -
   
   n= V
   
   Formula to find C.D. in an A.P.
   
   Divide the smn by !he number of terms (of an A.P.) and subtract the first term from this quotient. This result divided by half of the number S
   
   - (a -
   
   2
   
   ~) 2
   
   .]
   
   n S = - [2a + (n - l)d) 2 :. dn 2 + n (2a - d) - 2S = O.
   
   Bhaskaracarya's formula is the positive root of the above quadratic equation. He does not admit a negative d. Also n should be obviouslY positive.
   
   --a
   
   = (0 ~ 1) 12
   
   ~)2 d
   
   Comment:
   
   of tenns minus one is the,common difference. [Thus d
   
   CXX X1V II
   
   Add the square of the difference between the first term and half the CD. to the sum (of the given A.P.) multiplied by twice the C.D. Then find (the positive) square root of this result. subtract the first tertll from and add half the C.D. to this square root. The resultant divided by the C.D. is number of terms (in the A.P.). [ThUS
   
   o son, find the mouth (the first term a) of an A.P. whose S = 105, n
   
   'q
   .1
   
   Example
   
   Example
   
   g4'4Q"'4<:6:1 ~vA ~
   
   iiRtI:
   
   ~lI:JI~~~~~~1
   
   ~~~~\tl~~1 ;,j~~'(v,pJ
   Trffirli
   
   To capture enemy elephants, a king covers 2 Y (yojanas) on the firs Iday and then increases his distance by A.P. on subsequent days. If liJ~ travels 80 Y in 7 days, 0 intelligent boy, find out the extra distan€~ each day.
   
   ,
   
   A donor gave 3 0 (drammas) in charity to a Brahmin on the first daY· He continued increasing his donation each day by 2 D. If the total ,'mount paid by him equals 360 0, how many days did he give in ,;harity?
   
   Comment: n=
   
   Comment:
   
   ~ ~ ~ ~
   ~r-2-)(-3-60-X-2-+-:-(3-----,~)'--2 -
   
   ( 3-
   
   80
   
   l' -2 66 2 22 1 d=--=-x-=-=3-Y 7-1 7 6 7 7 2
   
   =
   
   JI440+4 - 2 2
   
   2 38- 2 =-2-= 18.
   
   ~)
   
   110
   
   LIliiva11 o/Bhaskaracarya
   
   Geometric Progression (G.P.)
   
   As is given in standard texts:
   
   ~ ~ ~ ~: ~: wlsfi'fa 1flf: I
   
   1'flOettnl'l~i'CI'" ~ ~ ··t ;jW,,,·,VJ ~.
   
   ~
   
   III
   
   I'rogressions (Series)
   
   a + ar + ar2 + ... + ar"-I
   
   rS
   
   = =
   
   S (r - 1)
   
   =
   
   ar" - a
   
   S
   
   ~I
   
   ~~O~'4MI~joi ~ ~ ~ II
   
   CXXXVIII
   
   ar + ar2 + ... + ar "-I + ar"
   
   :. S = a (r" - l) r -I
   
   If n, the number of terms in a G.P. is odd, then (n - 1) is called a . even, -n 'multiplier' (M) and'f' I It IS
   
   [Bbliskara _ , .
   
   is called a 'square' (S)
   
   2...
   
   .
   
   carya s termInology). Now begInnIng wIth n, continue thIS
   
   process (n - 1) [,0 r odd an d "2 n f or even until 0 is reached. Then keep the common . . . rati 0 (C.R . ) r agJanst 0 and start writing M or S in the opposite dIrection. Carry out the operations and then subtract 1 from the final result and d'IVI'de It . b y (r - 1). The result is the required sum.
   
   Comment: Suppose a = I, n method, we get the table'
   
   = 31 ,
   
   r -_ 2 Then following the above .
   
   eI~ I~5 I~ I~ I~ I~ I~ I~ I~ I 1
   
   Example 1
   
   Tf
   
   q,(lcgaj
   
   =
   
   Index 30 15 14 7
   
   6 3 2
   
   °
   
   M S M S M S M S M
   
   } 147483648 J073741824"' 32768 16384 128 64 8 4 2
   
   31 30 15 14 7 6 3 2 I
   
   To reach the
   
   31 It
   
   power, here B haskaracarya has given a shorter method by squaring at even numbers and multiplying by r at odd ones. In practice, multiplication might be easier than squaring.
   
   Qjuih,~ \I~$m,.t I
   
   ~~~~~~II CXXXVIIII
   
   A gentleman gave 2 C (cowries) in charity the first day and thereafter he gave everyday twice of what he gave the previous day. How many
   
   N (ni!jkas) did he give away in one month?
   
   Comment: Here a = 2, r = 2, n = 30. 30
   
   S = 2 (2
   
   - 1) = 2 (1073741824 - I) = 2147483646 C
   
   2 -1 = 104857 N, 9 drammas. 9 pal)as. 2 kakil)fs. 6 C.
   
   ..!.i
   
   Beginning with 31 we write 31 - 1 = 30, 30 = 15, 15 _ I = 14, = 7 r 2 2 , ... un II we reach O. Then in the second line we begin with M and . a Itemately write M an d S. · di cated operations' Now we perfo rm the In M = mUlti r . pIer, S square as shown below.
   
   lA
   
   ..:xample 2 ~ ~ ~: ~ ~!}ui'h1i(' I 'flU ~ ~ ~ ffiI
   
   f$ ~ II CXXXVIII II
   
   If a = 2, r = 3, n = 7, 0 friend, what is the sum?
   
   omment: S = 2 (3
   
   7 - \)
   
   3-1
   
   = 2 (2187 - l) = 2186 . 3-1
   
   Formula for the number of metres (in Prosody) ~~'(f21<'1arW !}Ulql1",t"i ~ ~ I ~ ~ qalqai,q II ~fqq~):ft Tlffi'ri ~ ~ ~ .. qlu,,'t II CXXXIX I I
   
   .. q,fJ'Ofi
   
   112
   
   Lil1ivatf of Bhaskaracarya
   
   [lr every metre there are four quarters. If the order of short and long v~'els is the &arne in each of the four quarters, the metre is known as 'e,en' (samavrtta).) The following formula gives the number of even metres. Take as n the number ofleuers in each quarter and r = 2 (for there are only two kinds of vowels, short and long). The total number of possible metres =t. [If the first and the third quarters are similar and so are the second and the fourth. it is called 'semi-even' (ardhasamavrt ta ).] In this case (le total number of possible metres = 2211 _ 2n. [If the number of letters is the same but all the four quarters are ditTerent' it is called 'odd metre' (vi $lJmavrtta).) In this case the total nU'lber =(2)4 - (2Y CO'1Iment: Suppose there are n letters in each line. Each letter can be S:lCrt or long and SO there are two choices. According to the fundal1ental theorem there are 2n possibilities in the even metre. th There are 2° x t possibilities in the 1st two (as well as 3rd and 4 ).
   
   nan these we subtract the even case 2 e.en cases: (1'''/ - t =2n (2n - I).
   
   n
   
   CHAPTER 28
   
   Mensuration Measurement of Sides and Areas
   
   to get the number of semi-
   
   :n the odd metre case, there are 2n x 2n x 2n x 2n possibilities from "b:ch we subtract the first two cases: 2n + 2n (2 n - I). So the number =(1r.)4 _ [2 n + 221t _ 2n) =(2n)4 _ 2m,
   
   In a right-angled triangle, one of the sides is called the base (bhuja or bahu) and the side perpendicular to it is called the altitude (kop). The same terminology holds for a rectangle.
   
   Example
   
   ~~~~I
   
   fWri ~ .q ~&jlqj"Sqj
   Comment: A
   
   0
   
   BC is the base. AB is the altitude.
   
   c
   
   AC is the hypotenuse.
   
   mere. Ortment: There are 8 letters. So number of even metres = i 256
   
   B
   
   a
   
   C
   
   =
   
   number of semi-even metres = (2')2 - 28 =65280 number of odd metres = (2')4 - (2')2 = 4294901760. }l3
   
   115
   
   Mensur-ation 1:4
   
   LiliivatI of Bhaskaracarya Example 2
   
   ~'tJq)<J"q< ~
   
   Gl:",11\4.i4)fhui\1
   
   -.p1 iilk: lfl1fhfi'l'fl~Hi\ ~ 1Ill: II
   
   'ffiRl~<Jfi4(1) ~ ~ ~ I
   
   CXLII II
   
   ffiI ."oiQq,vj ~ ~ ~
   
   In a right-angled triangle, square root of the sum of squares of the bw. and the altitude is the hypotenuse, and square root of the differere: between squares of the hypotenuse and base (respectively alti-
   
   Wei is the altitude (resp. base). [Thus b:: ~C2 + a 2
   
   , C ::
   
   ~b2 -
   
   a
   
   2
   
   . 3 '4 1 and the altl'tude 3.! h b If in a right-ang Ie d triangle, tease IS 4 find the hypotenuse.
   
   Comment: Hypotenuse =
   
   C~"ment:
   
   This is known as Pythagoras' Theorem. This was known in B.C.) whereas Py1I11~)ras published it in 560 B.C. Of course, no proofs were given at
   
   nh since the time of SulvasfiJrai<aras (3000-800
   
   W.lkaraclirya's times. Proofs of this theorem are found in textbooks.
   
   ma ¥
   
   ~ II CXLV II
   
   ,
   
   a= Jb2 - c2 .J
   
   witf"dwlol ~
   
   .q
   
   ~: I
   
   ~ ~ ~.".~'f(j~~: I
   
   169
   
   169
   
   13.fi.
   
   16
   
   16
   
   4
   
   -+-=--
   
   ~VJ ~ ~?lrq}c::i"m4UTtI I ~ ~~vor~7m ~ ~ II CXLVI II
   
   [In this stanza Bhaskarlicarya explains a method 0 f finding an approximate square root of a number which is not a perfect square.] To compute (approximately)
   
   ~ ~ tli 'ff1B ~ II CXLIIIII
   
   Jf
   
   choose a large square number x.
   
   Then compute approximately .Jabx and divide by bJ;. . Ift~ce the product of two numbers is added to the square of their
   
   dkence, the result is the sum of their squares. So also if the sum of t>\O lUmbers is multiplied by their difference, the result is the-differen:e of their squares. [Thus (1+» (a - b).J
   
   o'fc'Jqg'Cfi
   
   i
   
   + b2 :: (a - b)2 + 2ab and
   
   Comment:
   
   ~ =~. -Vb b.JX [l69 13.fi.
   
   i - b2 = In the above Example 2,
   
   = - 4- =
   
   13(1·414) _ 13(141·4) 4 400
   
   .= {4 476 + _l_} approx . 800 200
   
   til ~ i'r.t Cfi1 ~: I
   
   2
   
   By Bhaskarlicarya's method, choose x = 10000 = 100 and
   
   ~~:~~~~~IICXLIVII
   
   J169 = 8
   
   File the hypotenuse if the base is 3 and altitude 4. I f the hypotenuse <11J fie base are 5, 3 what is the length of the altitude? If the hypote-
   
   i$and the altitude are 5, 4, what is the base? 2
   
   VT
   
   Here we note that
   
   169x8x 10000 == 8 x 8 x 10000
   
   3677 2 = 13520329
   
   ~=4 477. 800
   
   800
   
   }
   
   and 169 x 8 x 10000 = 13520000 .
   
   2
   
   (ClR1Ient: 3 + 4 = 52 :. the hypotenuse = 5.
   
   Bhaskaracarya was aware of the fact that
   
   /2,.J3,.Jl7, ...
   
   are irra-
   
   tional numbers and so gave a method to compute their values approximately. In his times, 'decimal fmctions' were not known. They
   
   4
   
   3
   
   Uravali of Bhiiskaracarya
   
   116
   
   were introduced in Italy for the first time in the sixteenth century. Until then only vulgar fractions were made use of in all mathematical calculations. Bhliskarlidirya rationalizes the denominator.
   
   Comment: 1st Method. Choose y so that 12 and we choose y = 2.
   
   c = ~ = 2 x 12 x 2 y2 _ I
   
   Formula to find hypotenuse and altitude of a right-angled triangle given its base ~ 'P'l~ fti!~"4~G1'~*" 'f'~iIIfb~IJI'<',\1 ~: ~ ~~ ~~~IICXLVIII'
   
   ~ 'GI«'I¥~~~ ~: tlQl~~IiC).,gnlsfim1 m I 1ft!t ~ m q,q;,
   IT§¥l
   
   (f\ Cf))~qsoff
   
   a = base, b = hypotenuse, c = altitude (see fig. of stanza CXLl). Choose any convenient number y. Then given a to find b, c; c =
   
   ~ 2'
   
   Y -1 OR:
   
   b= a(y2 +1) 2
   
   Y -1
   
   .
   
   b=~( a; +y )~c=~( ~ -y J.
   
   Comment: (1) Here b2
   
   =
   
   a2 (y2 + 1)2 (y2 _ 1)2
   
   and
   
   117
   
   Mensuration
   
   4_ I
   
   = 16
   
   b '
   
   l- I divides 2ay. Here a
   
   = a (y2 + I) _ Y2 _ I
   
   -
   
   =
   
   12 (5) _ 3 - 20 .
   
   2nd Method. Here y must divide i and we choose y = 6. 2 b = ! ( 12 + 6) = 15 c =! (122 _ 6) =9 2 6 ' 2 6 . If we take y = 2 we get b = 37, c = 35. NOTE: These are known as (Pythagorean) triplets. Several solutiolls are possible: (i) (9, 12, 15), (ii) (12, 16,20), (iii) (5, 12, 13), (iv) (12, 35, 37). These are the only possible ones. The first two are deri ved from (3, 4, 5) by multiplying by 3 and 4 respectively and so are called secondary. (3, 4, 5), (5, 12, 13), (12, 35, 37) are called primary. Such distinction was not made at Bhaskaraclirya's times. There are wellknown formulae to find the triplets: [(2n + 1), (2n2 + 2n), (2n2 + 2n + I)) 2 - 1 -n 2-+ 1] . [2n, n2 -1, n2 + 1]; [n2 11,-2-' and
   
   Formula to find two sides when the hypotenuse is given
   
   ~ ~ ~'JUiliif iflVlf~Qf4 'fIffl4l'f3l1 ~I ~ m 'tllJfil'IC~"'"
   
   (fi'tlOh4h::('I(xn
   
   ~: I I CL II
   
   Given hypotenuse b, choose a convenient number y. Then a
   
   Eumple ~ ~ ltI ttl 4'I)~itioll ~ I Witi""'4i lIi: tani (f\ dlqiti'i(\lft·rd\ I I CXLIX II
   
   If the base of a right triangle is 12, find its altitude and hypotenuse in integers.
   
   c = b (1- y2) (y 2 + 1)
   
   =~ , y2 + 1
   
   LIlavati of Bhaskaraciirya
   
   118
   
   Mensuration
   
   119
   
   Example
   
   Example qtllfl)~~~ 1iVl lfI lJlq/fi'(C1\t'ltft I ~ ~ ffI ffI ~ ~ ~II CLlII
   
   ~1l'hQ(5j ~ ~~:
   
   If the hypotenuse is 85 find the other sides in integers.
   
   If al1 the three sides of a right triangle are unknown, 0 friend, tell (me) quickly their values.
   
   Comment: Here we choose y = 2, since
   
   l
   
   + I divides b.
   
   =2X85x2=68 c=8Sx3=51. We could have chosen y = 4 5 ' 5 2 2 and got a = 40, c = 75. By using (2n + I), (2n + 2n), (2n + 2n + I) a
   
   we find 2n2 + 2n + I = 85 :. n = 6 and we get (13,84,85).
   
   Another formula when hypotenuse is gi,·en ~V1 ~ ~: ASlfln Itt: I ~: JjfqVI: ~: 'fiafiN!:)oi ~: II CLlIII
   
   2b 2by a=b---,c=-r- . y2+1 Y +1 Comment:
   
   a=
   
   2by b(y2 - I) c=y2 + I ' y2 + I
   
   2 a +c 2 =b 2. These are the same (except for a sign) as in (CL), but a and c are in-
   
   mJl
   
   4t""QJ~~'4dt1 ~ 1.~ fitq~ II CLIV II
   
   Comment: Take any two unequal numbers, say x = 2, Y = I. Then a = 4, c = 3, b = 5. If x = 3, Y = 2, we get a = 12, c = 5, b = 13. One can construct many such triplets.
   
   q'IIJ1*lM'"'d~q.if q'il"d~ ~~ I ~ ~: ~ ~ tR '!t~ij)'~+iq II CLV II
   
   m
   
   X'~ B
   
   Y
   
   C
   
   Suppose a bamboo BAC is broken at A and the part AC touches the groWld at C. Given the length x of the bamboo and the distance BC = y, find XI and x 2 by: XI
   
   I
   
   y2
   
   2
   
   x
   
   =- (x - - ) ,
   
   1 y2 x 2 =-(x+-) . 2 x
   
   Comment: Here XI + X2 = x and x; + y2 = x~ . So 2_ 2 2 2 2 2 2 2 XI -x2 -y =(x-x l ) -Y =X -2xxl +XI -Y2
   
   terchanged.
   
   ... 2 XXI =
   
   X
   
   2
   
   -
   
   Y2
   
   Formula to constnlct right triangles with integral sides 1'C,k\lt~llE4l CII)~4·ri~~ ~: I tfil4l'1((1q)~ ,,«I(tQlij)'(oil"d: II CLlIIII
   
   Example 1 Given any two numbers x, y, there is a right trinagle with a = 2xy, 2 c:: x2 (x > y) and b = x + 2 2 22 2 22 422 2+ 2 Comment: b =(x + y) = (x - y) + x y =c a . This formula can be used whether or not x, y are integers.
   
   -l
   
   l
   
   ~ ~ ~~q~IJa:"o~ ~ qq.,a·Ii~41~" W 1f9: I ~ iqftfdf.'~.,q'1 w;i ~
   
   ~ ~~
   
   'OR:
   
   ~II
   
   CLVIII
   
   120
   
   Lrlavati of Bhaskaraclirya
   
   A bam,boo 32 C (cubits) high, standing on a level ground, was broken by strong winds. The tip of the bamboo touched the ground 16 C from the foot of the bamboo. Then, 0 mathematician, tell (me) the height of the point where the bamboo broke.
   
   Comment: Here x= 32, Y = 16.
   
   121
   
   Mensuration
   
   , 'here was a snake burrow at the foot of a pole, on the top of which was seated a domesticated peacock. The pole was 9 C (cubits) high. ,'he peacock saw a snake crawl towards the post at a distance of 27 C (rom the post. The peacock pounced on ihe snake at the same speed as the snake's crawl and caught it at a certain distance from the pole. Find the distance quickly.
   
   2
   
   XI
   
   = -I ( 32 - -16 2 32
   
   )
   
   = -I (32 - 8) = 12 2 .
   
   r
   
   Example 2
   
   ~ ~~~m"d~v, 'Iflm: ~ '4fR~,*,;d(if'~ I
   
   m
   
   ~: ~: ~ ~ &lIMCISrilf(4q).I: II CLV" II
   
   [Near the foot of a pole, there is a snake burrow. A peacock jumps from the top of the pole and pounces on the snake. To find the distance of the snake from the pole.] Given AB = x, AC = y, to find AD = r and SD = DC = s. S
   
   x
   
   r=
   
   s
   
   s:
   
   f(y _:2)
   
   s=y-r.
   
   y Comment:
   
   Comment: Here y = 27, x = 9
   
   Here ~ =S2 _
   
   x2
   
   =(y _ r)2 _ x 2
   
   :. ~ = / - 2yr + ~ - x 2 I x2 :. r= - (y - - ) .
   
   2
   
   Y
   
   =
   
   f(
   
   27 -
   
   ~: ) =f (27 - 3) = 12 C .
   
   ..be peacock and the snake travelled equal distances of 15 C. NOTE: Some modem mathematicians may object to the above solUtion on the grounds that the path of the peacock is a parabola due to the force of gravity. However it is wrong to conclude that Bhaskariidirya was ignorant of gravity as the following verse (Goladhyaya - Chapter Bhuvanako!jll, Stanaza 6) shows. ~'1''>!~ff
   That is the earth attracts inert bodies in space towa:rds itself. The attracted body appears to fall down on the earth. Since the space is homogeneous, where will the earth fall? Also the peacock does not fall down like an inert coconut falling down from a palm. He purposefully pounces on the snake and keeps his speed equal to that of the snake. Such conditions arise in many mathematical problems. So we cannot say that the problem is wrong. Also since the problem is on right triangles, we are justified in taking SD to be a straight line. Formula to find height of lotus and depth of a pond
   
   Example ~ ~ ~ ~ ~:e'I~Hj~ ~: ~ . l'~
   ..
   
   ~r:ffl("f(iJ)'Rq;vrt<'<'«f h 4';we4ivtkt~V'"1$1'11 u ~ ~ If»R4'l il ~ ~ ~~~II CLIXII
   
   LitavatT of Bhaskaracarya
   
   122
   
   Mensuration
   
   123
   
   I-
   
   [Height of lotus flower is AC = AD = x. When the lotus moves ~y wind and its top touches the water at D. BD = r. BC = y.]
   
   Comment:
   
   1(4 1) 15 Length of lotus stem = -1(.4 - - + -1) = -17 C .
   
   Y- y ) .
   
   C y
   
   1 (r2 Depth of water AB = 2" r
   
   I - - - - -_D
   
   Depth of water = - - - =- C 4 . 2 1/2 2
   
   Height of the lotus
   
   2 1/2
   
   1 (r2 AC=x= 2" -y+y ) .
   
   QI()('f'4"'l1""4»J'",,Qt4
   
   'P': ....)flIl>Vlf"'l'l{
   
   lPJ ~ I
   
   ~: ",)Q1i'if.:qn ~ ~ ~ ~ Wn4QH,{1I CLX"
   
   o friend. the distance between B. the point where the lotus stem me~ts the water surface and D where the tip touches the water level. ,i!! called the base. The height of the lotus above the water surface AD - AB = hypotenuse - altitude. Altitude is the depth of water.
   
   J'
   
   Comment:
   
   Comment: One monkey travels MD + DB = r +
   
   AS
   
   D
   
   2xy + y 2 + r2 = x 2
   
   :. x=I (r; +
   
   r
   
   M
   
   y)
   
   =x - y = 2"1 (r2 Y+ y ) .
   
   c
   
   s and the other MA + AS = c + x and r + s = c + x. Also s2 = x 2 + (e + r)2 S2 = x2 + c 2 + r2 + 2cr and r2 + i + 2rs = c 2 + x 2 + 2cx :. 2cx - 2cr - r2 = r2 + 2rs :. 2cx - 2cr = 2r2 + 2r (c + x - r) = 2~ + 2rc + 2rx _2r2 :. 2cx = 4cr + 2rx
   
   Example ~ql'!ifA(f,«~~ ~ ~
   
   4
   
   IA monkey jumps from a point M on a palm tree AM to D and then onto a well. B. To find MD.] ex r=-2c+x
   
   Here (x - y) ~ + r2 = x 2 • 2 •• X -
   
   2
   
   lt~dfi1)f.\;t~ ~som tr.t ~,rvrniql: I <1r~~*,'''''+''~\(Q11 ,,~lJqI:j ~ ~ ffit I I CLXIIII
   
   x hypotenuse
   
   mi
   
   1
   
   Here y = ~ • r = 2 . 2
   
   :. r = ~ = Height of Jump. 2e+ x
   
   tm1'I
   
   ~ ~'f('fq;ftlIq)fli fad~UGiol't I
   
   t sample
   
   ~ ~ ~'iR~'1I~ ~ dRt!:·.. :j ~ ~ ie1g'i'1gQlUI't I I CLXI' I
   
   '.'_«f"~"Q""'lJ(f9~ 1IJ'fl ~: -'~
   
   In a lake there were a large number of ruddy geese and cranes. A lOtUl 1 whose height above the water surface was one vIta C (Cubit)
   
   ~ ~ ~ ,Rt'fIQIMi'q ~ ~I IttIi W«n ~ 'Idl!ilq'i"; ~ Ib rii"§qfhrlilsfta 'Tfilm Rni ~~ -q I I CLXm II
   
   its tip. bent by a rustling wind. sank at a distance of 2 C. 0 mathe· matician! tell (me) quickly the depth of water.
   
   'nlere was a palm tree 100 C (cubits) high and there was a well at a 111.Iance of 200 C from the tree. Two monkeys were on the top of the
   
   (I
   
   1 ):-1
   
   ,I~
   
   124
   
   LiJavatl ofBhiskariicilftl
   
   125
   
   Mensuration
   
   ~
   
   tree. One of them came down the tree and walked to the well. Th other one jumped up and then pounced on the well along the hypote nuse. If both covered equal distances, find the length of the secon monkey's jump. Comment: Jump =
   
   Bhaskaracarya calls the smalIer a the base and larger c the altitude. This problem can also be solved by quadratic equations. Example 1:lI ~: qsoft'5Q~ ~:
   
   100 x 200 = 50 C. 2 x 100 + 200
   
   ,""cr,)f~1~4~ ~
   
   An objection may be raised that the second monkey traverses a para bolic path But as per the conditions on his movement, the path is t be taken a straight line. ~
   
   lRJ I
   
   a 1) ~~ 1P: II
   
   CLXV II
   
   If in a right triangle, the hypotenuse is 17 and the sum of the other two sides is 23, find the base and the altitude.
   
   Formula: To solve a right triangle given its hypotenuse (b) and th sum of the other two (a + c)
   
   Comment:
   
   ~ ~ ft'lVllftf~w) ~:
   and
   
   [To find the base and altitude when their sum and hypotenuse of J right triangle are given.] Subtract, from twice the square of the hy potenuse, Ihe square of the sum of the base and altitude. The squarc;J root of this result subtracted from (resp. added to) the sum of the sideiJ and divided by two happens to be the length of the base (resp. all" tude). [Thus given b and a + c = s,
   
   Example
   
   ~
   
   e=
   
   5 + J2b 2 - 52
   
   2
   
   anda=
   
   5 - J2b
   
   2
   
   -
   
   2
   
   52
   
   .J
   
   Note: If the hypotenuse and the difference of two sides are given the ' to find out the base (resp. altitude), this difference should be su tracted from (resp. added to) the square-root of {twice the square the hypotenuse minus the square of the given difference} and divide by two; (see also the comment of stanza CLXVl). Comment: b 2= i + c2 = (a + c)2 _ 2ac = s2 - 2ac and (c _a)2 = (a + e)2 -4ac = S2 + 2(b 2 _ 52) = 2b 2 _ S2
   
   :. c- a= J2b 2 - 52 . and c + a=s Solving these equations, we have the formulae.
   
   ~:4»~:::a~
   
   '11m:
   
   'q5lIjf
   
   Q
   
   ~I
   
   ~ ~~ ~SS?J iIUlcp)t1*111 CLXVI II
   
   If in a right triangle, the hypotenuse is 13 and the difference between the other two sides is 7, find the sides. Comment:
   
   Here if c - a = s = 7 and b c = 5+
   
   J2b 2 -
   
   52
   
   7+
   
   = 13,
   
   ~r-i-X-I-32-_-7':""'2 = 12
   
   2 J'2-b2::--'_-5':""'2 and
   
   a=
   
   2 5
   
   17 - 7
   
   =--=5.
   
   2 2 NOTE: Khanapur Shastri used the same formula as in (CLXV) and got a = -5. We use a different formula. Although Bhaskaracarya knew about negative square root, he says that a base of right triangle cannot be negative and this is accepted by modern mathematicians.
   
   LIliivatl of Bhaskaracarya'
   
   126
   
   127
   
   Afensuration
   
   Existence of Triangles ~~~q~:1
   
   ~...:~RIIt'1'1j'ii
   
   ~~~~.
   
   eB U~q>ii1I§
   ~ q~: ~~II CLXVIIII
   
   ~ ~ ~ ~~II CLXIXII
   
   [There are two vertical poles AB, CD. Strings AD and BC meet at E. To find EF = Y given AB =r and CD = s.] The product of heights of two poles divided by their sum is the perpendicular (EF). The products of the height of a pole (AB, say) and the distance between the poles is the distance (BF) between the pole and the perpendicular. [Thus sr y=-- , s+r
   
   In a triangle (or a polygon) it is impossible for one side to be greater than the sum of the other sides. It is daring for anyone to say that such a thing is possible.
   
   BF=~, FD=~.J s+r
   
   s+r
   
   "1ar :. -Y = BF. Comment: 6BEF, 6B CD are simi S
   
   6EFD and MBD are similar :.
   
   Example ~ ~"~jQq>t '!J""6lIa ~ I ~ lf5I ~ ~er-i
   
   faf.lflfict II
   
   CLXX I I
   
   If an idiot says that there is a quadrilateral of sides 2, 6, 3, 12 or a triangle with sides 3, 6, 9, explain to him that they don't exist.
   
   t
   
   I. = FD . r
   
   t
   
   y Y BF + FD sr . Adding - + - = I :. y = - - . Another proof can be given s+r s r t by co-ordinate geometry. Clearly t is not needed to compute y.
   
   Example ~.~,q. .qmj('~: I
   
   $('~i('l<"IV' ..'t{illhhl"hf"M~ II CLXVIII II
   
   Two poles have heights IS C (cubits) and 10 C respectively. Top of each is connected to the bottom of the other by two strings. Find the height of the point of intersection of the strings.
   
   Comment: y = ~= ISx 10 =6C . s+r IS+1O
   
   Comment: This is well-known result in Euclidean Geometry. It is unfortunate that ancient Indian mathematicians did not perceive the need to offer proofs for their assertions.
   
   ~ ~"~c:"tf'(3°~ ~ cr'(f) ~I ~:~ 'i'C'i"'l!)
   ~~~:I
   
   ~ ~
   
   m
   
   ~ ~ 1flIfciI11 CLXXII II
   
   lin 6ABC, AD .i BC. To find BD, DC.] In a triangle, (asswning a side as base) form the product of the sum and difference of (the remaning) two sides. Divide this product by the base. Add to and subtract from the base, this quotient. Divide the results thus obtained by two, and get the projections of the two sides (on the base). Square-root of the difference of the squares of a side and its projection on the base is the length of the perpendicular (drawn from the vertex on the base). The product of half the base and the perpendicular is the area of the triangle. [Thus if c > b,
   
   128
   
   UlavatJ of BhaskariiciiryQ
   
   Mensuration
   
   BD =
   
   f
   
   [a + (c + b)a(C - b)]
   
   Example 2
   
   DC =
   
   !
   
   [a - (c + b)(
   
   .2
   
   C-
   
   b) ]
   
   a
   
   2
   
   222b p=c-BD=
   
   Dc
   
   2
   
   .
   
   I
   
   Also Area ~ABC = - pa.] 2
   
   a
   
   129
   
   Cuttc<1CtlSl4\ ~ ~ lI?I ~ ~ I
   
   3J1N
   
   ~ ~ ~ ~ 'fI~i~ ffif
   
   .q II
   
   CLXXIV II
   
   A triangle has sides 10, 17 and base 9. 0 mathematician, find its altitude, the two projections (on the base) and the area.
   
   Comment: Here projection of Comment: These are given in school geometry. If C is obtuse, some modifications are necessary. See Example 2 below.
   
   A
   
   AB=BD=
   
   f[9+ 27;7]=15,
   
   projection of AC = CD Example 1
   
   =![27X7 -9]=6 2 • 9
   
   ~~~~~~
   
   <.f5f ~ql
   B
   
   fa I
   
   (mY ;;(
   
   f4q",)eNf~ ~II CLXXIIIII
   
   If the base of a triangle is 14 and the other sides are 13, 15, 0 friend, find the length of the perpendicular on the base, the lengths of two parts of the base into which it is dividedby.the foot of the perpendicular and the area of the triangle.
   
   Comment:
   
   BD = ![14 + (28)(2)] = 9 2 14 ' DC=![14- (28)(2)]=5 2 14 ' p=
   
   C
   
   '
   
   2
   
   P=J10 2 _6 =8 and area = Y2 x 9 x 8 = 36 sq. units.
   
   (UlqHi!I(f)~R1 ?fi~ ~
   
   9
   
   JIS2 - 9
   
   2
   
   = 12
   
   Here At8 is obtuse and so there is an appropriate modification. Formulae: Areas of (cyclic) quadrilateral and triangle ~fim;i ~:~ ~ ~ ~I '11."1'f~e'fi~ ~~ 'fli4iCa)q1~C; ~ II CLXXV II
   
   From half the sum of all the (four) sides subtract each side separately and take their product. Square-root of this product is imprecise area of quadrilateral. (However) the precise area of a triangle ascends in this manner. [Let a, b, c, d be the sides of the quadrilateral and s = a+b+c+d ---,------ semi-perimeter. Imprecise area of a quadrilateral is 2
   
   =~(s -a)(s- b)(s-c)(s - d) . Area of a triangle with sides a, b, c is obtained by
   
   and
   
   ~ABC =! x 14 x 12 = 84 sq. units. 2
   
   putting d = 0, viz. Js(s-a)(s - b)(s-c).]
   
   Comment: The formula for the area of a quadrilateral is exact for a cyclic quadrilateral. Fonnula for the area of a triangle was discovered by Heron (200 B.C.). The for-
   
   131
   
   Mensuration Liiavati of Bhaskaraciirya
   
   130
   
   mer. as also pointed out by Bhaskaracarya. is not correct for a noncyclic quadrilateral fcr which the correct fonnula is
   
   ~(s - a) (s- b) (s- c)(s- d) -abed cos2 (P;F). This derivation can be found in a standard trigonometry book.
   
   ~V1ff<=qdl dltad<'lI ot ~: I I CLXXVII II
   
   The (primitive) formula for the area of a quadrilateral is not accurate. This i& so because the lenghts of its diagonals are indetenninate. So how can we get an accurate value? Ancient mathematicians had fixed some values for the diagonals but they are not valid in all cases.
   
   Example I "~Ng~{I4'i(1 j(!Jitif>'ti&i
   
   ~ ::ttj)i( .. ~\t
   
   Even if all the four sides of quadrilaterals are equal (i.e .• a rhombus). diagonals can be different and consequently we get different areas·
   
   '"
   
   ~~ lP.iI <~:ri
   A quadrilateral has base 14. upper side 9 and (the other) two sides 12 and 13. If the altitude is 12. find its area by the primitive method.
   
   Comment: Clearly the quadrilateral is a trapezium since one side and the altitude are equal. Also we can break the trapezium into a rectan-
   
   ~: ClioiQ)4qi 31~gt
   gle and a right triangle. I
   
   So the area = 12 x 9 + - x 12 x 5 2 = lOS + 30 = 13S. Or using the primitive fonnula: Area = J<s - a) (s- b) (s- c) (s- d)
   
   =J(24- 12) (24 -
   
   9) (24 - 13) (24 - 14)
   
   =J19800 == 140.7
   
   .
   
   We get imprecise value since the quadrilateral is not cyclic.
   
   Example 2 We now find area of the triangle of stanza CLXXIII: a = 14. b = 13. c
   
   =15. s = 21.
   
   Comment: Nowdays high school students know that three independent elements of a triangle are needed to fix a triangle. Similarly, five independent elements are necessary to determine a quadrilateral. In stanza CLXXVI, besides the four sides we are given one angle to be a right angle.
   
   Area = ~rs(s--a)-(-s--b-)-(s--C) = J21x7 x6xS = S4.
   
   '1
   til
   
   f.!Rrn
   
   fitnrttt 111
   
   ~ ~ II CLXXIX II lfl!m 111
   
   f.mfl
   
   ~: I
   
   O'f ~ ~a)2f
   CLXXX II
   
   If a perpendicular or a diagonal of a quadrilateral is not given, then its area is indeterminate. So one who asks such a question. is a devil and one who can answer it must be a greater one because he does not know that even if all the four sides are given. the area is indeterminate. ~ ,fa~?:Iitgjvrt4' ~ ffiliitftaq~d' tnl ~UIJ "§l'fil'RI~ ~ ~'II40n"1Ior't I I CLXXXI I I ~qsvlfMf(lttOd6' ~ ~ g~qg,~ ~ I
   
   ~~ ~ 1!f ~~ d"'V1qM~+:Ild: II CLXXXII II
   
   LiliivalT (Jf Bhaskaracarya
   
   132
   
   Subtract the square of a diagonal of a rhombus from four times the square of its base, and the square-root of the remainder is the other diagonal. The precise area of a rhombus is half the product of its two diagonals. In a rhombus of equal diagonals and rectangle, the area is the product of the adjacent sides. The area of a trapezium is the product of half the sum of the parallel sides and the perpendicular (distance between them). [Rhombus of sides =a and diagonals d l, d 2 : (Given a and d l) d 2 = ~4a2
   
   -
   
   d;
   
   (Given dlo d2) Area = ..!.. d l d2• 2
   
   133
   
   Mensuration
   
   If we choose d l = 48 then d2 = 14 and area is 336 sq units. Thus a rhombus with given sides can have infinitely many pairs of diagonal s and areas. (2) Area = 25 x 25 = 625 sq. units. (3) Area =6 x 8 =48 sq. units. Example 1
   
   ehmI ~ ~ 'f<:"'~9('<j
   
   fimPm ~j~" ~ ~ I ~ ~
   ~~~~t'I'?I~~IICLXXXVII
   
   Area of a rectangle
   
   =
   
   Area of a trapezium
   
   = ..!..
   
   product of its adjacent sides. (sum of parallel sides)
   
   2
   
   x
   
   height.]
   
   Comment: These are standard formulae. Example
   
   The front of a field is eleven, its earth (base) is twice the front and the other two sides are thirteen and twenty. Twelve is the perpendicular (between the front and the base). Then find the area.
   
   Comment: We are given a trapezium whose parallel sides are 11 and 22, other sides are 13, 20 and the altitude 12.
   
   A
   
   ehmI "'fJft~C4Jq9'~ ~~~~lIlf~1
   
   ~~~~(fQ'I~
   
   II
   
   B
   
   1 Area =- ( 11
   
   .
   
   + 22) x 12::: 198 sq. UnIts. 2 [Here Bhaskaracarya h as given one more datum than is necessary. Given • the four sides we can compute the altitude, as is clear from the figure.]
   
   12~6 I~~ 5 .4---22
   
   ~~a. <'flflt(1I'1i!~ ~ ~II CLXXXIV II
   
   Example 2
   
   (I) 0 mathematician, in a rhombus of sides 25, find two diagonals and area. (2) If the diagonals are equal, what is its area? (3) Find the area of a rectangle if its base and vertical side are 6 and 8 respectively.
   
   Comment: .(1) Choose dl d2
   
   = ~4 X 252 -
   
   Area =
   
   30x 40
   
   2
   
   30
   
   2
   
   =30. Here a =25.
   
   =40
   
   600 sq. units.
   
   q.il"'I~q;'fl~('11 ~ ~
   
   'aJ.:
   In a quadrilateral, the upper side is 51, base 75, left side 68 and right side 40. Find its area, diagonals and altitude.
   
   Comment: Here only 4 elements are given and so we suppose that diagonal AC = 77.
   
   LWivatI of Bhliskaracarya
   
   134 MC=
   
   A
   
   K--=:"':'_~B
   
   Mensuration
   
   135
   
   [Sides of quadrilateral ABCD are given. Take some value for AC = 77.
   
   -.!.[(77+68) (77-68) +75]= 231 2 75 5
   
   Sides of ~ABC and ~ADC are known. Suppose BM 1. AC, DN 1. AC. Compute AM and MC. Similarly altitudes can be computed. Thus we get AN, CN and DN in ~ACD. MN = DR. BR = BM + MR = BM + DN.
   
   A
   
   MD = 75 _ MC = 144
   
   5 .'. AM= 308 5 Area of the quadrilateral = ~ADC + ~ABC .= 2310 + 924 = 3234 square units.
   
   Then BD2 = BR2 + DR2.]
   
   that the quadrilateral is cyclic. 2 2 If we take C = 90°, BD2 = 75 + 40 = 7225 :. B D = 85.
   
   Assume some convenient value for one diagonal. If we take this diagonal as the base, we get two triangles. Then we get perpendicular B M as well as AM, MC, ND, AN, NC. ~ ~ ~ 2 2 BD- = BR- + RD- = [BM + ND] + MN .
   
   Formula for a diagonal of a quadrilateral
   
   Formula to find area of a quadrilateral
   
   tk'?,(lIlii"1P341~
   1fiVTff~ ~ ~ ~ q:1 ~~sv:r (l1!I1044iof: ~:
   From the primitive formula ~49 x 77 x 42 x 66 = 3234 which shows
   
   mI
   
   ~q~'('Iq~a'fil q~"q.lw ~ ~ ~: II CLXXXVII II
   
   The projection of a side (on another side: = base of a quadrilateral) is square-root of the difference of the squares of the (given) perpendicular and the side. Diagonal (concerning with the two sides already considered) is square-root of the sum of the squares of the perpendicular and the difference of the base and projection.
   
   Comment:
   
   (See figure of stanza CLXXXVI.) DM=JAD2 - AM2 , CM=DC-DM AC =
   
   and
   
   ./Mc2 + AM2 .
   
   Formula to find the second diagonal ~S?I ~: ~~ g"'AhZJ~
   
   ~
   
   crm:
   
   ~
   
   9
   
   ri~: ~ 'l) I
   
   " ~ ~ ~ lffilU) II
   
   CLXXXVIII II
   
   ~Q)~..,,,,~qat1
   ri
   
   ~ II CLXXXIX II
   
   II CXC II
   
   M~t'i4l" t1!J'M~C: S1f~\iCqJvi: ~ ~: I ~
   
   9
   
   ri~: ~ 'l)
   
   crm: ~ ~ ~II
   
   CXCI II
   
   Take the sum of the two smallest sides of a quadrilateral as the base and the remaining two sides to form a triangle. Draw the perpendicular to the original base (of the quadrilateral). [Although the length of the diagonal is variable or uncertain, it has maximum and minimum values.] The length of the diagonal is at most equal to the above base and at least equal to the length of the prependicular. So the length of the diagonal is choosen to lie between the two above values. A smart (student) can easily understand this. Taking this value for the diagonal, compute the sum of the area of the two triangles and this ' is the (approximate) value of the area of the quadrilateral.
   
   Comment: ABCD is the given quadrilateral whose sides are given 75, 40, 51, 68 but the B' fifth data is not given. B haskaracarya says that we can take the diagoC 91 'A' nal AC to have a length
   
   ~8
   
   136
   
   LIlavatT of Bhaskaraclirya
   
   between the base 75 and the altitude A'M of ~A'B'C' where A'C' = 51 + 40 = 91. Obviously AC < 51 + 40 = 91. Now AM ::!:: 66. So we can choose AC = 77 and then compute BD = 85 and area ABCD = 3234 square units. Formula to find the diagonals of a trapezium 'tf'l •.,t'i .. tQ ~~ jd)'1~ q~qs(Ol4 ~ I
   
   ~ ~ 'SQ<.ijq~q mu) ~ c1"~~'k1('ttq II CXCII II
   
   ~ ~: I d<"t'4"q1f~
   
   '!Ifi':
   
   ~ II
   
   'tf'l•.,t'i~ ~: ~ I j6''''4<1: :rigfiM~qs, ~ II CXCIII II
   
   Draw a triangle with the difference of parallel sides of a trapezium. its base and the slanting side!> as other two sides. Then find lengths o. the altitude and segments of the base (into which the altitude divides the base of the triangle). Next from the base of the trapezium subtract one of the segments. To the square of the remainder add the square of the altitude, and square-root of this sum is one of the diagonals. (It is clear that) a trapezium cannot exist unless the base added to the smaller side is together greater than the sum of the upper (parallel) side and the other side.
   
   Comment: AM.l BC, AE II DC, BF.l DA, BE = BC - AD. Compute AM, BM, ME in~ABE.
   
   2 AC = ~""'AM-2 -+-M-C-
   
   BD B
   
   c
   
   M E
   
   = ~BF2 + FD2
   
   Mensuration
   
   137
   
   3iIMti .. ", ~ '146 <;1'11'{1 "<0'ql~,fj~l\!fq ~ lfiohlil'Qffll ~~~~"eI'~IICXC~11
   
   A quadrilateral has two opposite sides 39, 52, base 60 and upper side 25. Perpendiculars are unequal. Diagonals are 56 and 63. Taking these four sides but diagonals of different lengths, construct a quadrilateral. If this new figure is a trapezium, find the lengths of the diagonals. This example was given by earlier teachers.
   
   Comment: D
   
   BM= ![60_102X24]= 48 2 60 5 AM = ~AB2 _ BM2 = 189 5 BN= ![60+ 108X4]= 168 2 60 5
   
   DN = ~BD2 _ BN 2 = 224 5 :. AM~ DN NOTE: If the reader knows Ptolemy's Theorem, it is easy to recognize ABCD as a cyclic quadrilateral. AC x BD = AD x BC + AB x DC 63 x 56 = 25 x 60 + 39 x 52 3528 = 1500 + 2028 which is true. i.e. In the above quadrilateral if BD = 32 (rather than 56), then what is AC? BM = ![32 + 64 x 14] = 30 (AM lED) 2 32 2 2 AM2 = 39 - 30 :. AM == 25 .
   
   MC= ME+ AD FD= BM+ AD.
   
   If CN 1. BD, then
   
   Example (Scalene Quadrilateral) ~'t,f.=Q.,~q;""~'fft,f=qJl ~ I ~
   
   9
   
   qqfa~I('(I, ~ ~
   
   lftl
   
   ~ II CXCIV I I
   
   BN =
   
   ! [32 + 112 x 8] = 30
   
   52
   
   = 77.
   
   :. M = N . 2 32 That is. diagonals of ABCD intersect at right angles. Many commentators on Liliivati have missed this point. Now Mel = 602 302 .. CM ::!:: 52 AC
   
   ::!::
   
   25
   
   +
   
   Mensuration LIlavalf
   
   138
   
   ofBhiiskaracan~
   
   plied sums by the other (unmultiplied) sum. Square-roots of these quotients are diagonals.
   
   Here the older writers should have. taken approximate values rathe than actual square roots. Now consider the third example wherein A 25 B ABIICD. AN II BC, AM.l DC. CN = 25, ON = 35
   
   r0U
   
   o
   
   R N 25 C
   
   35 M
   
   AM=
   
   J39
   
   2 -
   
   ;5
   
   DM= '!'[35-
   
   2
   
   Comment: Let m and n be the diagonals of a cyclic quadrilateral (see the adjoining figure). Then n2 = (ad + be) (bd + ac) ab +cd
   
   and
   
   91X13]=~ 35
   
   3 297 MC= 60-5'=-5- ' 2
   
   = 71.
   
   BD =
   
   m2= (ab + cd) (bd + ac) ab+ be
   
   Since 0. + e = 7t, cos 0. + cos e = 0 c 2 + d 2..-::;: n 2 a 2 + b 2 _ n 2 i.e. + =0 2cd 2ab 2 i.e. ab(c 2 + d1 + cd(i + b2) = n (ab + cd). Simplyfying, we get Brahmagupta's formula.
   
   5
   
   = ~~38016
   
   AC = [AM2 + MC
   
   139
   
   (2S+%J + AM2 = 46.7
   
   Bhaskaracarya's Method
   
   1 195 493 Area = - (60 + 25) x = - square units, slightly different from 2 25 2 the exact value. Current methods of approximation differ from those of
   
   ~'QJ,ru2ijii'!$kQ: ~ ~ ~ ~ I
   
   ~ ~ ~ ~ ~ ffif ~"'&'«4''t'h'f: II CXCVII II ~u: ~ !jiN"Q'~"" ~: ~I ~ ~ ~ ~sft¥r-t ~ ~ (f'$I ~: II CXCVIII I I
   
   1J.f:
   
   B hiiskaracarya.
   
   (Here Bhaskaracarya gives his own simple method). Take two right triangles. Multiply the sides of one of the triangles by the hypotenuse of the other and the sides of the other triangle by the hypotenuse of the first. These are (the two sets of opposite) sides of a scalene quadrilateral. Further, sum of the product of the bases (shorter sides) of the triangles and the product of their altitudes is one diagonal (of the quadrilateral), The other diagonal is obtained by adding the product of one base and the other altitude to the product of one altitude and the other base. Thus the diagonals are easily obtained (by taking two right triangles). We do not know why (our) foremathematicians followed longer methods (for obtaining the diagonals of a quadrilateral).
   
   Formula The great Indian mathematician Brahmagupta has given this formula. He has shown that if a scalene quadrilateral is cyclic and its sides are given, then its diagonals can be calculated. h is a matter of pride for Indians that this formula was discovered in the sixth century whereas it took another thousand years for the west to rediscover this result. Bhaskaracarya has aptly praised Brahmagupta for this discovery. ~"'QI~4'Qj~Qls;:q);:q'iI~d ~ I ~ JU1sd~!J",q~: cr;vft ~ ~ II CXCVI II
   
   (Diagonals of a cyclic quadrilateral). Find the sum of the products of sides holding a diagonal of a quadrilateral, and obtain a similar sun!for the other diagonal. Multiply them by the number obtained on adding the products of the opposite sides. Divide each of the multi. 1
   
   140
   
   LWivat'i of Bhaskarocarya
   
   Comment: Take two right triangles of (generally integral) sides aJ, b), c), a2, ~, C2,
   141
   
   Mensuration
   
   Divide or multiply the smallest and biggest sides of the quadrilateral (hopefully) by any (convenient) positive number. The outcomes are the perpendiculars of one optional right triangle. Now di vide the remaining two sides of the quadrilateral by the hypotenuse of the first optional triangle. The quotients are the perpendicular sides of the other optional right triangle.
   
   =ala2 + b lb2. The other diagonal =a lb2 + a2bl.
   
   Example
   
   NOTE: Bhliskaracarya claims that his method is much simpler than Brahmagupta's and wonders how it was not discovered earlier.
   
   ~ ~ ~ fe)f(lfqf(l'Rh~=S9~ ~
   
   Exampie
   
   'I[iI>J
   
   Take al = 3, bl =4, CI = 5, a2 = 5, ~ = 12, C2 = 13. Then the quadrilateral has sides 52, 39, 25, 60 which is the same one in stanza (CXCV). With sequence of sides (25, 39, 60, 52), the diagonals are 5 x 3 + 12 x 4 = 63. 12 x 3 + 5 x 4 =56. However, with sequence of sides (25, 60, 52, 39), the smaller diagonal is the same and the bigger diagonal is the product of the hypotenuse of the two optional right triangles, that is, the bigger diagonal is CI C2 =65. There are no square roots nor big products. Bhliskaracarya's formula can be proved by using Brahmagupta's formula and substituting Cla2, C2aJ, 2 2 b2 d 2 2 b2 C = a + I an C = a + 2' I
   
   l
   
   2
   
   CI~'
   
   c2bl for the sides and nothing
   
   2
   
   REMARK: Is is evident that if four sides of a cyclic quadrilateral are given then, to compute the diagonals, one needs two optional right triangles. Suppose we consider two right triangles with sides 3, 4, 5 and 8, 15, 17. Then, by the above method. diagonals of the quadrilateral with sequence of sides (25, 39, 60. 52) should be 3 x 8 + 4 x 15 = 84 and 3 x 15 + 4 x 8 = 77. These are against the correct values 63, 56. Thus these right triangles are not compatible to a quadrilateral with sides 25, 39. 60, 52 (taken in any sequence), and we cannot choose two right triangles in an arbitrary manner. A pair of compatible right triangles to a cyclic quadrilateral can be choosen in the following manner:
   
   ~ ~tif'~f1l: ~ ~ ~: WfI ~f43df,(~r4
   
   ffi1
   
   ~I
   
   dR"iil.,,'
   
   ~ ~"
   ~~~~I
   
   m1IN If
   lR ~ ~~:
   
   wzn: wnu)
   
   'Il(
   
   .nf'ttlfflili ~~ ~ ~~ l:~sffl
   
   1(;
   
   1ht I I
   
   CC I I
   
   A (cyclic) quadrialteral has base 300, upper side 125, right side 260, left side 195 and the diagonals are 280 and 315. The perpendiculars are 189 and 224. Find the lengths of the following: L (1) GE, HF, DE. EC, DF, FC (2) PT, DT, TC (3) SR, DR, RC (4) SD. SC.
   
   s
   
   o o
   
   o::t
   
   D48 E RTF • 300
   
   132
   
   C •
   
   =300, CB = 260, AB = 125, AD = 195. Other values, AC = 315. BD = 280, AE = 189, BF =224, can be computed from the above and need not have been given. DE2 =(195)2 - (189)2 :. DE =48. EC =252 Comment: DC
   
   CF = ~2602 - 2242
   
   =132, FD =168, EF =120.
   
   1.t2
   
   LTtavatT of Bhaskaracarya
   
   From similar triangles AEC. YDC. CE = DC :. DY = 225. AE DY Similarly. we get RC = 300 - DR. 33 SR = 56 (300 - DR).
   
   143
   
   Mensuration 'f1~c1C161jVft f.\\ilf.\\iI(ijil''ii~ ~ ~
   
   m.em::
   
   w:zn: I
   
   gl~dd~rrq ~ II CCVI I I
   
   s
   
   16 SR = 63 DR. Solving RC = 3564 . GE = 64. GD = 80. HF = 99. 17 II 195 x 15 , SB __ 2600. HC = 165. RF= 77-. SA = - - 17 17 17 PT = 144. cr = 192.
   
   DR = DCXY2 xx \ Y2 X l
   
   +Y 1X 2
   
   x y x-'-+x 2
   
   Y.
   
   2
   
   ~~ l'iuI,oqq'fQ ~ I
   
   ~ 1J,: ~
   
   lmli
   
   ~ ~II CCI II
   
   ~:~~~I ~ ~~"tffi! "'fQffiPN: ~ II CCII II
   
   (Here Bhaskaracarya gives his method to solve the above intricate example.) DE and FC (projections of AD and BC on the base) are called sandhis. When these are subtracted from the base. the remainders are called pTrha or complements of the projections. Then GE = DE x BF ,GD = DE x BD . DF DF ~
   
   1J.vft
   
   A~4td~m ~
   
   m
   
   ~: I
   
   ~ ~ ~~: ~ Tfll CCIII II
   
   YD and LC are called
   
   ~poles '.
   
   They are at right angles to the base and
   
   . tops are on the d"lagonals. YD = DCx AE . Then we can get the th elr EC height of the point of intersection and the segments of the base by the use of stanza CLXVII. ~ ~: 't'«,.CljOI: ~ ~: I
   
   ~ ttt'('~"1'r.qdl
   
   'tt't4\('fI~
   
   1J.vft
   
   cfI ~ II
   
   CCIV II
   
   ~ ~ ~I
   
   ~:~:~~:IICCVII
   
   Comment: Two equations (for DR and SR) derived in stanza CC are given here by Bhaskaracarya in different fonns. Formula to compute the circumference of a circle given its diameter ~ ... .;c::lfi::n:a ~ (!JUlOI,«tl: ~ 11~: I
   
   'J~tlf(\&\ ~~
   
   M:
   
   ~S1llJ1 't4I&1146J,<41'4: II CCVII II
   
   For a given circle. the product of the diameter and 3927/1250 gives a good approximate value of the circumference. while the product of the diameter and 2217 gives a rough approximation of the circumference.
   
   Comment: Circumference
   
   ~
   
   3927 1250 22 7
   
   )(
   
   Diameter
   
   )(
   
   Diameter.
   
   circumference . . . IS a transcendental Irrational numdiameter her and is usually denoted by 1t • 1t is not a root of any algebraic equaIt is known that
   
   .
   
   .
   
   .
   
   n ,3m - ,3.1416,
   
   uon. For calculatIOns we use approximate values -
   
   7 1250 etc. It shows that Bhaskara and his (Indian) foremathematicians were
   
   144
   
   LrliivatI of Bhaskariicarya
   
   I'Ity 0 f (Circumference) · . . , that .IS that of aware 0 f the Irral10na dIameter Vedic term for 1t is trita (~gveda 1.52.5).
   
   145
   
   Mensuration Example
   
   1t.
   
   The
   
   q
   ~ ?liS,"Il"ia~ ~ ~ fiti ~ 1~ wi ~ ~ ~ ~ <.'ilC"ilq~'{11 CCX
   
   Example ~ ~ lf1
   wnui
   
   ~: ~~ I
   
   
   Find the circumference of a circle of diameter 7 and the diameter of a circle whose circumference is 22.
   
   Comment: Circumference
   
   What is the area of a disk of diameter 7? What is the area of a net which just encloses a ball of diameter 7? What is its volume? 0 wise friend, answer if you know pellucid UIavaH well. 22
   
   Comment: Area of the disk = -::;- x
   
   = 33.. x 7 = 22 . 7
   
   (7)2 77 . 2' =2" = 38.5 sq. umts.
   
   22 49 Surface area = 4 x - x -
   
   7
   
   22 = 7 . 7 This formula requires calculus for its proof. Diameter = 22
   
   II
   
   +
   
   4
   
   . = 154 sq. UlUtS.
   
   4 22 7 3 539 2 . Volume = - x - x - 3 == 179 - cft. umts. 3 7 2 3 3
   
   Formulae: Area of a disk, surface area and volume of a sphere
   
   ~ ~ "'''1ql~~ 1l~ ~ qq,«~«~ I ~ -qftfu
   ~ ~51!t4t ~ ~ ~
   ~~
   
   For greater accuracy, 3927. 2 d . Area of a dIsk = (DIameter) , an 5000
   
   ';;f
   
   ~ ~
   
   dm
   ~ ~ f.rq-ff ~ ~II CCIX II
   
   Area of a disk =
   
   .!.
   
   of a sphere = 4
   
   x
   
   4
   
   (circumference)
   
   x
   
   (diameter). Area of the surface
   
   area of its great circle. Volume of a sphere =
   
   liD' )2 for most practical purposes this may be - ( Iameter 14
   
   I 6
   
   (surface area) x (diameter).
   
   Comment: If the radius of the sphere is r, Area of great circle =
   
   .!. 4
   
   (2m) (2r) = m 2• Area of surface = 4m 2•
   
   I 2 4 3 Volume of the sphere = - (4m ) (2r) = - m .
   
   6
   
   D3 D3 I . + - x - where D = DIameter. 2 2 21 V = .-E.- D 3 ~ .!. 7t 8r3 =~ m 3 . 6 3 2 x 21
   
   Comment. Volume V = -
   
   3
   
   These formulae are proved in any calculus text.
   
   .
   
   146
   
   LTnivatJ of Bhaskaracarya
   
   BD2 :. ED= - - AD
   
   ;:tIJIllJIM1'lff:::aNkf*1c4 141't1'ffiC;::") ~: m: ~ I llJltlr
   ufun II CCXIII
   
   II
   
   ~ ~ 141't1SPfluj ~ ~ II CCXIV II
   
   (111ese two stanzae give three relations between chord. its arrow and diameter of a circle. Indeed. one of the three is computed when the olher two are known. BC = chord. MB = radius r. AE = Diameter 2r. AD = 'Arrow'.] Subtract the (postive) square-root of the product of the sum and difference of the (given) diameter and the (given) chord. The result divided 'by two is the arrow. Twice the (positive) square-root of the prodE~----~----~A uct of the (given) arrow and the difference of the (given) diameter and the arrow is the chord. Divide the square of half the (given) chord by the given arrow. The sum of this result and the arrow is the diameter. [Indeed. AD =
   
   .!. [AE 2
   
   J(AE + BC)(AE - BC)
   
   BC = 2BD = 2JAD (AE - AD) 2
   
   AE = AD + (BD) .] AD 2 Comment: FC = BF2 - BC 2 = AE2 - BC2 = (AE + BC) (AE - BC) TC =
   
   ~
   
   FC
   
   147
   
   Mensuration
   
   Formula: Perpendicular to a chord
   
   BD2 :. AE= AD+ ---. AD
   
   Example <:ttfil~'l'til"d4i1 vm ~ mJ I ~ lI'<=' 1111 0 II W
2. 
   o friend,
   
   diameter of a circle is 10 and its chord is 6. Find the length of the arrow. If the arrow is given. tell (me) the length of the chord. If the arrow and the chord are given. find the diameter.
   
   Comment: AE = 10, BD = DC = 3. MD = JMB2 - BD2
   
   = JS2- 32 = 4
   
   :. AD = I .
   
   In the second example. BD2 = BM2 - MD2 = 9 :. BD = 3 and BC = 6.
   
   In the third example. AE = AD +
   
   (BD)2 9 -w = 1+ "1 = 10.
   
   Formula: To find the sides of a regular inscribed polygon
   
   ~&iiq;lfi':o1., .. ,,,.eta"IU"<e9'1I~: I l\(~"IOI(1j .. tt~ ~: ~ I I CCXVI II 111"':1.,&1118\,,, fl~';~j't1"~: I '!1iClitc::t1U,,, ~ 141't1't1'fI~" I I CCXVII I I
   
   ~~~~:I 1'til"d'iU<*'lqI Oii, .,qldlod ~~ ~~ II CCXVIII I I
   
   ,'. AD = AM - MD =
   
   I
   
   '2
   
   BD2 = AD x DE = AD
   
   AE -
   
   I , -------------
   
   '2.J(AE + BC) (AE- BC)
   
   (AE - AD)
   
   :. BC = 2BD = 2JAD (AE - AD) ED x AD= BD2
   
   Sides of regular insribed polygons are: No. of sides 3 4
   
   5 6 7 8 ·9
   
   Side 103923a 84853a 70534a 60000a 52055a 45922a 41031a
   
   Diameter where a = - - - 120000
   
   149
   
   Mensuration
   
   148
   
   Lilavati of Bhaskaracarya
   
   Example
   
   Comment: For a regular polygon of n sides inscribed in a circle of 3I<erc::tlit)"i ~'f'"i'li'l't '1",r~~6t., "" ~ ~ I ~~ ~~ mf «Rj ~ (!fl$~~ ~ lif mill CCXX II
   
   radius r, side = 2 r sin ~ . n
   
   Using Trigonometrical tables, we get different values for n ... ,9.
   
   = 3, 4,
   
   5, Find the length of the chord of a circle whose radius is 120 and arc one-eighteenth of its circumference. Find the lengths when the arc length is doubled. trebled etc.
   
   Formula for the length of a chord: a rough approximation
   
   I Comment: Here d = 240, c = Is p.
   
   q,qHR&fQ~IQ: ~~: ~
   
   ~:~~:I
   
   3I,r.nRd'1 -mJ cR 'oIuli*t9tA I ~ ltt.l'f\Ol,
   Subtract the (given) arc (cut off by the chord) from the circumference. Multiply the remainder by the circumference. Call (this) product 'First'. (Now) multiply the square of the circumference by 5/4, subtract from it the First, and divide the First by this remainder. The result multiplied by four times the diameter is the (length of the) chord.
   
   Comment: d = diameter, p = circumference, c = arc length. 4dc (p-c) Length of the chord = 5 2 4"P -c(p-c) This fonnula gives an approximate value and it is difficult to trace its derivation. When the arc length c is zero, so is the chord length. The same is true if c = p, the circumference. So length of the chord should be proportional to c/(p - c). The third factor should be a multiple of the diameter d. Clearly the denominator must be of the second degree (one less than that of the numerator) in p and c. So the fonnula must be f.
   
   xc (p - c) . = y-c(p-c) . Now we have two specml cases:
   
   (i) f. = P when c =
   
   f.
   
   =
   
   I 17p 4d p x 18 5p2 _ 17p2
   
   18
   
   388
   
   42.06.
   
   97
   
   4 324 Its exact value = d sin 100 = 240
   
   x (0.1736) = 41.664. 2 When c = - p, f. = 84.12. 18 In Bhaskaradirya's time (A.D. 1150), there were no tables of logarithms or trigonometric functions and so the students were given ap-
   
   proximate fonnulae. Formula to find the arc length, given the chord ~Qld9dJftfifqscnfit"~ ~qift'Jqq1~: ~ ~: I
   
   ~~
   
   ~ ~ 'l",<::C'lINfM ~: ~ II CCXXI II
   
   Divide the product of the square of the 'circumference and 5/4 times the (given) chord by the sum of the chord and four times the diameter. Subtract this quotient from one-fourth of the square of the circumference. The (positive) square-root of this remainder subtracted from half the circumference gives the bow (the smaller arc).
   
   £. and 2
   
   (ii) t == £. when c = ..E.. 2 6 Solving these two equations for x and y we get the result.
   
   =68d = 17 x 240
   
   Comment: Arc length c =
   
   P
   
   2" ±
   
   2
   
   5
   
   -
   
   2
   
   P
   
   f.
   
   £.... __ 4_ 4
   
   4d + f.
   
   LWivali oJ Bhiiskaracarya
   
   150
   
   Solve the above fonnula (cf. stanza CCXX) for c and the two answers give the major arc and me mmor arc respectively. (Bhaskaraciirya intends to give only the minor arc.) Example ~ ~ ~ 301ltttfflll"l:' ~ ~I ~ a~ ~vlf4;ln"1VI~ ."fV1faq;~~~vI'{ I I CCXXII I I
   
   o arithmetician! If you know the fonnula for the arc length of a circle, find what proportion the arc length is of the circumference when the lengths of the chords are 42, 82, 120, .... The diameter of the circle is 240.
   
   CHAPTER 29
   
   Volume
   
   Comment: Here I. = 42, d = 240, P = 754.16. Since p is large, to facilitate the calculations, we compute 5 I I 4"1. -=-± - - - p 2 4 4d + I.
   
   c
   
   If I.
   
   = 82,
   
   c
   
   p
   
   =-I
   
   17 I =-or-forl.=42. 18 18
   
   8 or - . 9 9
   
   ilOlfq~I
   
   f?rnm- ~ ~ ~ I
   
   ~Hq;fqfQl ~qfllfd~ci ~ "If.}
   
   ~~
   
   T{
   
   wm q.,t!f<'It'4(§t11 ~ 11 CCXXIII II
   
   If it is an irregular ditch (or solid). measure the breadth at various points, add them and divide by the number of points. That is the (average) breadth. Similarly. calculate the (average) length and the (average) depth (or height). The product of the three (averages) will give the (average) volume. Example 'i""'HMI(H ~ <~?I14;","fq('1'{1
   
   fin! ~ Wi{Q·:ifj
   o friend!
   
   -q 11
   
   CCXXV II
   
   Measurements of an irregular solid were:
   
   151
   
   152
   
   Lflavatf of Bhaskaracarya
   
   lengths 10. 11. 12 e (cubits). breadths 6. 5. 7 e and heights 2. 4. 3 e. Find its volume.
   
   Comment:
   
   Average length
   
   =
   
   Average breadth
   
   =
   
   10 + 11 + 12 = 11 e . 3 6+5+7 =6e .
   
   Average height 2 + 3 + 4 = 3 e 3 . Average volume = 11 x 6 x 3 = 198 e 3. It should be noted that the measurements should be carried out at equal inte.rvals. It is clear that Bhaskaracarya gives practical methods.
   
   ~: I
   
   'tPHiJI(1Qi('1sDt,: ~ ~ ~ II CCXXVIII
   
   Comment: Volume of frustum MM' [pq + p'q' + (p + p') (q + q'»).
   
   Volume of a pyramid is
   
   ~ c::nii'<:n~«I9~ ~'k1I'(~uj
   
   9 ~ ~I q=;f
   
   ~ II CCXXVII I I
   
   There is a well in the shape of a frustum of a pyramid. (The base is smaller than the top.) Its top is a rectangle of sides 10 x 12 e 2 (cubits) and the base is half the size of the top. i.e. 5 x 6 e 2• If its height is 7 e, 0 friend, find the volume of the well.
   
   Examples 2, 3, 4, 5
   
   Find the sum of the areas of the mouth (top). bottom (base) and the rectangle whose (adjacer.t) sides are sums of the lengths and breadths (of the mouth and the bottom). The sum (of the three areas) divided by six is mean area of ditch (frustum). The mean area multiplied by depth (height) is the volume. (This) volwne divided by three is volume of pyramid.
   
   "6I
   
   Example 1
   
   Comment: Volume = - [120 + 30 + 15 x 18] 6 3 = 490 e •
   
   ~~~~,
   
   =
   
   These are standard formulae given in texts. Here ABeD and A'B'e'D' are similar rectangles.
   
   7
   
   Formula: Volume of a pyramid and its frustum
   
   It'i
   
   153
   
   ~: ~ 'fi'
   3
   
   j''&Vt(1('1'''(1tjf
   Volume
   
   ..!. 3
   
   ~S2I ~i"'4>'
   ~ ~ C:f,~~ciq~a ~ ~ ~ ~~ ~~'" IICCXXVIII I I
   
   (2) The top of a hole is 12 x 12 e 2 (cubits) and its depth is 9 C. Find its volume. (3) The circular top of a hole is of diameter 10 e and its depth is 5 e. .Find its volume. Find the volume of a pyramid (cone) whose base and height are: (4) as in (2); (5) as in (3).
   
   Comment: (2) This is a box (rectangular parallelo-piped) whose volwne = 12 x 12 x 9 = 1296 e 3.
   
   that of a prism
   
   with the same base and same height.
   
   ,
   
   (3) This is a right circular cylinder whose volume = 22 7
   
   = 2750 = 392.86 e 3 . 7 (4) Volume of the pyramid = I
   
   q
   
   ..!. . 1296 = 432 e 3. 3
   
   3
   
   (5) Volume of the cone = - (392.86) = 130.95 e . 3
   
   x
   
   (5)2
   
   x
   
   5
   
   154
   
   LillivatJ of Bhliskaraclirya
   
   Formula: Volume of a prism ~~~:~~tA~1 t~(WJ:*'
   
   tA
   
   ~: s~IIiNRI4Rhq ~ II 'CCXXIX II
   
   $~",)
   Volume of a prism equals the product of the area of its base and height. Number of (rectangular parallelo-piped) bricks in a heap . (prism) is obtained by dividing the volume of the heap by the volume of one brick. If the height of the heap is divided by the height of a brick. we get the number of horizontal layers of bricks.
   
   CHAPTER 30
   
   Comment: This is straight forward.
   
   Wood Cutting
   
   Example ~ICI'«(ICpM
   
   ~ ~
   
   fili
   
   taf
   
   fit'84) 4IGt'I¥til¥I(iJm:
   
   q)(1I4~Hi ~ ~
   
   ifjJ
   
   \tl ~ ~ I I
   
   hzhi !JlV'Ja+i !Jt'l1t4t4)4t I
   
   c:r'(i
   ~ ~ ~1Ii\()fa:t~ I
   
   CCXXXII I I
   
   [Log is a frustum of a cone. AB A rectangular prism 8 x 5 x 3 C (C = cubits) is to be constructed by lISing bricks of dimensions 18 x 12 x 3 A3 (A = arigu/as). Find the volume of the prism. number of bpcks required and the number of layers. (l C =24A). 3
   
   Comment: Volume of the prism
   
   . Volume of a bock
   
   = 8 x 5 x 3 = 120 C3
   
   =
   
   No. of bricks = 120 + No. of lavers .
   
   =3 +
   
   18xl2x3 24 x 24 x 24
   
   ~
   
   3 24
   
   -
   
   64
   
   = 2560
   
   =-3 C J 64
   
   =d], DC = d2• MN = h.J
   
   D~======::::lC
   
   = .!!
   
   (d1 + d2). 2 Multiply this by the nwnber of times the wood is cut to get the total area in A2. (A .= arigu/as). To get the answer in C (cubits) divide by 576. Area of a section (ABCD)
   
   N Comment: Section ABCD is a trapezium. If several sections are made. obviously they are not equal in area So this gives an approximate answer. I C2 = 24 x 24 A2 = 576 A2.
   
   = 24.
   
   155
   
   .
   
   . 156
   
   LrlavatiI of Bhaskaracarya
   
   Example
   
   1@ '1(!Ji3'C'ff'%)Slf
   
   ~sJ)
   
   ~: "(fijC'f~ ~ ~ ~ I (1(F(i<,'(o'~l ~ t:ffi,~
   frti ~ lid 11f1lm ~~ 1) II
   
   CCXXXIV I I
   
   A log of wood has base 20 A (ailgulas) long, top 16 A long and height 100 A. If it is cut at four places, o friend, find the total surface area of the wooden pieces.
   
   CHAPTER 31
   
   Volume of a Heap of Grain Comment: Area = 4
   
   100
   
   x -
   
   2
   
   = 12.5C
   
   2
   
   (20 + 16) -
   
   I
   
   576
   
   C
   
   2
   
   ~ ~S'j4Sij~Oi,c::tlitl:
   
   •
   
   ~:~~:I
   
   Formula: oblique cutting of a log ~ ~ ~
   
   R14j*'Cfi1. ~a~~~t:a: ~ 'fI1:'I1 ~~Oil~~ 'i1AJt~hJf<'W)Iq;t1QjqfS~ ~ ~ I q;4CM'(\if~'
   A log of wood is such that its vertical sections are squares. If it is cut obliquely, then the sections are rectangular and their areas = length x breadth. Cost of a heap of bricks or stones or carpentry depends on the number and skill of workers as well as softness or hardness of wood.
   
   ~~~~ \""1~C't0i'<': ~ ~ l!TVf: I I CCXXXVIII I
   
   If the grains are· big and spherical, their heap (right circular cone) has a height which is
   
   ~ th 10
   
   of the circumference. If they are small and
   
   I. . d·· I h spherical, the proportion is lIth and If they are pomte . It IS "9 t .
   
   . Volume can be computed by (height) x
   
   (
   
   • C' clrcumlerence 6
   
   )2 . TIus. .gives
   
   the answer in C3 (cubits); it also represents the number of Magadha khlirikas.
   
   Example q1rt~hk:::afiH'jijC'f114 ~'6ffi41 ~ lf5f ~ I
   
   tJ~ fq4a~"1q'tj lJ1'.J~ ~ ~ mI ~ 1) I I CCXXXVII I
   
   A log is 32 A (ailgulas) long and 16 A thick. It was cut obliquely at 9 llaces. Find the total area of the sections. Comment: Area =
   
   9 x 16 x 32
   
   576
   
   2
   
   = 8C .
   
   157
   
   158
   
   LI/avall of Bhliskaracfirya
   
   Comment: A heap of grains is in the form of a right circular cone. Its circumference can be measured but not its height without disturbing the heap. So Bhaskaracarya gives methods to compute the heights for 3 different types of grains.
   
   ~
   
   Volume = 27tr)2
   
   1tr2 h (r radius) as seen before. Bhaskaracarya's formula
   
   7t2r2h h = - 9-
   
   = (6
   
   = "3I1tr-. h.If we take 1t = 3. This is good for all
   
   ~ 'olume
   
   159
   
   ofa Heap of Grain
   
   (If the heap is in an outer comer, the base is
   
   2 th 4
   
   area of the disk.
   
   If it is against one wall, the area is half. If it is between the two walls, the area is one-fourth.] If the heap is kept in the outer corner, the circumference of the circle (of which the base of the heap is a part) is
   
   .!
   
   of the length of the base. If the heap is on one side of a
   
   3
   
   wall, multiply by 2 and if it is in the interior it will be 4 times. The
   
   practical purposes for the village folk .
   
   actual volumes of the heaps of grain will be Example
   
   2 ,.!. ,.!. 424
   
   respectively of
   
   the volumes so obtained. ~~mwd:~:~:
   
   ~~I ~ ~
   
   mti: fiMimT:
   
   Example
   
   ~ ~
   
   3rll ~~~ ~ it
   
   lftw{ II
   
   CCXXXVIII II
   
   On a flat ground there is a heap of big spherical grains of perimeter 60 C (cubits). 0 mathematician, find how many khariklis (K) are there. Find the same if the grains are small or pointed.
   
   r
   
   Comment: (i) Big, spherical: Circumference p = 60 C, h = 6 C, Vo!ume =
   
   (~O
   
   6 = 600 K;
   
   small spherical: h = 60 • Volume 100 x -60 = 545 -5 K . II 11' II 60 2 . d h = -60 Volume = (iii) pomte: 100 x - = 666 - K 9 ' 9 3' (ii)
   
   Formula: Heap in a corner ~~llt)q;~~ ~: ~ I
   
   fWq~"'f(u'vr~: ~ II CCXXXIX
   
   II
   
   ~ ,(1t)fAtl(Vi'(; ~
   
   aRt:cp)ulf~4(1t41~ ~1Mili,(: ~ I ~:~
   
   qq61 '1 q'fifi1'a:
   
   ~~ 1) ~ Q"5'R111 ~~ ~~ II CCXL II
   
   Find (i) (ii) (iii)
   
   the volumes of heaps of grain stored in front comer with length of the base f = 45 C (cubits), on a wall f = 30 C, inside comer f = 15 C.
   
   Comment: In all cases circumference = 60 C. Let h be the height (determined according to stanza CCXXXVII). (i)
   
   Volume
   
   = 2 (60)2 h = ~
   
   (ii)
   
   Volume
   
   = -1 x 100 x h =50 h C 3
   
   (iii)
   
   Volume
   
   = .!. x 100 x h =25 h C 3
   
   4
   
   6
   
   4
   
   100 h = 75 h C
   
   2
   
   4
   
   3
   
   CHAPTER 32
   
   Shadows Formul a: Length of a shadow tJYml): qiv'(J),<"'d~ ~ ~: q'f~,~«cml IC'tucfl .. q: I ~: ~ ~ qiulf('j'<'t 1fM~o~"9.m ~ ~: JA II CCXLI II
   
   Obtain squares of the (given) difference of hypotenuses and that of shadows. Divide 576 by the difference of these two squares. Add one to this quotient and then take the (positive) square-root. Multiply this result by the difference of hypotenuses, and write the product in two places. Add the difference of shadows to one place, and subtract the same from the other place. The outcomes divided by two are lengths of shadows. [Height of the vertical pole
   
   r
   
   =r, its shadows are t, t + q and the corre-
   
   sponding hypotenuses are c, c + p. We are given r, p, q and we have to find t when q > p. t
   
   = !2 [ _q + p
   
   I+
   
   576 .] q2 _ p2
   
   t q + q.] t is The other shadow
   
   161
   
   LIlavati of Bhaskaracarya
   
   161
   
   Comment: This concerns the altitude of the Sun. Some calculations are required. A pole (of length r) is fixed and a string is tied to its top. Marks are made on the ground where the shadows fall and the corresponding lengths of the string are noted. (c + p) 2 = r + (q + t) 2 and c l = r + 12 2
   
   163
   
   Shadows If BD = 3 C (Cubits = 72 A (arigulas)], AB = 3
   
   ~ C (84A) and CD =
   
   12 A. Find the length of the shadow.
   
   Comment· PD = .
   
   12 x 72 = 12 A. 84 - 12
   
   2
   
   :. 2cp = 2qt + m where m = q - p :. 4c 2p2 = (2qt + m)2 2 2 :. 4p2 (r + = 4q 2t2 + 4qtm + m :. 4t 2 (q2 _ p2) + 4mqt + m 2 _ 4p2r2 = O.
   
   Formula: to find the height of a lamp
   
   e)
   
   OI4)~ ~ ..,,
   Solve this quadratic equation and choose the root with positive sign and put r = 12. In LIlavatI. height r usually corresponds to a gnomon of 12 arigulas.
   
   Formula: To find the length of the shadow when the heights of the lamp and pole as well as the distance between them are given.
   
   ti'$\ 1fb'
   AB=
   
   BDxCD CD +. PD
   
   Comment:
   
   From the similar triangles PCD, CAQ, CQ PD. _ CQxCD AQ = CD .. AQ PD :. AB = BQ + AQ = CD +
   
   = BD xCD PD
   
   BDxCD PD
   
   [Given AB, DC = 12, BD, find PD.] Example PD =
   
   12 DB AS-CD
   
   Comment: By similar triangles, PCD, CAQ, PD CQ PD CD = AQ
   
   Given BD = 3 C (72A), PD = 16 A, find AB. DB
   
   :'IT= AB- BQ
   
   p :. PD =
   
   12 DB AB-CD
   
   Take CD = 12 A. 72x12 11 AB= + I2=66A=-C . 16 4
   
   Formula: Distance between the base points
   
   fctt'1~ t'I ..it
   Example
   
   ti~"'~'<~~' ~~: ~
   
   Comment:
   
   lhtl
   
   tiif)h'(1C:1S4iI!JCf;fl~ ~ ~ fl:n~qJl ~~ I I CCXLIII II
   
   BD=
   
   1411 ~~ ;flq.,'tid~ ~ I I CCXLVI I I "'~••;Ylm
   
   PD (AS- CD) CD
   
   '
   
   164
   
   LilavatT of Bhaskaraclirva
   
   ul(nu~~(f'(~:IU" 1fl U'<JIY""V~(f'(&,~
   
   1J.: I
   
   ~i~\mf: JNtn ~: ~ ~q~'lYiiZZlilq II eeXLVII II ~'
   [AB is the lamp. Pole CD = 12 A is kept at D and D' with lengths of shadows QD and PD'. DD' is also given.] PB - QB = PQ = d A PD'-QD= f. d QB=t XQD
   
   PB=~xPD' f.
   
   p
   
   D
   
   D
   
   B
   
   AB= QBx CD QD .
   
   All three are obtained by the Rule of Three. Just as the Lord occupies the universe in different forms, the Rule of Three pervades calculations involving shadows. NOTE: It is not clear as to what purpose is served by calculating the lengths of shadows and the distances of their tips from that of the lamp. Obviously, Bhaskaracarya knew the principle "Light travels in a straight line", but no reference is made to its use in Astronomy. Example
   
   wm
   
   ~$~ai1«'f(I ~ f$?lIC2i 1('f1 ~ 4i'<'4~~ ~ tll 'T": I
   ~
   
   l:f
   
   ftR«t
   
   'ZI'C:'
   
   ~ mmfiNj ~ lhtll eeXLIX II
   
   CD = 12 A, DQ = 8A, DD' = 2C (= 48 A), PD' = 12A. Find PB, QB and AB. Comment: QD = 8, DD' = 48, QD' = 40, PD' 52 x 8 PQ = 52. BQ = - - = 104 A 4
   
   PB = 52 x 12 = 156 A 4
   
   = 12,
   
   165
   
   Shadows
   
   _ 104 x 12 _ 156 A AB . 8
   
   tJ~~":Jv'''f''ff,
   CfI
   
   "ffil ~'
   ~
   
   fimlJ.1
   
   ~ ~s:P*'t~dNll1~«:41 ~~: ~ ~'I"1I~tJnJ ~ ~: }l<6jvliR4i't II eeL II
   
   Knowledgeable persons are fully aware that examples in Algebra and Arithmetic involving multiplications and divisions are tackled easily by the Rule of Three. To enable mediocre persons like us to master the Rule of Three, they have given us many different examples.
   
   Comment: In this stanza, the importance of the Rule of Three is described. Nowadays 'Unitary System' has replaced the Rule of Three with a consequent loss of speed. NaTE: It appears that this stanza is not Bhaskaracarya's'; 'it does not seem plausible that he considered himself as mediocre.
   
   CHAPTER 33
   
   Pulverization (Kuttaka) Before studying Bhaskaracarya's method, let us first see what is . . . , . ax + b c- = y where a, b, care mte'pulvertzatlon . Take an equatIon - gers. If x, yare also integers, then to find y we have to choose those values of x that will make ax + by a multiple of c. Such an equation is called 'pulverizer'. Pulverize means "beat the problem into powder". Indeed, this indeterminate analysis was designated in Indian Mathematics by the term kurrnka. Kurrn means 'to pulverize'. The above equation in the form ax + b = cy is called Diophantine equation of the first degree. (In this chapter a, b, c, x and y of the equation (ax + b)/c = y stand for dividend, constant, divisor, multiplier and quotient respectively.) trci~lR~~~
   l)"T'i ti m
   
   ~ ~ ~~II CCLlII
   
   o friend,
   
   one hundred is multiplied by an integer; 90 is added to or subtracted from the product; the results are exactly divisible by 63. If
   
   167
   
   168
   
   LIflivati of Bhaskarac3rya
   
   you are proficient in pulverization, tell me the multiplier correctly. [That is if 100x ± 90 is divisible by 63, find x.]
   
   Comment: Let us solve 100x + 90 = 63y or 100x - 63y = -90
   
   (I ).
   
   Let two consecutive solutions be (x" y,), (xr+J. Yr+I). Then 100xr+1 - 63Yr+1 = -90 = 100x, - 63y, :. 100 (Xr+1 - x,)
   
   = 63 (Yr+1 -
   
   y,).
   
   Since (100, 63) = I, Xr+I"" X, == 63 and Yrtl - y, = 100, i.e. (x,) is an A.P. with C.D. 63 and (y,) is an A.P. with C.D. 100. Now (I) is lOx 7y 7y == 10 - I. Clearly y and 10 - I must both be divisible by 10. y ==
   
   9"""
   
   30 satisfies this. So x =
   
   I~
   
   (20) = 18. So the solutions are (18, 30),
   
   (HI, 130), (144, 230), .... Similarly, solutions of lOOx - 63y = 90 are (45, 70), (108, 170), (171, 270), .... Now let us consider Bhaskaracarya's method and compare it with current ones.
   
   ~
   
   m:
   
   ~Ql$lfqIQqi4: 4;'11\211<' ~ '!'~"'tvfltl
   
   1f;{ f.m;it ~ ':f ~ ~ '5>\\l~\\lqq II CCLlIII
   
   ~ "1~?I
   f.tm
   
   ~: tBjqlAiar..11 CCLIV II
   
   ~~ ~~ ~ ~ ~: ~ ,lftl!)' .. 't I \JillIl ~ ~ tre: ~"T'1't ~ ~II CCLVII ~ ~ ~ ~: ~qNfhDf+('f
   ~ ~ flnn~ ~~~ ~
   
   tfr ~: II
   
   169
   
   First find the H.C.F. of the dividend and the divisor which will be I. For if it is not I, it will divide the constant and after division, the dividend, divisor and the constant will be in standard fonn. Go on dividing this standard dividend by the standard divisor till we get the remainder 1. Then write the quotients one below the other followed by the constant and O. This column is called the Downward Column or ·Creeper'. Multiply the last but one number by the number over its head and add to this product the lower number and write this final resulting number over the head of the last but one number. By striking off the top and the end numbers, reduce the column. Repeat the procedure several times until we are left with two entries (at the top of the column). Then divide the upper number by the dividend. The remainder so obtained will be the quotient; this gives the value of y. Similarly the lower number divided by the divisor leaves a remainder which gives the smallest value of x. In this procedure following changes may be necessary. If in the downward column, the number of divisions is even and the constant is positive, the remainders will be the values of x and y respectively. If the constant is negative and the number of divisions is even, then (Dividend - first quotient y) and (Divisor - first x) will be the desired y and x respectively. (If the number of divisions is odd, just interchange.)
   
   Comment: Now we'll solve the equations by Bhaskaracarya's
   
   ~~~~~~~I ~~
   
   Pulverization (KuHnka)
   
   CCLVI II
   
   If in a given equation the dividend, divisor and the constant all have a common factor, the equation should be taken as a proper one. Dividing by the H.C.F. we get an equation in which the coefficients will have no common factor. If the H.C.F. of the dividend and the divisor does not divide the constant the equation should be considered improper.
   
   method: 100x - 63y =-90. Dividend = 100, Divisor = 63, H.C.F. = I. Divide bigger number by smaller, I 63 100 I write the quotients in the right and I 2 37 26 left columns and remainders below. I 4 1\ 2 I 3 This way we get six quotients - an even number: I, I, 2 in the right I 2430 column and I, 2, I in the left one. I 1530 They are, in the order, I, I, I, 2, 2, I. I 900 Write these in a vertical column and 2 630 write below the constant 90 and O. 2 270 90 x I + 0 is written in front of I. I 90 90 x 2 + 90 = 270 is written above 90 x o x 90.
   
   Lr/avati of Bhaskarocarya
   
   170
   
   Continuing this way we write 630, 900, 1530 ending with 2430 at the top-most line. When 2430 is divided by 100, the remainder is 30 = y. When 1530 is divided by 63, the remainder is 18 = x. Now 100 x 1863 x 30 = 1800 - 1890 = -90. So (18, 30) is a solution. 81 144 207 Complete solution: x = 18 y = 30 130 230 I 330 I If the equation is loox - 63y = 90, we take y = 100- 30 = 70, x = 63 - 18 = 45. Complete solution: (45, 70), (108, 170), (171, 270) Next we'll see how to solve an equation when the number of divisions iii odd.
   
   I
   
   I
   
   ~ eWlq~'(~l ~ T.I ~ '111 ~: I
   
   t'Qra4lc:tlf':?ll f.l'(lJCflltii ~ ~
   
   Solve:
   
   .q
   
   ~ II CCLVII II
   
   -60x±3 .. = y In Integers. 13 60x + 3 = y. The table with 60, 13 is as 13 The sequence of quotients is 4, I, I, I, I odd in number. Here the constant is 3. We 60 4 write 4, I, I, I, I, 3, 0 in the 8 I left column and proceed from I 3 the bottom upwards as in the I previous problem. Here 69 divided by 60 leaves the remainder 9 and y = 60 - 9 = 51. 15 divided by 13 leaves the remainder 2 and so x = 13 - 2 = I I. If the equa60x- 3 tion is = y we get x = 2, Y = 9. 13
   
   Pulverization (Kurraka)
   
   171
   
   Example \!q;tatl~ ~ ~ ~
   
   qqqfieg4t I
   
   
   o arithmetician!
   
   221 is multiplied by an integer; 65 is added to the product; 195 divides the result exactly. Tell the multiplier quickly. 221x+65 .. [S olve = y In mtegers.] 195
   
   Comment: Here we can divide by H.C.F. (221, 195) = 13 and get 17x+5 y. Here the quotients for 17, 15 are I, 7. The top two num15 bers are 40, 35. 40 divided by 17 leaves remainder 6 and 35 + 15 I 40 leaves remainder 5. So x = 5, Y = 6. 7 35 Complete solution: (5, 6), (20, 23), (35, x 40), (50, 57), .... 5
   
   o
   
   x
   
   Comment: We'll first solve
   
   Example
   
   on the left side.
   
   ~ cg~qsfaQ~~"1~4): 'C'IQQqfi1d
   1 I
   
   13 5
   
   2
   
   4 I I I I 3 0
   
   69 15 9 6 3 x x
   
   m
   
   ' ax + b If In - - = y,
   
   c
   
   a. b have a common factor m, then
   
   a b -x+m m =1...=y' c m and solve this equation. Clearly values of x will be the same and y=my'. Comment:
   
   We'll take 100x + 90 63 Divide by 10 to get
   
   Y (stanza CCLI).
   
   10x+9 y , 63 ='iO=y . •
   
   I
   
   172
   
   LWivatT of Bhaskaraclirya
   
   Sequence of quotients for 10,63 is 0, 6, 3 (odd). 27 + 10 leaves the remainder 7; 27 0 171 6 171 + 63 leaves the remainder 45. 27 Since the number of quotients is odd, 3 x 9 x = 63 - 45 = 18, Y = 100 - 10 x 7 = 30. x 0
   
   Pulverization (Kurraka)
   
   173
   
   Five is multiplied by an integer; 23 is added to or subtracted from the product; 3 divides the results. Tell the multiplier. [Solve: ] 5x ± 23 3 =y . 5x±2 Comment: We solve - - = y =1= 7 = y'. 3 . f 5x + 2 ,. A solullon 0 - - = y IS
   
   ~ ~CjjulI*!4 ~ ~ ~II CCLX II
   
   The above methods are applicable when the constant is positive. If it is negative, take suitable complements (in a. b). Example
   
   y'
   
   = I.
   
   3
   
   . 5x + 23 So a soluuon of 3
   
   X
   
   = 2, y' = 4 and
   
   0
   
   f 5x - 2 - - = y ' is x = I,
   
   3
   
   y is x =2, Y = II
   
   and of 5x -3 23 = Y is x = 7, Y =. 4
   
   ~ ~ .. ~'
   ~ 'i~tlfi(ill ~ ~ ~
   
   .q
   
   1l~ ~II CCLXIII
   
   a arithmetician! an integer is multiplied by 60;
   
   Formula to solve the equation when the constant is zero
   
   16 is subtracted from or added to the product. The results divided by 16 become residueless in each case. Tell me (those numbers) separately. [Solve:
   
   1P:l:
   
   60x + 16 13
   
   If the constant (c) is zero or divisible by the divisor (b), a multiplier is zero and the (corresponding) quotient is the constant divided by the
   
   y in integers. ]
   
   Comment: Here choosing + 16 we get
   
   60x + 16 13
   
   y . Since 16 > 13
   
   . 60x + 3 = y - I = y' . As per (CC L VII), x = I I ,y ' = 51 . we consIder 13 6Ox-16 .. x = 11, Y = 52. To solve 13 = y, consider 60x-3 ' Here x = 2 13 = Y + I = y. ,y' = 9, ,'. the solution is (2, 8).
   
   ~:
   
   ~ ~ Vq) ~: ~II CCLXIII II
   
   divisor (that is one solution is x =0 and y = .£ ). b Example
   
   lA q"j~k": ~: qtq"~~(ij"tq aStm I '!Jf't4'lUi&di ~: ti "jVi ~ ~ .q I I
   
   ti'!f
   
   'i.n~ti~:dgdi: I
   
   ~ 1fi ~
   
   f.!mn:
   
   CCLXIV I I
   
   The product of an integer and five is added to 0 or 65. The outcomes divided by 13 leave no residue. a arithmetician! tell me that number • .I: I Imm~wate y.
   
   Example
   
   lA
   
   ~ ~ 1l?I e)q: ~ ~: I
   
   [S 0 Ive: (.) 5x y an d (11) .. 5x .+ 65 I -::::; 13 13
   
   Y. I
   
   Comment: I n (i), (0, 0) is one solution. Others are (13, 5), (26, 10),
   
   ~: lJ ~ ~: II CCLXO II
   
   In (ii), one solution is (0,5). Others are (13, 10), (26, 15), ....
   
   174
   
   LIllivatT of Bhliskaracarya
   
   Formula: constant is 1 etQ ~ q~2t>(Qj ~ 'i~ 'i~ 3or<"''«('t~ I ~~qffl~"A~ 'Mt:I'«(1~ ~q(1f(1.ufflll CCLXV II
   
   ax ± I First solve - - = y and get a solution (Xo. yo). Then a solution of b ax ± . der In . -c =y is: x is the remainder in -cXo- and y .IS the remaIn b b cyo
   
   a Comment: Suppose we wish to solve
   
   17x+ 5 15 =Y .
   
   . 17x + I FIrst solve - - - = y and get Xo. = 7, Yo = 8. 15 5 x 7 + 15 leaves the remainder 5 = x and 8 x 5 + 17 leaves the remainder 6 = y. Continued fractions provide a proof of Bhaskaracarya's method. To solve ax - by = I we form continued fraction for! , b 5 I I I IIIII eg -=--=--=---=-~-.. 13 13/5 2'+ ~ 2 + I + ~ 2 + I + 1+ 2 5 3 If we omit the last one, viz.,
   
   1
   
   !
   
   2
   
   and compute the remaining fraction
   
   2 - and we note that 13 5
   
   x 2 - 5 x 5 = I and thus x = 2+-I+I 2, Y = 5 is a solution of 13x - 5y = l. Similarly the penultimate con-
   
   we get
   
   I
   
   vergent of ! provides a solution to ax - by = I. Details can be found b in an algebra text. Formula: union of pulverizers with the same divisor ~ tt'(~C:§OI~ ~ tm ~ q~2t>(Qj ~ I 3J1l~ ~ \1'ffi1ro: ~: ~C:4J~0»~ II CCLXVI II
   
   175
   
   Pulverization (Kurraka)
   
   If an (unknown) integer is multiplied by two integers (separately) and the products divided by a (given) integer leave two remainders then (to find the unknown) assume the sum of the multipliers as dividend and the sum of the remainders as negative constant of a proper pulverizer, which is the union of two pulverizers.
   
   Comment: Ifax/c = y + b/c, i.e. ax - b = cy a'x/c = y' + b'/c, i.e. a'x - b' = cy' and then (a + a')x - (b + b') = c(y + y') is the union of two pulverizers (i) and (ii). Thus ifax ;: b(mod c) and a'x ;: b'(mod c) then (a + a')x ;: b(mod c). This is a well-known formula in the theory of congruences.
   
   ... (i)
   
   ... (ii)
   
   Example CI'i:
   
   tNf.Wi1
   
   Pt&dfa .. iIC~
   
   fl'<1lq~Sll ~ ~ ~: I
   
   ~: 'fQlfaaafa ..ec;lu ~9ttllm ~ ,(1~1~"1'{ II CCLXVII II
   
   If the product of an (unknown) integer and 5 is divided by 63 then the remainder is 7. If the same integer is multiplied by 10 and divided by 63 then the remainder is 14. Tell that integer.
   
   Comment: The united pulverizer is (5 + 10) x - (7 + 14) = 63y, i.e. 5x - 7 = 21y. One solution is (14,3), and x = 14,35,56, ....
   
   CHAPTER 34
   
   Concatenation (Permutations, Partitions etc.) ~,.,i(1~q;,~.,qio>Vh': ~ f?4il~'<1~$:
   
   I
   
   ~SOi'I~q;~iq;~q,,"~ ... : ~ ~ ftl~tig~: ~II CCLXVIII II
   
   To find the number of permutations of given (n) different digits (or objects), write 1 in the first place, 2, 3, 4, ... up to the number of objects (n) and multiply them. Divide the product of the number of permutations and the sum of the given (n) digits by the number of the given digits (i.e. by n); write the quotient the given number of times (i.e. n times) in a column but leaving one-digit place each time; add them; the result is the sum of the numbers fonned (by permuting the given n digits).
   
   177
   
   178
   
   LilavalT ofBhaskariiciirya
   
   Comment: The first part describes the familiar fonnula n Pn = nL = I x _ is the number of permutax2 3x ... xn. It is known that nP, =_n_'r)! (n -
   
   tions from n differen t objects taken r at a time. This is proved in alges bra texts. It is easy to derive the fonnula for the sum of the number fonned: it is (n - I)' (IOn -I) (sum of the digits) 9
   
   179
   
   has his hands. [Lord Siva has five heads and so it is presumed that he the ten arms.] Find the number of different Siva idols. Similarly, solve loa disc, a mace, a objects: problem for Vi~f.lu idols; Vi~f.lu has four tus and a conch.
   
   Comment: In the first exampl e, there are IO! = 362880 0 plossible
   
   in Siva idols. In the case of Lord Vi~f.lu, possible idols are 4! = 24 number. of [NOTE: In the daily ritual (sandhyavandanam) there are 24 names Lord Vi~f.lu.]
   
   Examp le taCiSllICCiSfO
   Concatenation (Permutations. Partitions etc.)
   
   Formulll: Permut ations with repetiti ons
   
   ~'1qlllli'!$cri f.mR iiillit=t q,qm: I
   
   ~&n~~tfI: ~ ~ (fffi&J$CF
   CCLXI X II
   
   s Using (i) 2, 8, (ii) 3, 8,9, (iii) 2, 3, ... ,9 how many different number can be formed? What is the sum of numbers, so formed, in each case?
   
   Comment: (i) Only two numbers 28 and 82. In this case ]p] = 3 x 2 x 1= 6 different numbers. (iii) Here the answer is sP s = 8! = 40320. s In, (i) n = 2, sum of the digits = 10 and so the sum of the number (ii)
   
   formed = 99 x 10 = I 10. 9 3, sum of the digits = 20 and = n (ii) 2 x 999 x 20 - - - - =4440 . 9 (iii) n = 8, SUm of the digits = 2 + 3 + . . . + 9 = 44. 8
   
   Answer = 7' (10
   
   -
   
   so answer
   
   =
   
   I) (44) = 2463999975360 .
   
   9
   
   n' t+ be n objects. Then number of permutations = r ,. s., 't., ... ,n = r + s + This is proved in standard texts.
   
   Examp le
   
   Examp le
   
   q"li,!'i1~:eij'l'lCiSCiSq,(iij'l~: qC{q'''f1''fh'''i'l1~ I
   
   aI....,)"'4lC'RIqs~~: ~ ~: ~
   
   ) (To find the total number of permut ations of given n digits or objects if certain digits are alike then form the produc t of the number. of perthat mutations of those places at which alike digits occur (assummg number each block of alike digits has different digits), and divide the t) of permut ations of all the given (n) digits (assuming them. differen . product thiS by Il) CCLVI obtained by the previous method (cf. stanza . And sum of the numbers formed is obtaine d by the previou s method Comment: Let al = a2 = . . . = 3" bl = ~ = .. . = b., CI = C2 •• • = Ct, . . .
   
   '1c;lfhi Uvnid: I I CCLXX II
   
   Lord Siva holds t~n different weapons, namely a trap, a goad, a snake, in a drum, a potsherd, a cIub,a spear, a missile, an arrow and a bow
   
   ~: ~~&l
   mm
   
   ~~t'I'~iffllijl"~~ : 0<
   180
   
   LrIlivatT of Bhaskaracarya
   
   Find quickly the number of different numbers that can be fonned with (i) 2, 2, 1, 1, (ii) 4, 8, 5, 5, 5. Also find their sums. 4! 24 C omment: (1.) --=-=6. 2!2! 4 In the unit's place 1 appears three times and so does 2. Their sum = 9. The same is true of digits in the ten's place, etc. and so the sum of the number = 9999.
   
   ") Nurn ber of pennutations (11
   
   51 = 20 =~ 3! .
   
   General fonnula for the sum n! (ra l + sb l + tel + ... ) n = r.I s.. I tl 9 (10 - 1). n In (ii), n = 5, r = 1, s = 1, t = 3. 51 5 Sum = 3! '5 (4 + 8 + 15) (10 - 1) = 1199988.
   
   Concatenation (Permutations, Partitions etc.)
   
   181
   
   (n - 1) ... (n - r + 1) . This method is applicable when n < r + 9. All (r - 1) ! this has been given briefly because this ocean of mathematics is vast. Comment: This is an example of partitions and the derivation will be found in books on theory of numbers. .
   
   (n - l) 1
   
   Fonnula can be WOllen as --.:..-....:...- (n- r)! (r-l)! NaTE: The great Indian mathematician (Late) S. Ramanujan made outstanding discoveries in the theory of partitions. It is said that he got inspiration from the above stanza. Example <{.,'W~~~$4¥jfj"'Q"< " I ~ 'ijaffi'6Ql ~ ~ PI-linn¥{ II CCLXXVI II
   
   Find the total number of different numbers of five digits whose sum is Example
   
   13.
   
   ~I~1~"'~~'<'3:$'(~ ~ ~: I
   
   Comment: n
   
   'lI>fa ~: ~ ~ ~'ltJ?I''{ II
   
   = 13, r =5 so the answer
   
   =~=495.
   
   CCLXXIII II
   
   8!4!
   
   Leaving aside 0, if the digits 1 to 9 are wrillen six at a time, then find how many different numbers are fonned. Comment: n P r
   
   = 9 P6 = 9
   
   x 8 x7 x6 x5 x4
   
   = 60480.
   
   f.H",&j$
   ~: w.rr ~: ~ ~~~ II
   
   CCLXXIVII
   
   "'4~~::N)+i&1"'fQ': ~~ ~ ~ ~ I ~ l1~ ' ~~ ~ 'Ifl1t?lIU\4fQ II CCLXXV II When the sum (n) of the digits and the number (r) of blank spaces are given, follow the method given below to get the total number of permutations:
   
   "" ~"" ~"" ~ tA: 'liIC$Q~ ~I "f~t"Iorq'i"'1i ~S4t'l&j",q'n~ I I CCLXXVII I I
   
   Although there is no multiplication, division, squaring or cubing in this concatenation, yet if it is asked to egotistical evil-minded lads of astronomers, their humbling is certain. 1)lri~ ~",f&C1C4ql~: ~ "";qj""d""''tI'' '<m;n1l t4lt·nqdlt: 'C(i('fM~j<'E''(~
   
   tftri
   
   rn
   
   'tl(!J
   This LmivatI clearly explains fractions, simple fractions, multiplication etc. It also beautifully describes problems i~ day-to-day transac-
   
   LTlavatf of Bhaskaraciirya
   
   182
   
   tions. Rules are transparent and examples are beautifully worded. Those who master this Lflavatf will be happy and prosperous. Alternate meaning: (Lass) Lflavatf is born in a respectable family, stands out in any group of enlightened persons, has mastered idioms and proverbs. Whomsoever she embraces (marries) will be happy and prosperous. ~ ~ (INTERPOLATED VERSE)
   
   ~ 41.'(O"~ ~ 1:1
   
   fll1ruri
   
   ~ 01: ~:
   
   Index of Verses
   
   ~~~~~~~~:I ~ ~ ~ 1:1 ~ o4lqi,"
   ff"II~'1'II~~ m~:~: II CCLXXIX II
   
   BEGINN1NG PART OF VERSE
   
   Bhaskaraclirya, the great poet and author of this book, had mastered eight volumes on Grammar, six on Medicine, six on Logic, five on Mathematics, four Vedas, a triad of three ratnas, and two (fore- and past-) Mimamsas. He understood that the Lord (Superme) cannot be fathomed. [Obviously he had not mastered only Mathematics but was well versed in many branches of knowledge.]
   
   STANZA NO. XL
   
   3Tftl:
   
   W?
   XXXVII
   
   "'9(04ol(I1\t!~fa~CMI
   
   CLXXXII
   
   "'9(04;:;iij";~
   
   CXCV XCIX L CCXXXVII CLXVII XXXI
   
   aNloe\JII(\QQ4ijl§cr»e4:
   
   CXCVII
   
   "'¥t(iiqsif(ii\(It)'t?Cilllq'Qi""ea:
   
   LV XIV
   
   31ta
   3II!C:'iwijc;
   
   ~uf ~
   
   31oel<:lllit).,. =tt¥tI;1i1t''t
   
   LXXVII CCXXXI CCXX 183
   
   Llliivati 0/ Bhiiskariiciiry(.
   
   18-t
   
   Index o/Verses
   
   185
   
   CCLXXIX XII ~ ~
   
   ttTit fu
   
   iI ~~:
   
   LXX
   
   ~ ~ ~ ~ stf;srf1rci~ ~:
   
   CLVIII
   
   ~: ~ ~: ~ "If'iU)~ mI mf
   
   CXXVIIl
   
   ~
   
   00
   
   ~: ~ ~~oih'l'(,
   
   am.i~~g-
   
   .q
   
   •
   
   CXXXVIII XXIX
   
   \:",6hQ~akI: ~ ~ ~
   
   CXIX
   
   to>1:'5IJ~'i"'Q@'
   CXXII
   
   ~."fatlfiT!rri ~ ~
   
   CCLVIII
   
   ~
   
   "'Iq''"?t,.,i
   
   ~~
   
   lf1IrItJ
   
   .tcrsn=.t1rA .q
   
   ~q;,d.,,'ft,,(1 ~ ZZffin ~: ~:
   
   CXXIV CXVIII
   
   CXXIX CCLXVI CXCIII CCLVI CLXXXIX LXXX, LXXXIV
   
   CCLXVlI ~ ~ ~jO,,~t~~
   
   CLXIV
   
   ~«!9i~'(oejfil'tt1'
   
   LXV
   
   ~~~
   
   CXCVI
   
   il «!
   CUll
   
   ~d"C"fj~qq!Jcfl
   
   CXC
   
   ~UJ~~:
   
   CUI
   
   ~ ~ Rtiit>1,,~~q;
   
   CVlI
   
   ~~~
   
   Lxvm
   
   ~: DS"'SdlSzM~~~
   
   XIII
   
   ~~~?N9~~
   
   CLXXXI
   
   ~~~
   
   LXXXI
   
   ~~It»'al~~:
   
   CCXXX
   
   q)U~'''9I1Cci ~
   
   CXUV
   
   ~
   
   CL
   
   ~ qq'li~9~vg~~
   
   CLXXXIV
   
   CLXXXVlII
   
   ehml~m
   
   CLXXXV
   
   r.61rC;fl :J0rr~ ~
   
   ~~~:~~~:
   
   lf ee)'11 tffi.,
   
   TR ~
   
   XVII
   
   CLXXIII
   
   ~~:~
   
   CXU
   
   CXCIX
   
   ~ ~\1I'k1tt~~«!1{CffiT
   
   CXLVlII
   
   ~ tfQtUllil!J
   
   ~ ~~ fljvl'1A61ltt
   
   CXLVII
   
   efti
   
   ~~~~:~
   
   CCXXIX
   
   ~
   
   LII
   
   li q"1· .. q~
   
   ~ qQltJl"lI ~ (f.lD
   
   ~ fiq:
   
   ~ ~
   
   (J3<jlji$ff.
   CCLX CCLXV CCLXIII XLVIII
   
   ccxvm
   
   LIliivatT of Bhaskaraciirya
   
   186
   
   187
   
   Index of Verses
   
   i(!IU~ii4'f4lfil~~~tAl
   
   XXI
   
   tmR~
   
   CC
   
   ~ "ill ~ '('~I~&1:
   
   XXVII
   
   dt't'ililfaiid:
   
   CCII
   
   ~~ tah'l'il>'
   CCXXVIII
   
   •
   
   ~ \f"iflfa~~';
   
   CXXXII
   
   dCI""n4 ..".
   
   ~~~~
   
   CXXX
   
   W4'~~~
   
   III
   
   ~~~~
   
   CCXXIII
   
   ~ ~
   
   CXXVII
   
   !JuiVI'1(i1'Hgd'f4
   
   LXXI
   
   ~ ill§""q;il ~
   
   !Ju4i t"QQq) ~
   
   XV
   
   t"Q~'~I~"ifl~
   
   Q4lTd4l1'flC:t"i
   
   CCXII
   
   ~iN Cf)lfhA,q4 taillulliCgiUiCPl:
   
   CCXVI
   
   t4S6sditll
   CLXI
   
   ~ 3\i141il:mC:"'d"!JQ~
   
   CLXXI
   
   ~\ift
   CLXXVII
   
   ~"IRlcMq 4~dScm ~
   
   CCXLVIII
   
   ~ ~"~'54Cf)i
   
   CLXX
   
   1m4UI ~~
   
   LXXXVII
   
   ~~ 'fIifl"t"i~
   
   CLXXXIII
   
   'l;"flrfit~~'ltll"'dd", U4I
   
   CCXV
   
   t4lct).,f.l&N~Iil:
   
   CCXIX
   
   cal'fl'denIA'"
   
   CCXU
   
   lm~
   
   tlI41"q)~d,,;ft !Jull 111
   
   CCXLVII
   
   c:,"tf!iGi ~
   
   9
   
   CCXUV
   
   C:flO1;qUli !JUJi''';:ifl''.:
   
   CXV
   
   CCXXXV
   
   c:tI"quti q'Cj~"'if''''
   
   cxm
   
   u1i ~ ~ u~ CfIf wi ~ ~
   
   XUX
   
   cn:q»C1ll,<"'d;( ~:
   
   CLXVI
   
   m: '(iIj"lj ~ ~ ~ m:
   
   XLII
   
   ~4:A~~
   
   CXXXV
   
   OC:&1i(i1~:S em ",.,uiQCf)'f4
   
   XXXV
   
   ctutalJ.,
   
   LXXXV
   
   ct ...t1tf~t'l4Q4'f4
   
   CCXIV
   
   ~~
   
   XCVI
   
   ~~~
   
   CCXIII
   
   ~~:~~:
   
   VIU
   
   dt¥i'4I41 i jqQ 1I5'Uif
   
   CXLII
   
   ~qsliCqsloqj ~"4'iC~4(
   
   CCLXIX
   
   tJ1'll"Cj): Cf) ohik'd ~ 1)
   
   tllcit'{1'ia
   
   f£Jm)
   
   d'lIl:
   
   .,,,
   9 ~ ~d~<mqtt
   
   vfm;ri ~ ~ ~ ~
   
   Ptf.t
   
   "i.Aftt..i..~~ t'lg'Cd'4
   lPl'll'i
   
   "~"Cffig*,
   
   CXCI
   
   CLXXVIII
   
   -,fit
   
   '!J'ifr
   
   XXIII
   
   CLXXIV CLXV
   
   1fI'i
   
   ~
   
   j
   
   G\l4.M
   
   ~ ~ ~ ~
   
   XCV
   
   IV
   
   LXXXIII
   
   wm
   
   XXXIV
   
   Index o/Verses LTliivatJ 0/ Bha.skarliclirya
   
   188
   
   189
   
   ~ ,"~aq)44'j4:
   
   CCLIII
   
   ~ ~ ~fltijcm
   
   CXXV
   
   ~RtM'''I+4
   
   CCXL
   
   ~(l)lI>~~f2I~h qs~'t164i1Sl:
   
   CCLXXII
   
   qli!l'<1?1) q ~
   
   LXIX
   
   f&At;f(1j(")~~
   
   CLXII
   
   tITGI~"CfTlC'I'I~ !]Olq.tlfic1 ~ ~
   
   CXXXIX
   
   ~1~<1
   CXCIV
   
   QICi)"'IGiUiilSgt"QC.:
   
   IX
   
   fU<:a~"'Itq;~::,,~ ~:
   
   CCXXXIX
   
   ,'IC,ft
   
   CLXIX
   
   ql'ljt'"ft'8.. 'tIiqsqsqIM'l~:
   
   CCLXX
   
   CCLXXVII
   
   ~"~'IGM"iI'1(iljm:
   
   CCXXXIII
   
   CVIII
   
   ~ 1)S4flilal~!JMI: fclst;y
   
   XCIII
   
   XXVIII
   
   ~ 4iClcilSS.j 1);f Ii!JU~fli
   
   CXXXVII
   
   "f"n:::SS!PQf4¥/ff&Jq;,ql:
   
   CCLXXV
   
   w'lICqs~·(qM~"EUi
   
   LXXXII
   
   f.\~qs".ctt4lilq
   
   CCLXXIV
   
   w~~4mIT~
   
   CI
   
   ~ :::q~fl¥~*k{I~L01t
   
   XXX
   
   wlQ'i'i'I'ic;'(ID lj);;R lj) ~:
   
   CXXXIII
   
   XCVIII
   
   W~Q,jqqd'Ctft"Ja8fdi
   
   CCXLV
   
   ~:~~:~:
   
   un
   
   w"iUiqsi~" ~
   
   XCVII
   
   ~-
   CLXVIII
   
   g'ilolf21ttJl ~ 't1'iI"1\711~:
   
   q.,'t1'<1"1",(I~I"'~~~ QS(4fiU<:l,
   
   LXXXIX
   
   lffitN fiB
   
   q1.-ol~~lltJtJ)'''rl<JR''1
   
   CCLXXVI
   
   ~
   
   XXXIX
   
   1ftfi!i ~iM\7I"'fQ
   
   llC 't'1jVf
   
   1«!
   
   n
   
   ., ~ ., ~ .. ~ 'fR:
   
   "f(&f<:;I1'I~"Cf1fl~R~ ~ ~ ~
   
   ~ ~
   
   wi
   
   wi
   
   19:
   
   ~
   
   ~ 't1",(4I"'d,(,{
   
   l(qitN,~ql~"dl~~af'"l~
   
   "t1'd
   
   lftui
   
   LXXVI
   
   lht
   
   JllnUi
   
   ~:
   
   LXXIX
   
   19:
   
   ,.):etICfffi"Ci
   
   CXXI
   
   at
   
   lj) ~
   
   LXXXVI
   
   fitu:i ~ ~:
   
   LX
   
   CCXVII
   
   en
   
   ~ 41(41!1~'I(1)(4..,q~ l'ft(4IQ~ ~
   
   XVIII
   
   CLXXXVI
   
   ~ '« Itfl!1<"f'F'<:(4IA ~
   
   LXXIII
   
   CXXXI
   
   ~Q: ~ 91ft<11~4i1
   
   CXCVIII
   
   cu
   
   '&f1ftit TJ: lj~ ~
   
   ~ 1) w~q)~\lf2I(1q)
   
   XCIV
   
   ~~~.
   
   ~: 'i4it"Qlf.\ ..JR4 .. ''':
   
   CV
   
   ~mi~~~
   
   Q'4ifllPttlila
   
   cPf 1:h fJ,qCf)'(uft·rd\
   
   a..
   
   XVI CIII
   
   LIliivali 0/ Bhaskaracarya
   
   190
   
   191
   
   Index o/Verses
   
   CLXXXVII CCUX
   
   "1~1"i(: ~~:
   
   XIX
   
   ~ m: ~q
   CCLII
   
   j\llqllh'll41
   
   'iii
   
   tuJ
   
   CCXXIV
   
   QlI ffldl(41)~ClSolf"'d\(jCa
   
   CUX CXUX
   
   *{fYttlQg(tl4I,(1 j(Jai4l;ftW
   
   CLXXVI
   
   infill4QIICClSIl\.,;: ~MGtlcP
   
   CIX
   
   ~ ~ ~ qllq"'CIII=tI~ ~
   
   XC
   
   ~ ~ 'iilltl~~I~
   
   CCUV
   
   j(JVjtfl"t\lfdg~~"qs~cN
   
   CCXXVI CCXXVII
   
   ~ ~ "" ~ om-ri
   
   XXIV
   
   ~ "(!Jj !JlifildlSlll ~SJ)
   
   CCXXXIV
   
   ,..,q@.,~c:r('j ~
   
   CXX
   
   "" ~
   
   V
   
   4fftGtfft~'
   U
   
   1R4lJRN4 ~~
   
   CCXXV
   
   lIRi 8ttcgrit4 '1t'1G flcP
   
   LXXV
   
   4Iq"':U":1
   
   CCLXXI
   
   ~ ~ q.,~lf'(lla
   
   UV
   
   W'I
   
   qlQjfiJmt: ~:
   
   CCLXIV
   
   W'I~:~
   
   CCLXII
   
   ~ f.!rsf;n ~"~14~Jtl4"'48:
   
   CIV
   
   .q,.j g
   CCLXXVIlI
   
   ~~6Qd ~WlI~ ~: ~:
   
   CUV
   
   ~~~1Pmft~~~:
   
   XLVI XXXVIII LXI
   
   CCL C ~ CII't",.,Sd: If ,(I~~."
   
   LXXll
   
   ~ ~ .'1~ql~lgiflvil
   
   CLVI CCLvn
   
   ~ "'lVf1IS "~I(~I
   
   CCLXI
   
   Q,,'4''8~I(''~: fitn;t ~
   
   CCX
   
   l4Q'f1~:
   
   CCXXXll
   
   qq¥,(I'It,.a
   
   4q)G''('':Jri''IC';~:
   
   m'(rRI,(ilCl(a~: ~: '(1t4h=tt,(q<1vl ~
   
   mit
   
   LXXVIII ~ (14):
   
   CXUll
   
   ,(ltQ)4
   LXVII
   
   "lt~4(J)~i~S1i!r t'1<'f't'Q),oq ~:tral
   
   LXIV
   
   ~ ~jv)ilCld
   
   LXVI
   
   ft
   
   ~ q"riqfft~
   
   XXXll CCI
   
   4fa~~af"tlll\di:Jrir~
   
   CCXXXVI
   
   ~:lM~~:
   
   LXll
   
   ~: ,*,vl~h; ~AgtQN,(: ~
   
   CLXXIX
   
   ~ ~: q,(H4jvr: ~ ~:
   
   CCIV
   
   ~
   
   1J.l'ft
   
   ~iIi~~a~ "" tfil ~:
   
   ccrn
   
   I
   
   192
   
   LTlivati oJ Bhliskaracarya
   
   .
   
   193
   
   Index oJ Verses
   
   ~(14&11"~~
   
   XXXIII
   
   ~~~~~
   
   M\(1I.I(1tj(166it(14SI(1eQI(1~(1IR1~
   
   XI
   
   1Rl
   
   ~ Ra;q~fifdl ~
   
   UX
   
   ilillfI(1ISf1,(~)'*"&1lq
   
   CXXXIV
   
   CLV
   
   ~: ~ l"f'<~"''
   LVII
   
   II
   
   XXII
   
   LXIII
   
   W ~ " '*9<',"",i r i QiQdutwr.,twll.,'ftIQ'1.
   
   CLX
   
   XUV
   
   1ft?4".. lrit., ~ "('f4I(¥(1I"f1~
   
   XCI
   
   CXLVI
   
   "'?4"'M~d~" ~
   
   XU
   
   CCXXlI
   
   "'?4"",qfad~"
   
   q..
   
   XLIII
   
   ~~~fil:n ,,,¢Cfiquil
   
   XLvmA
   
   '6'<1I.~'1lfA=t
   
   mft
   
   ~'i~~tQq~!l0n ~
   
   CCXLVI
   
   1ft ~: ~ 1II ~ 1II
   
   flh,$n ~"qoi1qoi ..",:
   
   CXI
   
   ..... ~qld"1R:~
   
   XXV
   
   CXXXVI
   
   ...q~qld: f~'(If4~S1II
   
   XX
   
   CCVIII
   
   ....N'< ... 411J.v:fI ~ ~~ ~
   
   CCV
   
   XCII
   
   ~ ~ ~: ~: ~i1ql""':
   
   CCXXXVIII
   
   CLXIll
   
   .,i4I.,i'iIiCif4 ~!I\ilif4
   
   CXCII
   
   ~~ ~1tvM4J'¥iQ'<: ~ l«l
   
   CCIX
   
   ~~~~
   
   ~
   
   CCXI
   
   fl~e)1~ ~:~
   
   CLXXV
   
   ~qld9d4'lfifqsc"fiNcm)
   
   CCXXI
   
   t1ifU'lf"'f'rtt) ~ ~ ~
   
   CXLV
   
   am\ .. ;jC:1ft.:t8~
   
   CCVII
   
   ~ ~ uifia ~
   
   XXXVI
   
   eaqsqC:&1"") ~
   
   CXXVI
   
   ~·~'1h"'Nqoi1 ~
   
   1i1I: MaflQd(1,i,!d(1id'C &1:
   
   CCXLII
   
   t1':4'iQlo,i
   
   'i'!1M~QI"fl'C~8'R11
   
   CCXLIll
   
   md
   
   ~4ffifc1ip'~ ~
   
   CCXUX
   
   'Cj~qunflC'l
   CX
   
   m~~~~
   
   CCU
   
   'C1~"'''&fjVft AG'lA"'c:f~"dI ~ ~:
   
   CCVI
   
   1J.1II
   
   cj'"fI'1(11I"t1'< ,.fil4.ft q,(lcCfiHi ~
   
   l«l m
   
   ~
   
   ~ '
   ~ l ' -Q
   
   !lUllfflffl
   
   ~"""I;j ~ ~
   
   Q
   
   ~ ~: 4S'
   1111
   
   101ft
   
   "''1ql~~&i
   
   ~
   
   ~:
   
   XLVII
   
   4S1(1lfaqRlIllIJl ~)4SMRI;a1
   
   ~
   
   R
   
   X
   
   LXXXVIII
   
   ~ ~
   
   1f4f
   
   f.!ta"'
   CLXXX
   
   • CXL
   
   \ft "' 4fVid 1
   
   ~ ~
   
   CXVI
   
   ,. fQ
   
   dV§ ......''14S'jqq'' ~ ~
   
   XLV CVI
   
   ..
   
   194
   
   LTilivatT ojBhlifkaracarya
   
   ~q;qG &tqGrcfo1~4Ilti4lSRt:
   
   CXXIII
   
   ~ 1ft1l~f1IHI"E'I~OI
   
   CLVII
   
   ''''i~''(8)M~{t<~ ~ ~:
   
   CCLXXIII
   
   ~1.,id¥)q;I~q .. i
   gm
   
   CCLXVIII
   
   tA: f41&4I('eq
   
   XXVI
   
   f4IQ)W' D»)"'tg1lC4'1
   
   VI
   
   +liQt44!1l1 06
   
   LXXIV
   
   ~'J'iftRJn""\('ia ~ ~:
   
   CLXXII
   
   l
   CXII
   
   ~dfqoqf.l&i1 gfilGtIMoj:
   
   ~
   
   Vli:HiI
   [Brackets containing number(s) stand rorthe verse number(s)]
   
   CXIV
   
   ~(H5'h"
   
   LVIII
   
   ~~~¥~
   
   CCLV
   
   ~'kf'If.q~: ~:
   
   ~le4>:J~~ ''''~'''=''401 ~ lRf ~ ltl'tffil'tffi'(\Olli ~"""'H~
   
   Subject Index
   
   4Ifttt.",..i
   
   vn Cxvn LVI
   
   Abja Acre Addition Addo Adhaka Adholi Altitude Angula
   
   (XII) (VI) (XII) (XXXIV) (Viii, LXXXViii) (VIII) (CXLi-CI..II, CLXIV) (cv, CCXXXII~CXXXIV. CCXXXVI, CCXLI, CCXLIII)
   
   Anguli Antya
   
   (XII)
   
   Anu~rupa
   
   (CXL)
   
   Ardhasamavrtta
   
   (CXXXIX)
   
   (VI)
   
   Area
   
   (CXLI, CLXXV,
   
   Arithmetic progression (A.P.)
   
   (CXXVI,
   
   cClx~cxl)
   
   CXXVIII~XXXII,
   
   CXXXIV, cxxxv)
   
   Arrow
   
   (CCXIV. ccxv)
   
   Asu Arhave Bahu
   
   (x)
   
   Bamboo
   
   (VI)
   
   (Ix) (CXLI)
   
   195
   
   196
   
   LTliivati of Bhliskarliciirya
   
   Base Bhagiratha
   
   (XXII)
   
   Bhuja
   
   (XI:I)
   
   Brahmagupta
   
   (cxcv. CXCVI. CXCVIll)
   
   Bhuvanko$O Champaka
   
   (CL VIIl)
   
   (CXLI-cLII)
   
   197
   
   Subject Index Exchange of jewels Existence of triangles Fractional residue Fractions - addo and dedo - compound - division - multiplication - simple - squares. cubes. square roots and cube roots
   
   (CVIIl. CIX)
   
   GadyliJ.la GadyliJ.laka Gal)Opati GaI.JeSa
   
   (Ill. IX)
   
   (II)
   
   Geometric progression (G.P.)
   
   (CXXXVI)
   
   (v)
   
   Ghanahasta Ghapka
   
   (VII)
   
   Gnomon GoHidhyaya
   
   (CCXLI)
   
   GuiJja Hasta
   
   (Ill. IX. LXXXI. LXXXVII)
   
   Hectare Hypotenuse
   
   (VI)
   
   (Ill)
   
   Diagonal of a quadrilateral Diophantine
   
   ( CLXXXVII--cXCVIIl)
   
   Iccha
   
   (LXXIX)
   
   (CCXLIX) (XVIll)
   
   Division
   
   (XIX) (II. XXXIV. LXXXII. LXXXIll.
   
   Infinite Infinity Interest Inverse proportion Jaggi
   
   (XLVII)
   
   Divakara
   
   XCIV. CVI. CVII. CXXIX.
   
   Jaladhi
   
   (XII)
   
   CXXXVII)
   
   Jffiinaraja
   
   (XLVII)
   
   (VIII)
   
   Kadamba Kakil)J Kala
   
   (LVII)
   
   Circumference of a --circle --disk Combination Compound fraction Concatenation Congruence Cowrie Cube Cube root
   
   Diima Dal}cJa
   
   (LVII) (CCVII. CCVIll) (CCIX) ( CXVIll--cXXI) (XXXIll) (CCLXVI. CCLXXVII) (CCLXVI) (II. XXXIV. CXXXVII) (XXVI-XXVIll)
   
   (xxx)
   
   Datta
   
   (xII)
   
   Davaifi
   
   (II)
   
   Dedo
   
   (XXXIV)
   
   Devanagarr Dhabfi Paisa Dhara1}Q Dharaka
   
   (XII)
   
   . Doa cynosuroides Dramma
   
   DroIJll Dvisapta Dwadasasra
   
   (II) (Ill)
   
   (XLIll)
   
   (IX) (VII)
   
   (CLXIX) (L V Ill)
   
   (xxxv) (XXXIll) (XLII) (XL) (XXXI) (XLIV. XLV)
   
   (Ill. LXXXVII) (Xl) (I. Xl)
   
   (x) (CLVIIl) (V-VII) (CXLI-cLII)
   
   (XLVII) (XCVII--c) (LXXXIV) (XII)
   
   (II. LXXXII. XCIV. CXXXVII)
   
   (x)
   
   K~a
   
   (LXXVI)
   
   Kar$O
   
   (IV)
   
   LWivatT of Bhaskaracarya
   
   198 K~Jhii
   
   (X)
   
   Kavadi Ketaki Khahara Khancfameru Khiiri Khiirikii
   
   (II, XXXIV) (lX) (XLVI, XLVII) (CXVIII-CXX, CXXII) (VII, VIII) (VII, lXXXIII, CCXXXVII, CCXXXVIII)
   
   Khan'a Khayuga Kosa Korl KroSa K$Ona Kucfava KuSa Kur/a KUJJaka Madhya Mahapadma Malati Afa$a
   
   (XII) (IX) (V, VI, XCIV) (XXXI) (V) (X) (VIII) (XLIII) (CCl) (CCl) (XII) (XII) (lX) (Iv, lXXXI, CXI, CXIII, CXV, CXVII)
   
   Maund Meru Method of transition Metre Alunamsii Mouth Nak$Otra Newton Nikharva Nimi$O Ni~a
   
   (IX) (CXXII) (lXI) (CXXXVIII-CXl) (CClXXIX) (CCXXVI)
   
   (X) (CXXlI) (XII)
   
   (X) (II, lVIII, lXXXI, lXXXII, lXXXVI, XC, XCII, XCIII, XCVIII, C, CII, CVII, CXXXVII)
   
   199
   
   Subject Index NiJhave Nivartana Piii Pala Pa1)a Parikti Pariirdha Pascal's triangle Piilali Piivali
   
   (IX)
   
   P~ali
   
   (VII)
   
   Pennutation Peshawa Phala Pirigalacarya PHha Pole PramiiI)a Prastha Progression - arithmetic - geometric Projection Prosody Ptolmey Pulverize Pythagoras Pythagorean triplet Quadratic equation Ramanujan Ratna Ratti Reverse process
   
   (CClXVIII)
   
   ~gveda
   
   (CCVII)
   
   Rukii Rule of - eleven - five
   
   (II)
   
   (VI) (II) (Iv. X, lXXXI, lXXXII, CVII) (II, lXXXII, lXXXIII, cxxxvlI) (XXIII, XXIV) (XII) (CXX, CXXII) (l VII)
   
   II
   
   (II) (lXXIX) (CXXII) (CCII) (vI) (lXXIX) (VIII) (CXXIII) (CXXIII see alsoA.p.) (C, CXXXVI see
   
   alsoG.P.)
   
   (ClXXIV) (CXXXVIII) (cxcv) (CCl, CClXV-CClXVII) (CXXIV) (CXLlX) (lXXI-lXXIII) (CClXXV) (CClXXIX)
   
   (III) (XLIX, l)
   
   (XCIV) (lXXXIX, XC)
   
   LWiva(j of Bhaskaracarya
   
   200
   
   - nine - seven - three
   
   (LXXXIX, XCIII) (LXXXIV, LXXXIX, XCII) (LXIX, LXXIX, LXXX, CCXLVIII, CCL)
   
   Rules for barter Samav!1(a Sandhi Sandhyal'andanam Sariku Sa5(rodi(a Seer Series Shastri Silindhra Simple fraction Singh Siva Square -root - transition Subtraction Sulvasfilrakara
   
   (XCV) (CXXXIX) (CCII) (CCLXX) (XII) (VII) (VIII) (CXXIII) (XVIII, XXII, LXXXI, CLXVI) (LX) (XXXI) (XII) (LV, CCLXX) (XX) (XXIII, XXIV) (LXIII, LXIV) (xII) (CXLlI)
   
   Suvar~ll1
   
   (IV)
   
   Tanka Ta(para Tola Trita Trup
   
   (IX)
   
   (IV)
   
   (X) (CCVII)
   
   (X)
   
   vaJa
   
   (III)
   
   Valla Vararaka
   
   (II, LXXXII)
   
   Vi~mav!1(a
   
   (CXXXIX)
   
   Vi~l)U
   
   (XLVII, LV, CCLXX)
   
   Vita Volume ofa -cone
   
   (CLXI)
   
   (III)
   
   (CCXXVIII)
   
   Subject Index -
   
   ditch or solid heap of grain prism pyramid and fmslllm
   
   201 (CCXXlII-CCXXV) (CCXXXVII--cCXL) (CCXXIX--cCXXXII) (CCXXV--cCXXVIII)
   
   Ylll'll
   
   (III, v)
   
   YOJona Zero - eighl rules
   
   (VI, LXXXIII)
   
   (XL VI, XLVII)
   
   f>IVJes.mr Narayana H. Pluu/lw was born in
   
   1902 at Nasik (Maharashtra). He graduated with a Master'sdegree in mathematics from Fergusson College of Pune. He taught in Government High Schools anel Colleges and after retirement ' sen'cel in private colleges as a Senior Profcssbr. He wrote several texthooks and a. hook on Indian , mathematicians. f>roJessOI' Phll.rflle spent a lot of time to construct proofs of \ mathematical methoosand fonnu'lae given in the original Liliwati. I-Ie useel his meagre ~avings to publish the Marathi version of Uliivali. He died in 1973.
   
   I'roPSS01' K17s/tnaji S. Palw(mliwll was horn in 1911 at Ratnagil'i (Maharashtra). HlHlid his B.Sc. · in ~ Mathctnatics-Php;i<:s from Bombay University. "Ie taught 1<,,- lhirty rears at Sheth Anandiial PodOll' Higll School. He has wr',lten . !\c\'cral texthooks un ~falhematics an.eI has lransl;lIt'd a hook Oil tobschevskia.n geollletl~ hOIll Fllgli,h illtO Marathi. He di{'eI in I ~IHI; .
   
   .
   
   I'lOjessor .~. A. Na;m/JIIlfy wa, hom ill 1.,:\1 :.. .. Bombay. He haswrittt'n onl: textbook. lind has' (co) author('d til ree: rc:llc.'III·ch· l/!vl.11 monographs and sewr:.1 I'c,,':,rl.·h pflpe .... He has t allllill ;1\ tlc\'c:r;d ("(t II Cl' donrd institutiolls ill India ani! :,bf'O~C1 .
   
   Pm/mar S. IJ. Sinth wu bem In lUfl1l village Chowkhnrll of Mi....pu ... III..... (U.P.) . He I M.A. In M.\h.", , Univendty of A1lahAbIcl In..... 'I !"'1111 ' ~ cll~"rcr. from
   
   Kum.un Unl
   
   (co) auth ' red two "nudent •. He hi ....U.... educational lnl,hudonl U. isnowProfe M..h....' • Kanari VI.hWl\,ldyalaya. H..._
   
   •
   
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