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Lilavati

Bhiskracarya

A Treatise of Mathematics of Vedic Tradition

510

Ipat-LId

Krishnaji Shankara Patwardhan

Somashekhara Amrita Naimpally

Shyam Lal Singh

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LILAVATI OF BHASKARACARYA

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LILAVATI OF BHASKARACARYA

Treatise of Mathematics of Vedic Tradition

with rationale in terms of modern mathematics

largely based on N. H. Phadke's Marathi translation of Lilavati

Translated by

KRISHNAJI SHANKARA PATWARDHAN

SOMASHEKHARA AMRITA NAIMPALLY

SHYAM LAL SINGH

510, PAT-L:7

135847

Oa 135847

236, 9th Main III Block, Jayanagar, Bangalore 560 011

8 Mahalaxmi Chamber, Warden Road, Mumbai 400 026

120 Royapetah High Road, Mylapore, Chennai 600 001

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Printed in India BY JAINENDRA PRAKASH JAIN AT SHRI JAINENDRA PRESS, A45 NARAINA, PHASE-I, NEW DELHI [10 028

51 0 0 PAt- L :.

Foreword

The names Bhaskaracarya and Lilavati are well-known; If a student displays outstanding talent in Mathematics, the school teacher lovingly calls the student "Second Bhaskaracarya". Many legends about Lilavati are in vogue. It is but natural that Indians have a special interest in Bhaskaracarya and his works, especially the Lilavati. Bhaskaracarya was born in Maharashtra. Many translations of the Lilavati as well as commentaries on it are available. However, there is no work in Marathi which studies this subject thoroughly from all angles in a modern twentieth century fashion. Professor N.H. Phadke has taken great pains in the preparation of the present work A New Light on Lilavati; thus filling a void.

Eighteen years ago Professor Phadke wrote a booklet on Indian Mathematicians. Therein we find a lot of authentic and interesting information about Bhaskaracarya and Lilavati. Doubtless the readers of that booklet will look forward expectantly at the present treatise.

In the Lilavati, Arithmetic is presented as an enjoyable playful activity. Professor Phadke has admirably succeeded in preserving this spirit in the A New Light on Lilavati. Since Professor Phadke studied many books in Sanskrit, Hindi, English and Marathi, he could have easily given scholarly notes and references all over. But the author has resisted that temptation and has given the relevant supplementary information.

Lilavati of Bhaskaracaya

formation in six appendices. So the book is easy to read and the readers enjoy the playful atmosphere.

The Lilavati 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 Visnu 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 Visnu there are a conch, disc, mace, and lotus. They can be placed in 4x3x2x1 = 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 10! = 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: "O, you intelligent girl Lilavati," "O, friend," "My beloved," "Deer-eyed," "Fickle-eyed," etc. Scholars may draw varying conclusions from these regarding the types of students studying in the "Fun with Arithmetic" classes of Bhaskaracarya. Yet, one thing is crystal clear to the readers of the Lilavati; 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 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 1, 2, 3, 4, 5 or 6 at a time from six substances which are respectively sweet, bitter, astringent, sour, salty and hot. O, arithmetician, tell me how many different relishes can be prepared."

Foreword

pared?" One does not know whether boys and girls prepared such relishes but the readers relish, time and again, such gourmet Arithmetic!

Lilavati is such an ancient Arithmetic! Today we know many more techniques and results. In this regard, Professor Phadke has been quite alert. While explaining stanza CLXXV, he discusses (i) cos(t) = 0 and (ii) cos(t) # 0. This is an example of a result which was not known in the times of Bhaskaracarya but well-known now even to conscientious high school students. There is nothing surprising about this. Undergraduate students of Mathematics are now required to study several results which Newton did not know.

For those who are interested in both Sanskrit and Mathematics, Lilavati is an attractive illustrated arithmetized book of stories. But it would not be fair to merely consider New Light on Lilavati as a cultural or literary treatise. In the history of Mathematics, Lilavati occupies an honoured place. It shows the extraordinary prowess of Bhaskaracarya who was not only a top Mathematician but also an excellent teacher. Plato says in Republic: Arithmetic has a very great and elevating effect, compelling the soul to reason about abstract numbers, and if visible and tangible objects are intruding upon the argument, refusing to be satisfied.

If there be any truth in the above statement, then without getting distracted by interesting problems, poetic fancy and attractive descriptions, a child will be influenced by Lilavati and hopefully a new Bhaskaracarya will emerge.

In the history of Mathematics one does find such unusual incidents: The well-known Mathematician Madame Kowalewski was inspired by pages of an old Mathematics book which were used as wall paper! I'll not be surprised if some "Lilavati" inspired by the present treatise New Light on Lilavati, will blossom forth into a top-notch researcher in Modern Mathematics.

Pune University

VV.Narlikar

23-5-1971

Lokmanya Tilak

Professor of Applied Mathematics

NO_CONTENT_HERE

Foreword by V.V Narlikar

Roman Transliteration of Devanagari

Translators' Preface

xiii

Bhaskaracarya: His Life and Work

Lilavati

1. Definitions and Tables

2. Place Values of Digits

3. Addition and Subtraction

4. Methods of Multiplication

5. Division

6. Methods of Finding Squares

7. Square Root

8. Method to Find the Cube

9. Cube Roots

10. Eight Operations on Fractions

11. Addition and Subtraction of Fractions

12. Multiplication of Fractions

13. Division of Fractions

Lilavati of Bhaskaracarya

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

15. Eight Rules Concerning Zero

16. Reverse Process

17. To Find an Unknown Quantity

18. Method of Transition

19. Square Transition

20. Quadratic Equation

21. The Rule of Three

22. Inverse Proportion

23. The Rule of Five

24. Rules for Barter

25. Simple Interest

26. Combinations

27. Progressions (Series)

105

28. Mensuration

29. Volume

151

30. Wood Cutting

155

31. Volume of Heap of Grain

157

32. Shadows

161

33. Pulverization

167

34. Concatenation (Permutations, Partitions etc.)

177

Index of Verses

183

Subject Index

195

Roman Transliteration of Devanagari

VOWELS

Shorl		7	4	(and #)
Long	311	3	311	311
		ai	au	Anusvara	Visarga	Non-aspirant

CONSONANTS

Classified	3	7	8	<
	kh	g	gh	n
M	4	6
j	jh	< J
3	9	dh	7

Lilavati

Bhaskaracarya

Un-Classified

xii	9	5	b	I
P	ph	7	bh	m
Compound	y	7	j	4
ks	Ur

Translators' Preface

It gives me great pleasure in placing before you this book, largely a translation of the Lilavati Punardarsana written in Marathi by my revered Guru, the late Professor N.H. Phadke in [971. He took great pains in writing this Marathi translation with comments and explanations.

I must thank Smt. Subhadrabai Phadke for directing me to her eldest son Shri G.N. Phadke, Chief Engineer, Metropolitan Railway, Calcutta for the authority to publish this English translation, made by me, of Professor Phadke's Lilavati Punardarsana. I acknowledge the kindness shown by Mr. G.N. Phadke in giving me the authority.

It was some time in June 1982 that my student Somashekhar Naimpally expressed to me his desire to publish the English translation of the Lilavati. By the grace of the Almighty, I have been able to complete this translation before I completed 71 years of my life. I thank Somashekhar. I also wish to thank my wife Laxmibai (alias Mai) without whose cooperation I would not have been able to complete this work so soon.

Pune

KS. Patwardhan

November 2, 1982

xiii

xiv Lilavati of Bhaskaracaya

In 1977, I was inspired to translate Lilavati 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 gratitude.

Thunder Bay S.A. Naimpally

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 I owe 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, certain renderings of certain stanzas 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 Lilavati can give pleasure and insight to every class of readers, both from children to senior citizens.

Rishikesh S.L. Singh

24-11-1991

Bhaskaracarya: His Life and Work

Rarely does one come across a person, at least in Maharashtra, who has not heard about Bhaskaracarya or Bhaskara II, the great poet and mathematician. His book Siddhantasiromani, especially the first part known as the Lilavati (Slate Mathematics or Arithmetic), is well-known all over the world. Because of the poetic name Lilavati and the excellent problems contained therein, Bhaskaracarya has earned the respect of scholars for the last eight hundred years.

The great nineteenth-century German mathematician Weierstrass said,

“A mathematician, who is also not something of a poet, can never be a complete mathematician.”

One finds, in the West, many mathematicians who have a flair for writing poetry. Omar Khayyam, famous for Rubaiyat and sitting under a tree with wine and women, was primarily a mathematician. In the same way, we find in Bhaskaracarya an extraordinary combination of

Bhaskara was a great Astronomer of the seventh century:

Quatrains:

Lilavati of Bhaskaracarya

Bhaskaracarya was born in 114 AD. He has made this clear in the fourth part of the Siddhantasiromani titled Goladhyaya (Astronomy). In the 58th stanza in the chapter of problems he says:

I was born in Saka 1036 (A.D. 114) and wrote Siddhantasiromani 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 Sandilya 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:

Bhaskaracarya 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 AD. 1078. He wrote two books Karana-Grantha and Jataka-Tika-Grantha. As Bhaskaracarya studied under the guidance of such a competent teacher, he became an expert in many branches of learning.

Location of Vijjalavida

In the first stanza quoted above, Bhaskaracarya says that he hailed from Vijjalavida. But no definite information is available regarding its location.

Bhaskaracarya: His Life and Work

The last part of the name ('Vijjalavida') is corrupted into Bida and some think that Bhaskaracarya hailed from Bida. But Bida is 64 km far from Ahmednagar in Marathawada and is not anywhere near the Sahya mountain. Besides, there is no descendant of Bhaskaracarya traceable in Bida. Some opine that Vijjalavida is the town Bedara near Hyderabad. The reason for this goes back to the great poet Faizi of Emperor Akbar's court. At the suggestion of Akbar, Faizi translated Lilavati into Persian: Therein he says that Bhaskaracarya hailed from Bedara. But Bedara, which is 80 km far from Sholapur, does not have a single small hill even nearby: Hence, Vijjalavida is not Bedara.

In Bhaskaracarya's times, a town Kalyana was ruled over by the Calukya dynasty. There is neither any reference to this dynasty in Bhaskaracarya's works nor a reference to him is made in the history of the Calukya dynasty. Is Vijjalavida the town of Vijapura? Some experts consider Bhaskaracarya to be a Vaisnavite Brahmin from Karnataka; Pandit Sudhakar Dwivedi also takes him to be Vaisnavite on the basis of his frequent quotes from the Visnu Purana in the chapter 'Bhuvanakosa of Goladhyava.' But then Vijapura is at least 160 km east of Sahya Mountain. Also, "Vijapura" seems to be derived from "Vidyapura" or "Vijayapura" rather than "Vijjalavida." So Vijjalavida cannot be identified with Vijapura.

Further, there is evidence that Bhaskaracarya was not from Karnataka. Nrsimha in his commentary Vasanavarttika on Lilavati writes:

sfHTrTATd: | TERTRTTIIY)

i.e. Bhaskaracarya was from Maharashtra. In another commentary, Maricika Munisvara describes Vijjalavida thus:

It seems from the above quotations that Vijjalavida is situated in Khandesha or Nasika district which is north of the river Godavari. There is another Bida near Kolhapura which is close to the Sahya Mountain but not to the north of Godavari. Bida (of Marathawada), Bedara, Vijapura are no way near the ranges of Sahya Mountain.

Xviii

Lilavati of Bhaskaracarya

Munisvara wrote his commentary in 1608 which is approximately 500 years after the composition of the Siromani. We consider that he must have had reliable source of information regarding the Bhaskaracarya'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 km 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 km from Calisagaon must be the former Vijjalavida.

Dr. Telanga discovered the 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. Krsnaraja, Indraraja, Govana and Suryadeva. Further, there is a reference to a grant given to Cangadeva, grandson of Bhaskaracarya, 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 Cangadeva alone belonged to Patana. It is possible that Cangadeva 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 firm 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 Bhaskaracarya'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:

Bhau Daji: Journal of the Royal Asiatic Society (1865), Pp. 392-418 and Journal of the Bombay Branch of the Royal Asiatic Society, 8 (1868), P. 231.

Bhaskaracarva: His Life and Work

Bhaskaracarya's Education

It is needless to mention that Bhaskaracarya must have been an exceptionally intelligent student. He gives his genealogy in his Algebra and Question 61 of 'Goladhyaya'. In both the places he records that he received his education from his father: In Algebra he says:

3rararfradi fdgti 447: 1

From the above it is clear that Bhaskaracarya's father was his teacher: Scholars had bestowed on him the title of "Acarya" (Preceptor). He says that he learnt Algebra also from his father and mastered it sufficiently to write a book on it. He was well versed in Poetry, Grammar, Mathematics, Astronomy etc. and kept himself up to date. In stanza 279 of the Lilavati we learn about the subjects one had to master to earn the title "Acarya". Ganesa Daivajia calls Bhaskaracarya "Ganakacakracudamani" (A Jewel among mathematicians). Bhaskaracarya excelled in teaching Mathematics and Astronomy. It is clear from the Copper Plate Inscriptions of the time that the scholars hesitated to debate with Bhaskaracarya's students.

Works

Bhaskaracarya wrote his Siddhantasiromani when he was 36. It is in four parts: (1) Lilavati; (2) Algebra; (3) Planetary motions; and (4) Astronomy. The Lilavati mainly deals with Arithmetic but also contains Geometry, Trigonometry and Algebra. Those days (as it is true even today) both educated and uneducated were interested in future events, and naturally every Astronomer had to study Astrology. For this purpose it was necessary to master Planetary Motions and Astronomy and the Lilavati provided prerequisites for this study. Bhaskaracarya wrote his Lilavati by selecting good parts from Sridharacarya's TriSatika and Mahaviracarya's Ganitasarasamgraha.

Lilavati of Bhaskaracaya and adding material of his own; All the prerequisites for the study of Astronomy are in the Lilavati. In order to solve problems concerning periods in Planetary Motions, Bhaskaracarya has added solutions of what are known as Diophantine Equations.

Bhaskaracarya's Algebra too is an excellent book on which several commentaries have been written in several languages. A well-known commentary on Algebra, the Navarkura, was written in 1612 by Krsnadaivajia who was an Astrologer in the court of the Emperor Jehangir. In 1634, a Persian translation of Algebra was made by Ataulla Rasidi, an Astrologer in the court of Shahjahan. Strachi translated it into English in 1813 and Khanapur Shastri rendered it into Marathi in 1897. Pandit Sudhakar Dvivedi wrote a Sanskrit commentary on Algebra wherein Diophantine Equations, Squares, Pellian equations are included.

Being more involved compared to the first two parts, neither the Eastern nor Western scholars have paid any attention. However, they are worth reading by teachers and students of Mathematical Astronomy. In his Astronomical book, Bhaskaracarya maintains that the earth is stationary. In his work on Planetary Motions, he considers Lunar motions, Revolutions, Eclipses and such topics which are profound and involved.

Besides these, Bhaskaracarya wrote several works: Karana-kutuhala, Sarvatobhadrayantra, Vasisthatulya, Vivahapatala. It is our misfortune that the original manuscripts are not available. Maharashtrians do not generally have an awareness of the necessity of preserving historical artifacts. However, the handwritten copies of Lilavati with commentaries which are as old as two hundred years are still available.

Valiant Bhaskaracarya

Readers will notice that Lilavati is a monumental work. According to the prevailing customs, Bhaskaracarya wrote it in verse form. To explain a scientific topic like Mathematics in a verse form is indeed a difficult task. But a still more difficult job is to infuse qualities poetic.

It appears that Bhaskara and the ancient Indian astronomers considered the earth stationary just for observations and subsequent calculations only.

Bhaskaracarya: His Life and Work

But Bhaskaracarya overcame all these hurdles and wrote the famous work Lilavati. If the art of printing existed then, several editions of the Lilavati might have been published. However, despite the non-availability of the printing machines, the handwritten copies of Lilavati were used from one end of India to the other. The (whole) Lilavati reached other countries a little later, in the sixteenth century. All those who read this book were overwhelmed by Bhaskaracarya's skill. It seems Bhaskaracarya himself loved his first creation, i.e., Lilavati more than his other works. Beautiful poetic flights of Lilavati are not found in any other mathematical or scientific compositions.

The Poet Bhaskaracarya

From his works, it is clear that Bhaskaracarya had mastered Grammar, especially that of Panini. Also, like the poets, Bhaskaracarya was fond of alliterations and metaphors; e.g.

He was also quite witty. In one problem, a wealthy person is supposed to be giving one "kavadi" in charity to a poor man! We give further evidence of Bhaskaracarya's poetical skills by examples from Astronomy. There are vivid descriptions of seasons. It is all beautiful.

See verse 160, Lilavati

Verse 131, ibid:

Verse 133, ibid.

Verse 73, ibid.

Verse 75, ibid.

Verse 60, ibid:

Verse 43, ibid.

Cowric.

Lilavati of Bhaskaracarya

adding material of his own. All the prerequisites for the study of Astronomy are in the Lilavati. In order to solve problems concerning periods in Planetary Motions, Bhaskaracarya has added solutions of what are known as Diophantine Equations.

Bhaskaracarya's Algebra too is an excellent book on which several commentaries have been written in several languages. A well-known commentary on Algebra, the Navarikura, was written in 1612 by Krspadaivajia, who was an Astrologer in the court of the Emperor Jehiangir. In 1634 a Persian translation of Algebra was made by Ataulla Rasidi, an Astrologer in the court of Shahjahan. Strachi translated it into English in 1813 and Khanapur Shastri rendered it into Marathi in 1897. Pandit Sudhakar Dvivedi wrote a Sanskrit commentary on Algebra wherein Diophantine Equations, Squares, Pellian equations are included.

Planetary Motions and Astronomy being more involved compared to the first two parts, neither the Eastern nor Western scholars have paid any attention. However, they are worth reading by teachers and students of Mathematical Astronomy. In his Astronomical book, Bhaskaracarya maintains that the earth is stationary. In his work on Planetary Motions, he considers Lunar motions, Revolutions, Eclipses and such topics which are profound and involved.

Valiant Bhaskaracarya

Readers will notice that Lilavati is a monumental work. According to the prevailing customs, Bhaskaracarya wrote it in verse form. To explain scientific topics like Mathematics in a verse form is indeed a difficult task. But a still more difficult job is to infuse poetic qualities.

[ appears that Bhaskara and the ancient Indian astronomers considered earth stationary just for observations and subsequent calculations only-

Bhaskaracarya: His Life and Work

But Bhaskaracarya overcame all these hurdles and wrote the famous work Lilavati. If the art of printing existed then, several editions of the Lilavati might have been published. However, despite the non-availability of the printing machines, the handwritten copies of Lilavati were used from one end of India to the other. The Lilavati replaced most other texts that were used earlier. The (whole) Lilavati reached other countries a little later; in the sixteenth century. All those who read this book were overwhelmed by Bhaskaracarya's skill. It seems Bhaskaracarya himself loved his first creation, i.e., Lilavati more than his other works. Beautiful poetic flights of Lilavati are not found in any other mathematical or scientific compositions.

The Poet Bhaskaracarya

From his works it is clear that Bhaskaracarya had mastered Grammar, especially that of Panini. Also, like the poets, Bhaskaracarya was fond of alliterations and metaphors, e.g.,

See Verse 160, Lilavati

Verse 131, ibid.

Verse 133, ibid.

Verse 73, ibid.

Verse 75, ibid.

Verse 60, ibid.

Verse 43, ibid.

Cowric.

xxii Lilavati of Bhaskaracarya

The late Balakrsna Janardhane Modaka rendered into Marathi the following:

In the Hemanta (winter) season the earth looks more enchanting because of the plenty of stored grains. The drops of water, in the form of dew, appear like pearls and add to the beauty. The well-fed contented cows and bulls move about in flocks and the farmers enjoy this beautiful sight.

Teaching Ability

In the Lilavati, Bhaskaracarya has displayed his various skills. While teaching Mathematics he also wanted to convey information on religion, the Vedas, Puranas, Epics etc. In the stanza 76, the information is given regarding a chariot and the fact that Karna, though a brother of Arjuna, was his enemy. The entire battle-scene is graphically depicted. This stanza can be compared with Kalidasa's Sakuntalam, Grivabhangabhiranam: The graphic descriptions of a drove of swans; flock of elephants; a colony of bees; the attack of a snake by a domesticated 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 contemporary tradition. 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 skillfully than his predecessors. Although he took earth to be stationary and the center of the universe, it did not adversely affect his calculations. Bhaskaracarya was familiar with many results that were later discovered by Copernicus and Tycho Brahe. He also had some idea of limits which was (500 years) later discovered by Newton.

He moulded difficult language like Sanskrit to express scientific subjects like Mathematics. By using terms from fine literature, he...

Bhaskaracarya: His Life and Work coined many equivalents for numbers. Verily he opened a great storehouse of technical terms. If one wants Marathi equivalents for scientific terms in English, one may easily find them in Bhaskaracarya's works. These could be useful to teach Mathematics in Indian languages. Further, his Lilavati has been used as a textbook for the last eight hundred years in India. Even now it is used in some places in some Indian provinces.

He was quite practical too as he was of the opinion that if eclipses, conjunctions etc. do not occur as given in ancient almanacs, then the almanacs must be changed: Orthodox people used to stick to false beliefs, like demons Rahu and Ketu swallow the Moon and the Sun during eclipses. What actually happens is that at the time of Lunar Eclipse, the Moon enters the shadow of the Earth and so it becomes invisible. Bhaskaracarya knew this fact but found it difficult to convince the public at large. He used to say:

Well, Rahu is not a demon. It is the shadow of the earth that makes the Moon invisible. If you still believe in Rahu, at least say that he entered the shadow of the Earth and then swallowed the Moon.

This is how he found a way to reconcile the old and the new discoveries. Though a practical scholar, he was not to compromise with his principles. There was a great Indian Astronomer Lalla (A.D. 768) who wrote a textbook on Mathematics which was used until Bhaskaracarya's time. That text contained some incorrect propositions and formulae. In his book on Astronomy, Bhaskaracarya corrected some of Lalla's formulae like the correction of the formula for the measurement of the surface area of a sphere. He has given this in Lilavati too. Though he claimed himself to be a student of Brahmagupta, he never hesitated to criticize the latter's long or wrong or tedious methods. Sometimes he also criticized Aryabhata.

The late Mr. Dixit conjectured that Bhaskaracarya might have visited Baghdad as some anonymous Indian astrologers had visited Turkey and the middle-east as honourable guests according to some Arab sources.

Indeed all other languages derived from Sanskrit.

See stanza 18Sb, Lilavati.

xxiv Lilavati of Bhaskaracarya

If this view is acceptable, then there should have been some reference to his works to the Arabian mathematicians. Hence, the above view that Bhaskaracarya had visited the middle-east is not convincing.

Social Conditions

It is rather difficult to draw a picture of the social conditions of Bhaskaracarya's period by way of a book on Arithmetic. In Bhaskaracarya's times, scholarship was assessed by the study of ten 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 risky. Niska, a silver coin, was the principal currency. Peacocks were common pets. There was abundance of lotus flowers, bees, elephants, and birds. Siva, Visnu, and Ganesa were universally worshipped. The Ramayana 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 patronized by rulers, Bhaskaracarya seems to be an exception: He was 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 mathematics merely studied Bhaskaracarya's books, still no research was done until the 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 chords would have been considerably simplified. He knew that the earth attracted other bodies but he missed Force, Mass, Acceleration. For the computation of daily motion of planets, he introduced the concept of instantaneous velocity by dividing the day into large number of small intervals. The formula <sin 0> = cos 0 is implicitly available in Siromani. He was aware that the differential coefficient vanishes at an extreme value of the function. He appears to be the first mathematician who could perceive the ideas of differential calculus 500 years ago.

Bhaskaracarya: His Life and Work

before Newton and Leibniz He had little knowledge of eilipse, parabola or hyperbola. Had he proved theorems in Algebra and Geometry, he would have achieved greater status as a mathematician. Nonetheless, his achievements put him on the top of the list of mathematicians.

We are fortunate to have had such a great mathematician in India. But we are so negligent that we forgot even to celebrate the 800th anniversary of the composition of his Siddhantasiromani. Let us pledge to remember to celebrate Bhaskaracarya's eight hundredth death anniversary in 1993.

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Lilavati (Pati Mathematics)

Praise of Lord Ganesa

[First] I offer my salutations to the elephant-faced Lord who creates love in His devotees, by remembering whom all obstacles are destroyed and whose feet are revered by the community of gods. Here I give methods of state mathematics Lilavati which is loved by discriminating people because of its clarity, brevity as well as its literary flavour.

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CHAPTER

Definitions and Tables

Twenty varatakas make one kakini. Four kakinis make one pana: Sixleen panas make one dramma: And sixteen drammas make one niska:

Comment: Bhaskaracarya gives the following label concerning the coins used in the eleventh and twelfth centuries:

20 kavadis (Cowries)	=	1 kakini (davad)
4 kakinis	=	1 pana (paisa)
16 panas (paisas)	=	1 dramma
16 drammas	=	1 niska

Until the twentieth century, the above coins were more or less used in India. During the British regime, coins pai, ruka and dhabu paisa were in circulation. Such coins were also in use during the rule of the Peshawas. In 1910 it was possible to get 40 Cowries for one Paisa and they could be used to procure sundries such as hot peppers and coriander. The word dama for price has evolved from 'dramma'.

Lilavati of Bhaskaracarya

Measure for gold

2 yavas = 1 guija (ratt), 3 guijis = 1 valla

8 vallas = 1 dharana, 2 dharanas = 1 gadyanaka

14 vallas = 1 dhataka:

This verse III gives various measures for weighing silver or barley-corn (with its outer shell) is called "vava". Gold and silver are weighed in grams and so guijas are not in use. Gunja is a fruit with black, red or blackish red colour. It is shaped like green lentil but twice its size. Gunja, being tough and durable, has been in use since ancient times; village goldsmiths use guijas for weighing silver and even now. One guija weighs 5 gold 16 troy grain.

Presently, gadyana or gadyanaka and dhataka are not in use. But valla or vala is still in vogue. One dhataka = 42 guijas; 1 gadyanaka = 48 guijas.

5 guijas = 1 masa, 16 masas = 1 karsa

Those who know measures for gold call gold karsa 'suvarna'.

In the times of Bhaskaracarya, the principal measure for weighing gold was a karsa: 80 guijas made one karsa. The modern measure tola equals 96 guijas: tola = 113 gms.

Units of Length

Definitions and Tables

8 yavas = 1 angula, 24 angulas = 1 hasta (forearm)

4 hasta = 1 danda, 2000 dandas = 1 krosa or kosa:

Comment: If eight grains of barley-corn are kept bellywise close to each other, the length is one argula, i.e., the phalanx of the fingers. There were no standard measures in India. There were many kingdoms and principalities, each having its own weights and measures. So the unit of length was not the same all over India. However, assuming that the height of a man is 3½ hastas (forearms), the above units could be uniformly used throughout India. Ordinarily, hasta = 20 inches, 1 inch = 2.5400 cms. So the height of a man equals 5 ft. 10 inches. A standard unit of length came into use all over India only during the British Rule. Old measures remained in the literature as well as in unimportant transactions. Now miles have given way to kilometers. One kosa is two miles but now both kosa and mile will go.

vojana = 4 kosas; 10 hastas = 1 bamboo.

nivartana: Area of a square with sides 20 bamboos.

Comment: A bamboo or a pole is easily available all over for measuring lengths. The English used pole (bamboo) = 5 yards. In the fifth verse, arguli is taken as a unit of length which we can compare with foot in the English system. One nivartana is approximately two acres (1 acre = 0.4089 hectare).

Measures for Grains in Volume

Ghanahasta (Unit Cube) is a solid which has twelve edges each of length one hasta; its volume is the unit. In Magadha state, quantity of grain of unit volume is called khari.

Comment: 'Dvadasasra' means a solid with twelve edges and not (twelve) corners as the late Khanapurkara Shastri interpreted: A cube has twelve edges but only eight corners (vertices). The measure khari from Magadha must have been the standard measure for grains.

Lilavati of Bhaskaracarya

Varanasi, Aliahabad, Gaya, Patna which are in southern Uttar Pradesh or Bihar, (ored an influential part of ancient India. Hence measures from this part of the country were taken as standard. One khwri equals eight pavalis or 32 kg: 'Sastrodita must be taken to mean Government order.

16 khari	=
drona	=	adhaka
adhaka	=	prastha
prastha	=	kudava

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 frustums 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 kudava, atave, nithave are still in use in Konkana: The current Government measures are cylindrical and grain is now measured by weight.

(The order of the digits is from right to left. Thus) dvisapta 72, kha-yuga 40 where kha denotes 0 and yuga denotes 4. % gadyanakas 36 guijas = tanka, Seer 72 tankas, 40 Scers Maund. These are Turkish measures.

Comment: Since these measures are called Turkish, this verse must have been inserted in the Lilavati later on. In Bhaskaracarya's times, there was no Muslim influence either in the north or south India: Only the north was attacked by Mahamud of Gazni. Of course, these commercial terms must have been known to businessmen but it is unlikely that they were in common use.

Definitions and Tables

Remaining measures of time etc. are well-known. E.g: nimisa, the unit of time, measures the time for a wink (of an eye) of a wise man.

tatpara	=	1/30 nimisa
truti	=	1/100 tatpara
18 nimisas	=	kastha
30 kasthas	=	kala
30 kalis	=	naksatra
ghatika	;	ghatikas
ghatikas	=	1 ksana
60	=	Day.

Alternate measures of time:

asu	(one complete breath)	Time taken to recite 20 long vowels
6 asus	=	pala
60 palas	=	ghatika
60 ghatikas	=	Day
30 Days	=	Month
12 Months	=	Year

This is the end of the chapter on ters

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CHAPTER 2 Place Values of Digits

XIM

(In this verse Lord Ganpati is invoked again)

I offer homage to that Lord Ganesa whose neck is adorned by amusing black and most poisonous snakes and who has unctuous lustre like a blue lotus;

Taferrartr yui Teffafa

Positions of the digits from right to left are unit, ten, hundred, thousand, ten thousand, hundred thousand (lakh), million, ten million (crore), hundred million, billion (abja), kharva, nikharva, mahapadma, Sanku, jaladhi, antya, madhya, parardha: The value of each digit on the left is ten times that on the right. This decimal system

Lilavati of Bhaskaracarya

The system of place values was conceived by ancient scholars to make practical calculations simpler.

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 to 9) to represent any given number made mathematical operations (addition, subtraction, etc.) easy. What Europeans call "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 Devanagari numerals have been in use in various parts of India since A.D. 400 and the English numerals are their modified forms. Although this verse goes up to parardha (107), there are terms for numbers up to 1014 in Sanskrit.

See Datta and Singh History of Hindu Mathematics; and O.P. Jaggi's Indian Astronomy and Mathematics.

CHAPTER 3 Addition and Subtraction

Explain the method of addition and subtraction in half a stanza. I shall

Addition or subtraction should be done wise from right to left or vice-versa.

First write down the given numbers one below the other (taking care to see that the digits match the places, that is, for example, digits in ten's places should be in the same vertical line). Then beginning with the unit's place add (or subtract) the digits, then move to ten's place. In this procedure two things have been taken for granted viz. (i) If the sum of the digits in the same place is greater than ten, then one has to carry forward the digit in the ten's place of the sum to the left; (ii) The number to be subtracted must be smaller than the other number. These things are not explicitly stated. Although addition and subtraction can be done from left to right also, this method is not in vogue. The method is given in brief because it is clearly explained.

Lilavati of Bhaskaracarya

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.

Comment: As shown in the adjoining figure, write the numbers to be added one below the other, keeping their proper positions (places) of digits. Then draw a horizontal line below the last number, viz. 100. Add all the digits in the unit's place to get 20; write 0 in the unit's place below the line and carry 2 to the ten's place. Now add the digits in the ten's place (along with the 2 carried over) and the sum is 6. Write 6 in the ten's place below the line and carry over to the hundred's place. The sum of the digits on the hundred's place, together with the carried over, is 3. 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 subtracted below 10,000 and draw a horizontal line below 360 and the sign before 360. In the unit's place, 0 subtracted from 0 yields 0. In the ten's place, 6 cannot be subtracted from 0, so bring from the hundred's to make it 10. 6 subtracted from 10 gives 4 and we carry on the hundred's. Next in hundred's place, 4 subtracted from 10 yields 6 with 1 carried over. Finally, 1 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.

CHAPTER 4 Methods of Multiplication

Now [ tell the method of multiplication in two and hall stanze

4r 3u:Je@a 47 *7 joangora:57 &l gora:{a: / / XVI //

321T3A73T7

Direct Method

First multiply the digit in the unit's place of the multiplicand by the multiplier, then the digit in the ten's place and so on up (to the last digit on the extreme left.

Comment: This method can be used when the multiplier is a small number and the user has memorized the multiplication tables. Of course, if the product at any stage consists of two or more digits, one has to carry over the part without the digit in the unit's place. There is

Lilavati of Bhaskaracarya

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

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.

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 multiplier is 45, first multiply by 9 and then the result by 5.)

Place Method

Multiply by each digit of the 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 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 first one. This procedure is carried on until all the digits of the multiplier are used. Finally add all these products.

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.

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: To elucidate them Bhaskaracarya gives the following example.

Methods of Multiplication

O! you auspicious girl with lovable eyes of a young deer; if you have well understood the above methods of multiplication, what is the product of 135 and 12? Also tell me what number will you obtain if the product is divided by 12.

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 x 12 = 36 to which we add 6 to get 42. We write 2 in the ten's place and carry over 4. We add 4 to 1 x 12 = 12 to get 16 which is written to the left of 2. (The explanation given by the late Khanapurkar Shas concerning this method appears to be incorrect.)

To use the second method, we split 12 = 8 + 4. Now 135 x 8 = 1080 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 this method cannot be used. If the multiplicand has factors we can interchange the multiplicand and the multiplier.

The fourth method uses the digits 2 and 10 in unit's and ten's place respectively. We first multiply 135 by 2 and get 270. Next we multiply by 10 and get 1350 which is written by shifting it one place to the left. Then we add the two numbers 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, 135 x 20 = 2700 and 135 x 8 = 1080. 135 x 12 = 2700 + 1080 = 1620.

In stanza XVIII, "the number formed by 5, 3 and 1" means 135 written from right to left since the digits are reversed. The multiplier is "Divakara" which means the Sun representing 12. In this stanza, there are some vocatives used to address Lilavati and we conclude that Bhaskaracarya must have been teaching a female student.

"siaii 7iml m;" = "Digits move in reverse order."

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CHAPTER 5

Find the largest integer whose product with the divisor can be subtracted from the extreme left hand digit(s) of the dividend. This integer is the first digit of the quotient: If the divisor and the dividend have a common factor, then the common factor can be cancelled and the division is carried out with the remaining factors.

Comment: To explain the process of division, we consider the example at the end of Stanza XVIII: i.e., divide 1620 by 12. The first method is:

Lilavati of Bhaskaracarya

1620 (135 12 can be subtracted from 16 once only and so the first digit of the quotient is 1 and the remainder is 4, We then have to divide 42 by 12 and 12 * 3 = 36 can be subtracted from 42 So the next digit of the quotient is 3. Now the remainder is 60 which is 12 x 5. So the last digit of the quotient is 5 and the division is complete. The quotient is 135 and remainder 0.

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.

3)	1620	(540
540	(135	15
12	14
42		12
000	20
00	20
00	00

In the above example the remainder is 0. If the remainder is not zero and we follow the second method, we may end up with different remainders. For example:

In 427 the quotient is 15 and the remainder 7.

and 61 the quotient is 5 and the remainder 1.

CHAPTER 6 Methods of Finding Squares

First Method

The product of a number with itself is called its square. To square a number use the following procedure: First write the square of the extreme left-hand digit on its top. Then multiply the next [i.e., second] digit by the double of the first digit and write the result on the top. Next, multiply the third digit by the double of the first digit and write the result on the top. In this way arrive at the unit's place. Next cross the first digit and shift the number so formed one place to the right. Then repeat the same procedure. Finally, add all the products written at the top and the sum is the required square.

Second and Third Methods

p@:

Lilavati of Bhaskaracarya

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.

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 now use the above-mentioned three methods and give the algebraic theory behind them.

Problem

Friend, if you know the method of finding squares, find the squares of 9, 14, 297 and 10005.

We find the squares of 297 and 10005 only.

First Method

In row A write 297. We make five vertical columns to facilitate the work: Square of 2 is 4 which is written above 2. Twice 2 is 4 and its product with 9 is 36 which is written above 4, taking care that 6 is above 9. Then we multiply 7 by 4 and write 28 above 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 written above line C. Add all the products above the row A and the sum is 88209.

Methods of Finding Squares

Some writers take the sums of numbers between A and B which is 788, between B and C which is 936 and C and D which is 49 and add thus:

788 936 49 88209

The late Khanapurkar Shastri and Pandit Hariprasada Bhagiratha have not explained this point properly.

This example can be solved by other methods:

Second Method: 297

(7)? = 49 and 2 290 x 7 4060. The sum of these three: 84100 + 49 + 4060 = 88209.

Third Method: We choose 3 since 297 + 3 = 300. 297 - 3 = 294. Then

(297)?	(300)	(294)	(3)?
88200			88209

The first method uses the formula:

(c)? = a? + 2ab + Zac + b? + Zbc + c, (a + b +

In the above examples these are 4, 36, 28, 81, 126 and 49, (and with proper place values 40000, 36000, 2800, 8100, 1260 and 49). It is interesting to note that the square of 97 is 8100 + 1260 + 49 = 9409.

Thus we get 7? = 49, 97? = 9409 and 297? = 88209.

The formula used in the second method is (a+b)? = a? + 2ab + b2 and the third method uses a (a + b) (a - b) + b}. Of course, the second and third methods are not so useful when big numbers are involved. In the first method care should be taken to write the products in proper places.

By the second method, (10005)? = (10000 + 100000000) 10000 25 100100025.

We also show the first method which is self-explanatory.

510	125847

Lilavati of Bhaskaracaya

22
2		2	5
0	B
A	5		0
0	5	8
0	5	X
8	8	0	5
S	X	8	8
8	5

CHAPTER 7 Square Root

Tarpfai T

(Starting from the unit's place, mark alternately vertical and horizontal bars above the digits so that the given number is divided into groups of two digits each with the possible exception of the extreme left group. The extreme left group will contain either one digit or two digits and will have a vertical bar on its top or on the right digit respectively.) From the group of the extreme left, deduct the highest possible square of a1 (say): Then write 2a1 in the neighbouring column; this is called parkti (row). To the right of the number obtained as from the above subtraction, write the digit from the next group with a horizontal bar. Now divide the number so obtained by 2ap; this quotient a2 should not be more than 9. Now write 2a2 below 2a1 after shifting it one place to the right and add. The result is the second parkti. Write the next digit to the right of the remainder so obtained.

24 Lilavati of Bhaskaracarya

and from that subtract the square of the second quotient az. Now to the right of the remainder s0 obtained, write the next digit and divide this by the second parkti. 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 pankti after shifting it by one place to the right. The result is the third parkti. Then write the next digit of the given number to the right of the remainder and subtract from it the square of the third digit of the square root. Repeat this process: The result is the required square root.

Tai 7gul d 72 TaNi 4ᵖʳⁱ XXIV | |

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.

Comment: We'll find the square root of 88209 by Bhaskaracarya's method. The first thing is to make horizontal and vertical bars 88209. From its first group; viz. 8 subtract the highest possible square which is 4. We get the first remainder 4 = 8 - 4. Now write (from the given number) to the right of the remainder 4 to get 48.

4	48	18
36	58	2nd
122	14	3rd
81	594
58	410
406	297	297
049
49
00

2 * 2 = 4 is the first parkti. 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 below 4. Add the two to get 58 which is the second paikti. Then subtract 36 = 9 X 4 from 48 to get 12. To its right write the next digit 2 and we get 122. From this subtract the square of 9 to get 41. To the right of 41 write 0, get the next digit from the given number: Divide 410 by the second parkti viz. 58 and get 7 as the quotient and 4 as remainder. Next we write

Square Root

this number 7 in the root column and 7 * 2 = 14 to its right with below 8 Add the two (0 get 594 which is the third pankti: Write the last digit 9 of the given number to the right of 4 to get 49. From this subtract 72 49 to get the remainder 0. The required square root is the number obtained by writing the digits from the root column in the order in which we derived them. Thus it is 297 We can get the same number as half of the third pankti.

In the above method of Bhaskaracarya, we take one digit at a time. In the current method, we take two digits at a time.

Below we tabulate the method of finding the square root of 100025

i001d0d25	rool	parkti	1st
00	00		2nd
Q0_			20
00			000	3rd
00			200
20)	001		000
00			2000	4th
			10
200)	100	10005	20010
Oo0
			1000
QQQ0
2000)	10002
10000
			25
			25
			00

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CHAPTER &

Method to Find the Cube

Cube of given number is its product with itself thrice over. If we want to find the cube of a number of two digits, say, 10a + b, write a first. Below this, write 3alb by shifting this result one place to the right. Below this write 3ab² after shifting it one place to the right. Below this write b³ after shifting it one place to the right. Add all the results, and the result is the cube. This procedure can be modified by starting from b but then each time the shifting should be made to the left. If there are more than two digits, then find the cube of the two digits.

28 Lilavati of Bhaskar

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

Second Method

Or split the given number into two parts. Multiply their product by 3 times their sum. Add this to the sum of the cubes of the two parts and we get the required cube.

Another Method

The cube of a given number is the square of the cube of the square root of the given number.

Comment: The method given in stanza XXV depends on the algebraic identity: (a + b)3 = a3 + 3a2b + 3ab2 + b3.

This further leads to (a + b + c)3 = (a + b)3 + 3(a + b)2c + 3(a + b) c2 + c3.

which can be conveniently used. The methods given in stanza XXVII depend upon the identities (a + b)3 = a3 + b3 + 3ab(a + b) and (a2)3 = (a3)2.

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

Comment: (9)3 = 9 x 9 x 9 = 81 x 9 = 729. We'll find (27)3 by the first two methods.

Method to Find the Cube

First Method

3(2)	84	84
3(2)	294	294
	343	343
		19683

So (27)3 = 19683.

Second Method

Take 27 = 20 + 7

203	8000
73	343
3(20)(7)(27)	11340
19683

In this method there is no shifting as in the first method.

To cube 125 we group (12) 5.

123	1738	1728
3(12)25	2160	2160
3(12)(5)2	900	900
(5)3	125	(125)3	125
		1953125

Now we apply the last method to find (9). Evidently

(9)3	((3)7) = ((3)2)	(27)3	729

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CHAPTER 9 Cube Roots

Draw a vertical bar above the digit in the unit's place of the number whose cube root is wanted. Then put horizontal bars on the two digits to its left, vertical bar on the next and repeat until the extreme left-hand digit is reached.

From the extreme left-hand section, deduct the highest cube possible and write to the left side the number (a) whose cube was subtracted. Write to the right of the remainder; the first digit of the next section to get new sub-dividend. Now divide by 3a and write the quotient b next to a. Then write the next digit from the section to the right of the remainder obtained above. The next divisor is 3ab². In the next step take b' as the divisor: Continue this procedure till the digits in the given number are exhausted.

32 Lilavati of Bhaskaracarya

Comment: Although 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.

Root Parikti	19683	cube root of 19683
First we put the bars like 19683	First group is 19	and the other 683.
3(2) ()	84	84
302) (7)?	328	Subtract 28 8 from 19.
294	294	The remainder is 11 and we write 6 next to it to get 116.
The new divisor is 3(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	302)? (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)? = 294 which is subtracted from 328 to give 34.
We write 3 to the right of 34	and 343 = 7' which when subtracted from 343 gives remainder 0.	The required cube root is 27.

By the same method we now find the cube root of 1953125. The cube root is 125.

Root Pankti	1953125	The current textbooks on Arithmetic don't deal with square roots and cube roots.
These are now found by the use of logarithms or by the tables.	Such means were not available in the days of Bhaskaracarya.	The above method of Bhaskaracarya is quite simple; this simplicity disappeared in later methods.
This method may be extended to obtaining nth root of positive integer.

CHAPTER 10 Eight Operations on Fractions

Reduction of Fractions to Common Denominator: Simple Fraction

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 vice versa. 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.

Lilavati of Bhaskaracarya

Examples

Friend, tell me 5, 3 and 63

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 1

7	5	3
	45	3	5	45 + 3 + 5	53
		15			15	15

We could have added two fractions first and add the sum to the third: 3, 3, 5, 3, 3 and 3, 48, 3, 53

Example 2

In 14, 63 we note that 14 and 63 have a common factor and the quotients (after dividing by) are 2 and 9 respectively.

So 14, 63 = 126, 126, 126 = 18

Note that L.CM: {3,5} = 15 and LCM {14, 63} = 126.

Compound Fractions

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.

Eight Operations on Fractions

This is an example of a fraction which involves "of". Fraction of a fraction is also a fraction.

Example

~minded guy! miser gave beggar good to

2	3	5	16

If you know compound fractions well then, my dear child, tell me how much money the miser gave to the beggar?

The products of the numerators and the denominators are respectively 6 and 7680. So the result is

7680 1280 dramma 1 kavadi or Cowrie.

So the miser gave the smallest coin to the beggar. This shows the sense of humour that Bhaskaracarya possessed.

Addo and Dedo Fractions

Method 1

An addo fraction is one which consists of an integer plus a simple fraction whereas a dedo 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.

Lilavati of Bhaskaracaya

is an addo fraction which is nowadays written as

Similarly 7 2 is a dedo fraction, 2 Jx3+2 22 which is a simple fraction. In current textbooks a 'rational fraction is defined as where a, b are integers and b ≠ 0.

Method 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.

For example, consider the problem of adding ⅓ of 5. The above method gives:

1	(7 + 1)	8
35

The current method is:

5	5	7 + 4
35

Similarly, 5 2 3 35 5 1

35

By way of the present method, it would be:

2	7 - 2	5	7
5	35	35	35

0 friend! 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.

Eight Operations on Fractions

2 + 1 3 Alsosimplify +; (4 (ii)3+} (): (iii)3 8 and 3-4(H)]-43-4()]: 44 of this result to itself:

(iv) 2 andadd

9,3-4 42 Comment: 2 + = =

4	3	12	12	3
3	1x(2+4=2-1	2	3	2 *(8_1 =14 =12
8	3x8	24	12	1X(7_3) =28
3	12x 7	84	3

(iv) 3 x(8 - 4) 7

8 2x8 16

3 Tx (7 + 9) 112=[

9 16x7 112

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CHAPTER 11 Addition and Subtraction of Fractions

First reduce them to the common denominator. If one of the terms is a integer, take 1 as its denominator and proceed: Next add (or subtract) all the numerators so formed to make up the numerator of the result. The common denominator is the denominator. Cancel common factors, if any, and the final answer is obtained.

Comment: 13 + 4 = 13

15 + 91 + 156 = 262

which is also the current method.

Example

XXXLX / /

Lilavati of Bhaskaracarya

friend! Tell mc the sum of 2 6 and the result of subtracting this sum from 3.

Comment: L.CM, of the denominators is 60.

2	12	15	20	30	10	87	29
60						60	20
3	29	60	-	29	20
20	20

CHAPTER 12 Multiplication of Fractions

XLII

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

Comment:	3	2	5	35
	6	72

Example

0 friend! If you are an expert in multiplication of fractions, tell what is the product of 21 and 2 as well as of 3 and 2.

Comment:	2	1	2	15 = 5
	2		6

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CHAPTER 13 Division of Fractions

Division of one fraction by another is equivalent to multiplication of the first by the reciprocal of the second which is obtained by interchanging the numerator and the denominator.

Example

friend, if your intellect is as sharp as the sharp point of a blade of kusa grass, then give me the answers to 5 + 23 and 6 3.

Doa cynosuroides:

Lilavati of Bhaskara I

5 + 21 5 3

6 3 { 3 2

CHAPTER 14 Squares, Cubes, Square Roots and Cube Roots of Fractions

To find squares, cubes, of fractions, find respectively the squares, cubes, of the numerators and divide them respectively by the squares, cubes, of the denominators.

Comment:	7x7	49
	3x3
81	√81	9
16	√16	4

Ti a T

Lilavati of Bhaskaracarya

Find out quickly, if you know how to find squares etc. of fractions, the square and the cube of 3 and the square root and the cube root respectively of the numbers obtained.

Comment:	(63) = (3)	49
49	√49	7
64	√64	8
343	√343	7
8	√8	2

Readers would have noticed that in Bhaskaracarya's times, there were no symbols for exponents nor for radicals.

CHAPTER 15 Eight Rules Concerning Zero

If zero is added to a number, the result is the same number; the square etc. (i.e., square, square-root, cube, cube-root) of zero is zero; any (non-zero) number divided by zero is khahara, i.e., infinite; the product 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 and multiplier zero in rest of the operations (until the final operation is reached). This is because if a number is multiplied by zero and divided by zero then the result is the (former) number; (For example;

Lilavati of Bhaskaracarya

7 * 0 should not be written as 0 because 7 * 0 + 0 is 7; and if one writes 7 x 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 remains immutable.

Comment: a + 0 = a for all numbers a; 0? = Vo = 0' = Jo 0. For any non-zero number a, a/0 = 00, a * 0 = 0 for all a. For non-zero a, a x 0 = a; a $ 0 = 0. Especially for the rule a/0 = 0, Bhaskara emphasizes in his Algebra that whenever such a situation occurs, it remains immutable in form and concept both, that is, it must be written as a/0, and that any (finite) number added to it or subtracted from it will not alter its value. He points out that this type of mathematics is used in astronomical calculations.

The example given by Bhaskara in Lilavati (see Stanza XLVIII) and mathematical situations occurring in his astronomy suggest that by 1/0 is 'infinity' he means 'lim h → 0 through positive values'. However, the symbol & was not in vogue in ancient Indian mathematical works. Not only this, but several astronomers and mathematicians did not agree to Bhaskara's concept of infinity. For example, Jianaraja (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. Khahara 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 khahara-form (such as 5/0) was used 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).

agrlq (There is no change in infinite (khahara) figure if something is added to or subtracted from the same. It is like: there is no change in infinite)

Eight Rules Concerning Zero

Visnu (Almighty) due to dissolution of creation of abounding beings

Example

Find (i) 5 + 0, (ii) 0}, 0 ', Vo_ Jo , (iii) 0 * 5, (iv) 10 + 0. (v) If a certain number is multiplied by 0 and added to half of the result, if that so obtained is first multiplied by 3 and then divided by 0, the result is 63. Find the original number. It is obtained by 'reverse process'.

Comment: (i) 5 + 0 = 5, (ii) 0? = 0' = Vo = Jo = 0, (iii) 0 * 5 = 0, (iv) 10 + 0 = 10. (v) If the number is x,

(xxo + Xxo) x 3 = 63

that is 9x = 63. Thus, x = 14. This is the answer Bhaskaracarya would expect: In modern terminology this is equivalent to lim (rxh + rXh)x; = 6 as h approaches 0.

This gives x = 14. It is unfortunate that this modern concept of limit in traditional language could not be understood properly even by traditional savants. As a consequence, the theory of modern differential was delayed by about 4-5 hundred years.

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CHAPTER 16 Reverse Process

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; addition as subtraction and vice-versa: Again if by adding to or subtracting from given number its part, we should find the indicating fraction, viz N or N as the case may be and then apply the reverse process. Of course, the given number is to be multiplied by the indicated fraction. Next, subtract or add this result to get the unknown.

This will be made clear by examples: (i) Add 10 to a certain number. If 4 is subtracted from 2/5th of the sum so obtained, the result is 12. Find the original number.

Lilavati of Bhaskaracarya

Problem Solving

(i) Find the number which when added to 3/5 of itself yields 40.

Solution: 12 + 4 = 16. 16 + 2/5 = 40

40 - 10 = 30 Ans.

(ii) Find the number which when added to 3/8 of itself yields 33.

Solution: We have to add 3/8 of the number: So multiply 33 by 8/3 to get 33 x 8/3 = 88. So the answer is 88 - 33 = 55.

(iii) Find the number which when 3/5 part of it is subtracted yields 15.

Solution: 8 - 3/5 = 9.

15 + 9 = 24 Ans.

Example

A certain number is multiplied by 3. To this product add its 1/3 part:

Divide the sum so obtained by 7 and then subtract 2/5 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, O 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 * 2 = 20

20 - 8 = 12

12^2 = 144

144 + 52 = 196

Reverse Process

V196 14

14 + 14 21

21 * 7 147

147 _ 147 3 84

3+4

84 + 3 28 Ans

By Algebra, in the equation last but one above, *-xx =84 implies *X(7 _ 3) 84

Hence, x = 84x7 = 147.

Bhaskaracarya takes z+zx3 =147. Here also zX1 =147 Or

2 =147*4=84.

So the required number is 84/3 = 28

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CHAPTER 17 To Find an Unknown Quantity

(Subject to certain conditions)

To discover the unknown number; begin with any convenient number x. Then according to the conditions given in the problem, carry on the operations such as multiplication, division etc. Next multiply by the result given in the problem and divide this product by the number obtained above. Thus we get the unknown by the method of supposition.

Example 1

Lilavati of Bhaskaracarya

A certain number is multiplied by 5, and of the product is subtracted from the result: This is divided by 10 and to this quotient, half, one-third and one-fourth of the original number are added. If this is 68, find the original number.

Suppose the original number is 3. Then

3 * 5 = 15.

15 - 15 * 1 = 10.

10 + 10 = 1.

Since the final result is 68, the unknown number is:

68 x 3 = 68 12 48 Ans

In this method of supposition, we have to add or subtract only parts of or multiples of the original number: Otherwise the answer may not be correct as the following example shows:

When 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 * 5 = 50.

50 - 15 = 35. 35 = 7.

Since the result is 29, the unknown number is 29 x 10 which is not 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,

29 x 5 + 15 = 32.

To Find an Unknown Quantity

Example 2

Of a group of elephants, half and one-third of the half went into a cave. One-sixth and one-third of one-sixth were drinking water from the river. One-eighth and one-third of one-eighth were sporting in a pond full of lotuses; The leading of elephants was leading three female elephants. In this situation, how many elephants were there in the flock?

Suppose the number is 1.

Comment: Then the last group of elephants is which is [T - +] part of the group.

Now the number in the brackets is 2 21 5 36 252

So the number of elephants 1/252 = 4x252 = [008 Ans]

Example 3

From a bunch of lotuses, 1/4 are offered to Lord Siva, 1/3 to Lord Visnu, 1/6 to the Sun, and 1/12 to the goddess. The remaining 6 were offered to the guru. Find quickly the number of lotuses in the bunch.

Lilavati of Bhaskaracarya

Comment: Suppose the total number of lotuses is 1. Then the number of lotuses left is

1 - (4) - 1 - 20 + 12 + 0 + 15 = 1 - 57 = 1 - 12

So the total number of lotuses is 6 x [ = 120

1/20

Example 4

In 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 pearls in the necklace.

Comment: Suppose the total number of pearls is 1. The final number is 6.

Remaining pearls = 3 + 8 + 10

20 + 12 + 10 + 6

T - 48 = 1 - 4 = 1

the total number of pearls = 6 x [ 30.

1/5

From this example it appears that amorous problems were not taboo.

Example 5

434n: qredilg

To Find an Unknown Quantity

0 friend! One-sixth of the bees in a colony entered patali 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 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 remaining one bee is which represents the

	1	+	5	30
10	+	20	+	15	+	12	+	2
	60
		59
	60	60

So the number of bees in the colony is 1/60 60.

Example 6

A pilgrim carried a certain amount of money. He gave away half the amount to Brahmins at Prayaga. He spent two-ninths of the remaining amount in Kashi. One-fourth of the remainder was paid as duty. He then spent 6/10th part of the remainder in Gaya. Finally, he returned home with 63 niskas: If you know the fractional residues, find the amount he carried.

Comment: Suppose he had 1 niska to begin with. He spent 2 in Prayaga and so is left.

In Kashi he spent 2 2 1

and he was left with 1 1 18

Lilavati of Bhaskaracarya

Dutypaid 18 72

and so he has 7 72 3

Amount spent in Gaya 16 40

When he left Gaya he had 1 60

By the method of fractional residues, he had 63 9 x 60 = 540 7 /60

Example 6A

A lover gave his fiancee some jewels for making ornaments. She used one-eighth of them for an ornament for the forehead. She used 3/7th of the remaining for a necklace. Half of the remainder were used to make armlets. 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: The jewels were used for an ornament for the forehead.

So remained.

Those used for the necklace 3=3

Balance 3 2

Armlets used 2 2

Balance was

To Find an Unknown Quantity

Belt needed: 3 3

Balance was: 3

th of the total jewels: 16 jewels; and so the total number of jewels were 16 = 256.

Example 7

deer-eyed one! there was a colony of bees. One-lilth went to the kadamba tree, one-third to silindhra tree. x3 were moving around kutaja tree. One bee was attracted by the fragrance of ketaki and 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 and the bee was unable to choose. Consequently, the bee went once towards ketaki and then towards malati and vice-versa like a lover being called simultaneously by two messengers of his two beloveds.

Suppose the number to be found is 1. Then:

1 - ( = 3) - (3 - H) * [ 1 - 8 6 1u part is + part is [, i.e. 15 15 15

So the number of bees = 15.

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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.

Let the two numbers be a and b. We are given c = a + b, and d = a - b. Clearly, a = (c + d)/2 and b = (c - d)/2.

My dear child! If you know the method of transition, find two numbers whose sum is 101 and whose difference is 25.

By the above method:

a = (101 + 25)/2 = 63; b = (101 - 25)/2 = 38

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CHAPTER 19 Square Transition

CASE - [

If the difference of the squares of two numbers is divided by their difference, we get their sum. Then using the method of transition, we get the numbers:

In Algebra we have a² - b² = (a + b)(a - b)

a² - b² = a + b.

a - b

Example

0 mathematician! tell me two numbers whose difference is 8 and the difference of whose squares is 400.

66 Lilavati of Bhiskaracarya

Comment: 400 50

a + b = 50, a - b = &

a = 50 + 8 = 29, b = 50 - 8 = 2

CASE-II

LXVI

The second line of slanza (LXVI) poses the problem: Find two numbers a, b such that a2 + b2 = 1 and b2 are both squares (a < b).

First Method (LIV):

Take any number X (say). Compute a = (8x - 1)/2x.

Then compute b = a/2 + 1.

Second Method (1st line of LIVI):

Take a number x. Then the two numbers are 1/2x + x and 1.

Comment: 1st Method: Of the two methods, the first one is rather tedious. Algebraic explanation is as follows:

a = 8x2

b = + 1

Now a2 + b2 = 1 = a2 + + a + 1 - 1 = a2 + 2a.

(a2 + 8)

(8x2 - 1)2 = 64x - 16x2 + 1 + 32x

16x

(8x - 1)2 (8x + 1)2 = 64x2 - 1.

64x

8x2

Square Transition

Similarly,

b2 _ a _ 1 = a + a + l - a7 ] = (a)

Znd Mehtod:

If x is any number; then

1 = 1 + 2x2 b = [

a + b _ [ = (1 + 2x2 + I - 1 = 1 + 21?

4x 2x

a2 _ b7 _ [ = 1 + 42 + 4x' 8x2 [-2x

The second method is easier than the first.

If we take x = 4, we get a = b = 57

b7 _ 1 = 63 6} _ a _ [ = 49 Readers can check with x = 2, 3, 4.

(ii) x = 1, a = 2 b = 1,

b67 _ 1 = a67 _ 1 =

NOTE: We can extend the method to get b2 _ & = d2 and a2 b7 = e2

Another Derivation of the First Method:

Let the two numbers be b and a + 1, (a + 1) _ 62 _ 1 = a7 + 2a _ b2 and so it is a square if we choose 2a b2

Now (a + 1)2 + b2 _ [ = a2 + 2a b2 must also be a square,

i.e., by choosing a = b2 6 1 262 b2 (62 + 8) must be a square.

It is sufficient that b2 + 8 is a square. So b = 2t _ for some t, since

b2 + 8 = 4 _ 4 + + 8 = 21 + 4)_ Now put t = 2x_

Lilavati of Bhaskaracarya

Then b = 4x - 2. 81? _ [ ad a + [ = b? 1ae thc two numbers.

Example

LXVI

Find two numbers such that whether it is subtracted from the sum of their squares or the difference of their squares, the result is a perfect square. O friend! even intelligent mathematicians who know the six algebraic methods fumble in trying to solve this problem:

Comment: By the First method, choosing we have

a = X8 _ [=] and b = +1=3

2 *

By the second method, choosing x = 1,

a = x2 + 4 = and b = 4.

Example

LXVII

Choose any number x. Then a = 8x2 + 1 and b = 8x are the two numbers.

Comment: b2 - 1 = (8x2 - 1)2 = 64x4 - 64x2 + 1

(2x2 - 1)2

a2 - b2 = [4x2 (2x2 - 1)]

Square Transition

A derivation of this method is given below:

Suppose the two numbers are a + [ and b. Then

(a + 1) + b² = a² + 2a + b² and

(a + 1) - b² - [ = a² - 2a - b² must both be perfect squares. So we put

2a and b² = 2an which makes them squares. Thus b² = 2an = n

and n = 4x² Then b² = 64x⁶, b = 8x In the same way, a = 8x

and a + [ = 8x + [.

LXIXI

Some find algebraic rules as difficult as the arithmetic ones. However; many would not find any difficulty in the above six or other rules.

LXX

In arithmetic, the rule of three is the essence whereas in algebra, clear intellect is necessary. Many mathematicians may grasp easily but others would require explanations.

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CHAPTER 20 Quadratic Equation

Suppose we are given x # bx and we are required to find x. Add to the above b and find the square root of the result: Next subtract or add (as the case may be) b and square the result. This is x2.

If we are given x2 * 1 bx, then multiply by and proceed as before.

Tal

Tr 7u1 Tarjg

Lilavati of Bhaskaracarya

(LXXI)

This is the standard method in quadratic equations x² + bx = c is given: Then x = -bx + b² = c + b² Suppose c + b² = p

Then x = b = p and x² = p

In current practice we take ax² + bx + c = 0. Bhaskaracarya takes a = 1 and -c > 0. He also takes only positive square roots, whereas we take √16 = +4. So in the above working x > 0, p > b > 0. Also instead of solving for X, Bhaskaracarya solves it for x².

(LXXII)

Here the equation is x² + bx + c. When simplified this becomes x = bx and then (LXXI) can be used.

Example 1

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 amorous pair of swans was playing in water. How many swans were there?

Comment: The quadratic equation is x² - 2x = 2

	2	X + 49
	2 + 49	81
16		16
(-4)	(-2)

Quadratic Equation

Example 2

0 learned one, find the square number x? such that x2 + 9x = 1240.

Comment: We have to solve x2 + 9x = 1240.

9x + 81 = 1240 + 81 = 5041

x = 71, 2 = 31

= 961

Example 3

A flock of swans contained x members. As the clouds gathered, 10x went to Manasa Lake, and flew away to a garden of Hibiscus Mutalis: The remaining three amorous 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 x + 10x = 8, x = 6,

10x = 6

8x = 48

Lilavati Bhaskaracwya

80x (40)? 48 (40)2 1936

40 49 49 49

12 and x? = 144

Example 4

Arjuna became furious in the war and, in order to kill Karna, picked up some arrows (say x). With half of the arrows (Vz x"), he destroyed all of Karna's arrows. He killed all of Karna'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 Karna. Finally, he cut off Karna's head with another arrow. How many arrows did Arjuna discharge?

Comment: Here the equation is x² - 4x = 6 + 3 + 1 10

i.e. x² - 8x = 20

8x + 16 = 20 + 16 = 36 = 62

x = 1.1 = 100,

Example 5

From a group of black bees (2x²), the square root of the half went to the malati tree. Again, 8th of the bees went to the malati tree. Of the remaining (WO), one got caught in a lotus whose fragrance captivated him; he started wailing and his beloved responded. Then, O beloved, how many bees were there?

Quadratic Equation

Comment: Here the equation is 2x + 8(2x2) = 2, i.e. 2x2 = 2

2x = 9

3 81 = 9 81 225

2 16 16 16

15 '-(

X = 6 2x = 72

Example 6

To a certain number (x), the [8 times of the square root of the number (viz, 8x) is added. When one-third of the number viz: 42 is added to this sum, the result is 1200. If you know arithmetic well, tell the number.

Comment: Here the equation is x + 18x = 1200

54x = 3600

2X = 900

27 729 900 + 729

2 16 16

15129 42)

X = 123 41 - 24

= 576

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CHAPTER 21 The Rule of Three

There are three quantities involved herein: The first one on the left (a) is called pramana (scale), the second (b) phala (result or fruit) and the third (c) iccha (desire or requisition). The answer to be found (d) is called iccha-phala (desired result). Here a and c must be of the same kind and b should be different from a and c. The formula is d = b x c which is of the same kind as b. In a reverse case, the inverted rule of three is applied.

Lilavati of Bhaskaracarya

Comment: 2 = € in case of direct variation. and (ii) ab cd in case of inverse variation. So in the first case d = b xc and in the second case, d =axb

If, in 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.

Example 1

If 22 (Pala) of saffron costs N(niskas) you expert businessman, tell me quickly what quantity of saffron can be bought for 9 niskas.

Comment: 5 guijas = 1 masa and 64 masas = pala: So pala = 320 guijas 22 pala 800 guijas: The rule of three gives 9*2 105 palas: This is direct proportion, since more money buys more saffron.

NOTE: Here kumkuma means saffron and not the red powder that is applied to forehead, as the late Khanapurkar Shastri takes it to mean. In 1937, (Wo rupees bought 2 tolas of saffron. The red powder is cheap even now.

The Rule of Three

Example 2

63 palas of purified camphor is sold in the market for 104 niskas. Friend! tell me, after giving a thought, what will 12 palas fetch.

Comment: This is an example of direct proportion.

63 P 104 N : : 12 P : d

So d = 49 x 104 / 63 = 82.292 N,

So the answer is 20 niskas; 3 drammas; 8 panas; 3 kakinis and 119 varalakas.

Example 3

If 8 kharikas of rice can be bought for 2 drammas; then how much rice can be bought for 70 panas?

Comment: This is also an example of direct proportion.

32 P : 8 K : 70 P : d

d = 70 x 9 x 32 / 8 = 315 / 128 = 2.59 K

(Here P is for pana 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.

NO_CONTENT_HERE

CHAPTER 22 Inverse Proportion

If, in a given situation, the desired result (d) decreases (resp. increases) as the requisition (c) increases (resp. decreases) then the inverted rule of three is applied by an adept mathematician.

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, the 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.

Lilavati of Bhaskaracarya

Example 1

If the price of a sixteen-year-old female slave is 32 niskas; find the price of a twenty-year-old one. An ox which has been yoked for two years costs 4 niskas: what will an ox which has been yoked for six years cost?

Comment: (i) Price decreases as the age increases.

So d = 32 x 16 / 20 = 25.2 N.

(ii) Here the price is 4 x 2 / 4 = 1 N.

Example 2

One 'gadyanaka (=48 'guijas') of 10 carat can be bought for 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 x 48 / 15 = 32 guijas.

Example 3

With a measure of 7A (adhakas), 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 is 100 x 7 / 5 = 140 units.

CHAPTER 23 The Rule of Five

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 [

There are three problems in this.

Lilavati of Bhaskaracarya

(1) If 100 niskas (= N) fetch 5 N interest per month, find the interest on 16 N for one year.

Interest = 16 x 12 x 5 = 48

Interest = 100 x 1 = 9-N

In a tabular form, the rule of five is:

100 N Principal	16 N Principal	5 N interest
1 month	12 months

(2) The above problem is altered: at the same rate as in (1), find the period to fetch interest on 16 N.

Let the period be X months

100 N: 16 N 1M inverse

5N: 48N direct

In the above, the first proportion is inverse since if the principal is less, the period is longer to gain the same interest.

So X = 100 x 1 = 48 = 4 x 3 = 12 months

16 x 5

(3) Suppose we are given the period and interest and we have to find the principal (x). Here

SN: 48N 100 N direct

5 inverse

1M: 12 MJ

X = 100 x 48 = 16 N

5 x 12

Example 2

If the interest on 100 for months is 55 what will be the interest on 62 1 for months?

2 5

Comment: The rule of five is:

The Rule of Five

85

100	62	2	26	direct
	1M:¹⁶N	5	direct
5			100*4= 3=7;	5
26	x125	x 16		5
	2	5		5

Example 3

For 100 N (niskas) 8 pieces of multi-coloured embroidered (clothing) material, each measuring 3 cubits x 8 cubits, are available. O businessman, if you are proficient in business, tell me quickly the price of piece 3 2 cubits cubit.

		3 cubits	cubit
8 cubits:	cubits	100 N	x	direct
		X=100 X2X	X3x8x8
		175N	192

Example 4

fqoa

9 & XCM |/

Lilavati of Bhaskaracarya

30 planks of wood, each measuring 14 cubits 16 fingers 12 fingers cost 100 N (niskas). Find the cost of 14 such planks whose dimensions are each 4 measures less than the former.

Example on the rule of nine:

14 cubits	10 cubits	16 fingers	12 fingers
12 fingers	& fingers	30	14
10 x [2x8x 14x10O	50=162N
14xl6xl2x 30x [	3

Example 5

If & drammas are transport charges for carrying the planks, of the previous example, through a distance of kosa, find out the charges for transporting the second set of planks (measuring units less in each dimension) through a distance of 6 kosas.

Comment: This is to be solved by the rule of eleven.

14 cubits	10 cubits	16 fingers	12 fingers
12 fingers	& fingers	8 D		All are direct proportions
30	14	1K	6K
X=10x 12x8x 14x6x8=8 drammas:	14xl6x12x30x [

CHAPTER 24 Rules for Barter

In different kinds of goods are to be exchanged, use the rules of three, explained earlier. However, the price and the quantity of goods are in inverse proportion.

Example

In a market, 300 mangoes can be purchased for 1 dramma 16 paise. However, 30 pomegranates of good quality are available for paisa. Find out quickly how many pomegranates can be exchanged for 10 mangoes.

Comment: First Method: As per Bhaskaracarya's method.

88 Lilavati of Bhaskaracaya

Second Method:

First we determine the number of pomegranates equivalent to 300 mangoes:

300 mangoes	10 mangoes	30	direct
16 Paisc	Paisa		inverse

X= [0 x 30 x [6 16 Pomegranates

300 X [

300 mangoes = 16 * 30 = 480

300 mangoes = 480 pomegranates

10 = 16 pomegranates.

10 mangoes = 480 X

300

CHAPTER 25 Simple Interest

(To compute simple interest and principal.) Multiply the standard principal (100) by the standard period (1 month or 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: amount; P = principal, I = interest; R = rate of interest; Y = Period. Po = standard principal (usually 100) Yo = standard period (1 year or month)

P = A x Po * Yo

Po Yo + RY

Lilavati of Bhaskaracarya

According to the present method, first calculate the interest on Rupees (Rs.) 100 for 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 interest.

Example

When the interest rate is 5% per month, the amount after one year is 1000 N (niskas). Find the principal and the interest.

P = 1000 x 100 x [ 625 N

100 x 1 + 12 x 5

[ = 1000 x 12 x 5 = 375 N

100 x 1 + 12 x 5

We could have got I = A - P = 1000 - 625 = 375 N. The interest 375 N on 625 N for one year appears high.

Additional Note

(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 principal and divided by the sum of the quotients written separately are the desired parts of the given principal.

Comment: Using the notation of the previous example with suffixes to denote the parts (we consider three parts):

Simple Interest

(P1 + P2 + P3) x 100

(P1 + P2 + P3) x R1 Y1 etc.

Interest in each case is:

P1 R1 Y1 > P2 R2 Y2

100

R1 Y1 R2 Y2 R3 Y3

I from which we get the formulae for P1, P2, P3.

Problem

94 N (niskas) 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.

Comment: Here P1 + P2 + P3 = 94

R1 Y1	35	R2 Y2	30	R3 Y3	20
1	+	35	1	20
R1 Y1	R2 Y2	R3 Y3	30	420

P1 = 9 420 x 1 = 24, P2 = 94 x 420 x 28

47 35

P3 = 94 x 420 x 1 = 42.

47 20

Profit Sharing

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 multiplied by the total output and divided by the total investment.

92 Lilavati of Bhaskaracarya

Comment: Il a, b, c ae investments and is the output, then the shares are ax/(a + b + c), bxl(a + b + c), cx/(a + b + c) respectively.

Three grocers invested 51, 68, 85 N (niskas) respectively. Skillfully they increased their total assets to 300 N. Find the share of each.

Comment: Total investment = 51 + 68 + 85 = 204 N. Their shares are:

Investment (N)	Share (N)
51	75
68	100
85	125

Profits are 24 N, 32 N, 40 N:

We have now a formula concerning filling up of reservoirs (pools, lakes, tanks).

CII

One divided by the sum of the reciprocals (of the times taken by the sources to fill up a pool) is the time of filling the pool when the sources are used simultaneously.

Comment: Suppose the sources take times (1, t2, ...) to fill up a reservoir. If they are simultaneously used then the time taken is:

We use the same method at present; Of course at Bhaskaracarya's time there were no taps.

Example

R geryeria TI:

ad 4 fjhtt:

Simple Interest

Four streams flow into a pool and separately they take 2, 3, 6 days respectively. If all the four are used simultaneously, find the time required to fill up the pool.

As explained in the previous stanza:

Time = 1 + 2 + 3 + 6 = 12th day.

The four streams can fill up 1, 2, 3, 6 pools in one day and so together they can fill up 12 pools in one day. So time for one day = 12 pools.

Formula concerning parts of grain

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).

For the ith grain, suppose, gi measures cost d; and we need quantity %i. Then for money worth d units,

cost of j5 grain = Bi

d1 * g1 + d2 * q2 = 81 * g2

Lilavati of Bhaskaracarya

Now to find the quantity of grains in the mixture substitute the proportion of the grains for d9, in the above formula:

Example 1

3 units of rice and 2 units of mung beans can be bought for one dramma. A grocer, I have 13 kakinis and want rice and mung beans (a kind of kidney beans) in the proportion 2:1. Quickly give me the grains so that we can cook, eat and start on our next journey by caravan.

Cost of rice: 1L2 2X13 dramma

Cost of mung beans = (7/2) 2 X 1

Cost of mung beans = (1 x 8) - (7 + 4 - 192)

Quantity of rice = (2 x 4) - (+4) - 3 measures

Quantity of mung beans = (1 x 9) - (+) measures mung

We can prove this formula by Algebra. Let x be the measure of beans. Then rice = 2x measures.

Cost of mung beans: 8 drammas

Cost of rice: 4 drammas

So the total cost: 8 + 4x + 39X drammas = 56

Simple Interest

which is given as 13 dramma:

64	So 39 x 13 1= 24 mcasurcs	56	64

Example 2

Good quality camphor costs 2 N (niskas) for P (pala), sandalewood costs D (drammas) for P and aloe costs D for P. You grocer's son (who pleases his mother) what quantity of each is available in the proportion 1:16.8 for N?

Comment: Noting that [ N= 16 D we get

cost of camphor	32 X [ 32 x 1 x 16	8X16 + 8x8x2
cost of sandalewood	XI6x [6 8 D	36
cost of aloe	x8x2x16 8 D	36
quantity of camphor	142 4P	32 = 3 + 8-&
quantity of sandalewood	P
quantity of aloe	8 4-9P_

We can work out this problem by algebra. Suppose camphor is x P. Then sandalewood = I6x P, aloe &x P_

Lilavati of Bhaskaracarya

Total cost = 32x + 16x + 8x = 36x = 16 D

X = P

Formula for Exchange of Jewels

(Suppose n persons possess a1, a2, ..., an jewels respectively. Suppose each one gives q jewels to each of the others and then their jewels are worth the same. From the numbers of jewels, subtract the product of the number of persons and the number of jewels donated (uniformly) to each of them. An assumed number (which could be the L.CM. of the remainders or any multiple of the LCM) divided by the remainders will give the prices of the corresponding jewels.)

Compute ai nq the price of a jewel with the person is nq where x is the supposed worth of each. It can also be found by IT-, ( ; 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) | i < n} is divided by ai nq, the prices will be integral.

Example (Exchange of Jewels)

Four merchants had 8 rubies, 10 sapphires, 100 pearls, and 5 diamonds respectively. They became friends on their journey and each one gave...

Simple Interest

To each of the others one jewel from his lot. This made their jewels worth the same prices of these jewels.

We give three methods.

Bhaskaracarya's: Jewels 8, 10, 100, 5. Number of gifts = 1, number of persons n = 4. So the remainders are:

8 - 1 * 4 = 4
10 - 4 * 1 = 6
100 - 4 * 1 = 96
5 - 4 * 1 = 1

If we suppose x = 96, the prices of jewels are:

ruby: 96 - 24
sapphire: 96 - 16
diamond: 96

Here 96 is the L.C.M. of the remainders. If we follow the second method we get the product of the remainders = 4 * 6 * 96 * 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 types of jewels. 2nd merchant had 6 S and one each of the four types of jewels. 3rd merchant had 96 P and one each of the four types of jewels. 4th merchant had 1 D and one each of the four types of jewels. Since their worth is the same, we get 4 R = 6 S = 96 P = 1 D. We choose the price of pearl to be 1 Rupee and we get the answer as before. 3. Algebraic: If the prices of the jewels are a, b, c, d respectively, the equations are:

5a + b + c + d = a + 7b + c + d = a + b + 9c + d = a + b + c + 2d

4a = 6b = 96c = d. If we choose c = 1 Rs, we get the same answer as before.

Formula to find the Weight of Pure Gold

(To find the weight and fineness of gold if two or more kinds of gold are melted together) The fineness of the mixture equals the sum of the products of weight and fineness of the constituents, divided by the total weight. If the same sum is divided by the weight of pure gold,

Bhaskaracarya

Lilavati of

one gets the fineness of pure gold and if it is divided by the fineness, the result will be the weight of pure gold,

Comment: If the fineness of the mixture is X and weight y then xy

Note that y = Y1 + ... + Yn:

Example

golden mathematician, four types of gold 10 M (masas) of 13 C (carats), 4 M of 12 C, 2 M of 11 C and 4 M of 10 C are melted together to form a new one. Find its fineness. If this is purified and 16 M gold is obtained, what is its fineness? If the mixed gold when purified has 16 C fineness, what is its weight?

Comment: Sum of the products

13 * 10 + 12 * 4 + 11 * 2 + 10 * 4

130 + 48 + 22 + 40 = 240.

Fineness of the mixture = 240 / 20 = 12 C

which naturally lies between 10 C and 13 C.

Fineness of the purified gold = 240 / 16 = 15 C

If the fineness is 16 C, weight = 240 / 16 = 15 M.

Here 5 M of impurities were burnt out.

Formula: From Mixture to Component

Multiply the weight of the mixture and its fineness: Subtract from this, the sum of the products of fineness and the corresponding weights of

Simple Interest The given components. The remainder divided by the weight of the component (whose fineness is not known) gives the desired fineness.

Comment: (Using the notation of the previous stanza):

xy XiY ,

XJ Yj

Example

8 M (masas) of 10 C (carats), 2 M of 1 C, 6 M of unknown C are mixed together to form 16 M of 12 C. What is the fineness of the unknown?

Comment: Here x = I6x12 - 8x10 - 2x1

192 80 _

22 15C _

Multiply the fineness of the mixtures by the sum of the known weights of the components of the mixture. Obtain the difference of this and the sum of the products of fineness and the corresponding weights of the components. The outcome divided by the difference of fineness of the component (whose weight is not known) and that of the mixture gives the desired weight:

[In the above notation, Y, (y, X, - X

Comment: x(Y1 + Y2 + Y3) = XiY1 + X2Y2 + X3Y3:

Lilavati of Bhaskaracarya

3 M (masas) of 10 C (carats), M of 14 C are mixed with some quantity of 16 C. If the fineness of the mixture is 12 C, find the weight of 16 C gold.

By Bhaskaracarya's method

Y3 = 12 (3 + D) - 10 x 3 - 14 x 1 = 48 - 44 = 4.

16 - 12

Weights of Components

(If fineness of a mixture and its two components are given, to find weights (indeed the ratio of the weights) of the components) Obtain: the difference of the fineness of the mixture and that of the component with greater fineness; and the difference of the fineness of the mixture and that of the component with smaller fineness. The two differences multiplied by an assumed number will give weights of the components.

Suppose Xp > X2:

Y1 = X - X2

Y2 = Xp - X

Y = X - 42 (Y1 + Y2)

X - X2

Comment: x(Y1 + Y2) = X1Y1 + X2Y2:

Example

Friend, two small balls of gold of 16 C and 10 C were melted together to form a mixture of 12 C. Find out the weights of the balls.

Comment: If the mixture weighs y M (masas), then

12 - 10 = Y

16 - 12 = 2y

Y1 = Y x 16 - 10 3, Y2 = y x 16 - 10

So the proportion is 1.2.

CHAPTER 26 Combinations

Starting with the number n write down n, (n - 1), (n - 2), Divide them by 1, 2, 3, to get n - 1, n - 2. Then the number of combinations of things taken 1, 2, 3 at a time are:

n (n - 1) n (n - 2) respectively. Or the number of combinations of n things taken r at a time are:

[n (n - 1) (n - 2) (n - r + 1)] + [1 x 2 * 3 * ... * xr]

This can be used to solve the problem when r = 1, 2, 3.

Lilavati of Bhaskaracwya

This is useful in prosody to discover all possible meters, in architecture, medical sciences, chemical compositions etc. I am omitting these (applications) for the sake of brevity.

Modern notation is n (n - 4) (n - r + 1). The first line of the stanza is not clear, as no mention is made regarding starting number in the product.

Example 1

Friend, there are six letters in the fourth line of gavatri meter. If we choose only 1 g ('guru' means long vowel in Prosody), how many meters are possible?

Comment: If the fourth line contains six 'laghu' (short vowel), then the only combination is eeeeee. If we have one g and five l, then the combinations are:

(i) egeeee
(ii) eegeee
(iii) eeegee
(iv) eeegee
(v) eegeee
(vi) eeeeeg

As seen above, if there is 1 g, the number of meters = 6.

For 2 g, the calculation is: 6 x 5 = 15

For 3 g, the calculation is: 6 x 5 x 4 = 20

For 4 g, the calculation is: 6

For 5 g, the calculation is: 6

For 6 g, the calculation is: 6

Khandmeru is Pascal's triangle.

Combinations

Also (8) + (9) = + (#) = 2"

If we take the combinations of all the four lines as above, the total number of combinations will be 64 * 64 * 64 * 64 = 16777216.

Example 2

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, 1 door, 2 doors, 3 doors, are opened, how many different types of breeze arrangements are possible?

How many kinds of relishes can be made by using 1, 2, 3, 4, 5 or 6 types from a sweet, bitter, astringent, sour, salty, hot substances?

Comment: As per the above formula, the possibilities are respectively. These are 1, 8, 28, 56, 70, 56, 28, &.

Total number = 28 + 256 + 2 + 42 + 30 + 20

The number of different relishes (chutneys) = 64 = 26

Pascal's Triangle

Bhaskaracarya has referred to 'khandameru' which is known as Pascal's triangle. This was known to ancient Indian mathematicians. Pingalicarya 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 Lilavati of Bhaskaracarya

Pascal's triangle

	(x + y)	(x + y')	(x +	(x + y)	(x + y)'	(x + y)S
10	10		6	15	20	15
	Y)G

For example, (x + y)' = x"' + 4xy + 6xy" + 4xy + y

Also note 5 = 1 + 4, 10 = 4 + 6, 10 = 6 + 4, 5 = 4 + 4 in the 6th line.

This is based on the formula (2) + ( = , ) - ( "; " ) : In Bhaskara-carya's times, the above was written like a 'meru' (mountain).

CHAPTER 27 Progressions (Series)

MI 0947 427 @A5 Rnq Faal MI CXXII

1 + 2 + 3 + ... + n = nn + l

1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + ... + n)

n (n + 4) (n + 2)

Example

Find 1 + 2 + ... + 9 and 1 + (1 + 2) + ... + (1 + 2 + ... + 9).

Comment: As per (CXXIII) the answers are 9 x 10 = 45 and 9 x 10 x 11 = 165 respectively.

105

106 Lilavati of Bhaskaracarya

In current notation we write i (i + )

The first one is an Arithmetic Progression, and is a special case of a + (1 d) + (a + 2d) + ... + (n - 1)d = n/2 (2a + (n - 1)d).

Formulae for En can be established by Mathematical induction and are found in current books;

6754gyai M CXXV/ !

12 + 22 + ... + n² = (n + 4)(2n + 4)

13 + 23 + ... + n³ = n(n + 4)

Arithmetic Progression

If the first term is a and the common difference (CD) is d, the n-th term of the A.P. is given by € = a + (n - 1)d. The sum of the n terms is given by S = n/2 (a + €). Here a + € is the middle term.

Example 1

M CXXVI

Tell me the sums of 12 + ... + 9²

1 + ... + 92 = 9 (40)(192) = 285

and 13 + ... + 93 = 9 x 10 = 2025

Progressions (Series)

Example 2

If the first term of an AP. 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: e = a + (n - 1)d = 7 + 7 * 5 = 42 and middle term = 7 + 42 = 42 which is not a term in the A.P.

Sum = 8 * 49 = 196.

Example 3

A gentleman gave 4 D (drammas) as charity (to 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.

Comment: Here a = 4, d = 5, n = 15.

Last term e = 4 + 14(5) = 74.

Middle term m = a + e = 4 + 74 = 39.

Total charity = 15 x 39 = 585 D.

Formula to find the first term of an AP

To find the first term of an A.P, divide the given sum by the number of terms. Multiply the number of terms minus one by half the common difference, and subtract this result from the quotient already obtained. [Thus a = S / n - (n - 1)d / 2]

Lilavati of Bhaskaracarya

Example

son, find the mouth (the first term a) of an A.P. whose S = 105, n = 7, d = 3 -

Comment: a = 105 6x215_9 = 6

Formula to find CD in an AP

Divide the sum by the number of terms (of an A.P.) and subtract the first term from this quotient. This result divided by half of the number of terms minus one is the common difference. [Thus d = (n - 1)/2]

Example

To capture enemy elephants, king covers 2 Y (yojanas) on the first day and then increases his distance by A.P. on subsequent days. If he travels 80 Y in 7 days, intelligent boy, find out the extra distance each day.

Comment: 80 _ 2 3-2-4y. d = - 66 x

Progressions (Series)

109 Formula to find the number of terms of an A.P.

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 CD. Then find (the positive) square root of this result. Subtract the first term from and add half the CD to this square root. The resultant divided by the CD is the number of terms (in the AP).

Thus,

n = 2Sd + (a d) - (a

Comment: S = 2 [2a + (n - 1)d]

dn + n (2a - d) - 2S = 0.

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.

Example

A donor gave 3 D (drammas) in charity to a Brahmin on the first day. He continued increasing his donation each day by 2 D. If the total amount paid by him equals 360 D, how many days did he give in charity?

Comment:

n = 2x360x2 + 3 - 6 - 2

V1440 + 4 - 2 = 18

110 Lilavati of Bhaskaracarya

Geometric Progression (G.P.)

If n, the number of terms in a GP, is odd, then (n - 1) is called 'multiplier' (M) and if it is even; it is called 'square' (S) [Bhaskaracarya's terminology]. Now beginning with n, continue this process (n - 1) for odd and (n / 2) for even until 0 is reached. Then keep the common ratio (CR) against 0 and start writing M or S in the opposite direction. Carry out the operations and then subtract from the final result and divide it by (r - 1). The result is the required sum.

Comment: Suppose a = 1, n = 31, r = 2. Then following the above method, we get the table:

31	30	15	14	2
M	S	M	S	M

Beginning with 31 we write 31 - 1 = 30, 30 - 1 = 15, 15 - 1 = 14, 14 - 1 = 2 until we reach 0. Then in the second line we begin with M and alternately write M and S.

Now we perform the indicated operations: M = multiplier, S = square as shown below.

Index	To reach the 315th power; here Bhaskaracarya has given shorter method by squaring at even numbers and odd using
2147483648	31
1073741824	30
32768	15
16384	14
138	6
8
2

In practice, multiplication might be easier than squaring:

Progressions (Series)

As is given in standard texts:

Lilavati of Bhaskaracarya

[In every metre there are four quarters. If the order of short and long vowels is the same in each of the four quarters, the metre is known as 'even' (samavrtta).] The following formula gives the number of even metres. Take as n the number of letters in each quarter and r = 2 (for there are only two kinds of vowels, short and long). The total number of possible metres = 2n.

[If the first and the third quarters are similar and so are the second and the fourth, it is called 'semi-even' (ardhasamavntta).] In this case the total number of possible metres = 22n 2r.

[If the number of letters is the same but all the four quarters are different, it is called 'odd metre' (visamavrtta).] In this case the total number = (2n)4 - (2n)2.

Comment: Suppose there are n letters in each line. Each letter can be short or long and so there are (2n) choices. According to the fundamental theorem there are 2n possibilities in the even metre.

There are 2n x 2n possibilities in the first two (as well as 3rd and 4th). From these we subtract the even case 2n to get the number of semi-even cases: (2n)2 - 2n = 2n (2n - 1).

In the odd metre case, there are 2n x 2n x 2n possibilities from which we subtract the first (2n) cases; 2n + 2n (2n - 1). So the number (2n)4 - [2n + 22n - 2n] = (2n)4 - 2n - 22n.

Example

Find the number of even, semi-even and odd metres in 'anustupa' metre.

Comment: There are 8 letters. So

number of even metres = 28 = 256

number of semi-even metres = (28)2 - 28 = 65280

number of odd metres = (28)4 - (28)2 = 4294901760.

CHAPTER 28 Mensuration

Measurement of Sides and Areas

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 (koti). The same terminology holds for a rectangle.

BC is the base.

AB is the altitude.

AC is the hypotenuse.

Lilavati of Bhaskaricarya

In right-angled triangle, square root of the sum of squares of the base and the altitude is the hypotenuse, and square root of the difference between squares of the hypotenuse and base (respectively altitude) is the altitude (resp. base) [Thus b = √(c² + a²), C = √(b² - a²), a = √(b² - c²)].

This is known as Pythagoras' Theorem. This was known in India since the time of Sulvasutrakaras (3000-800 B.C.) whereas Pythagoras published it in 560 B.C. Of course, no proofs were given in Bhaskaracarya's times. Proofs of this theorem are found in textbooks.

MI CXLII

If twice the product of two numbers is added to the square of their difference, the result is the sum of their squares. So also if the sum of two numbers is multiplied by their difference, the result is the difference of their squares [Thus a² + b² = (a - b)² + 2ab and a² - b² = (a + b)(a - b)].

Example 1

Find the hypotenuse if the base is 3 and altitude 4. If the hypotenuse and the base are 5, 3 what is the length of the altitude? If the hypotenuse and the altitude are 5, 4, what is the base?

3² + 4² = 5²

Comment: the hypotenuse = 5.

Mensuration

Example 2

In a right-angled triangle, the base is 3 and the altitude 3 find the hypotenuse.

Hypotenuse: 169

Comment: 43J2

1ad M CXLVII

[In this stanza Bhaskaracarya explains method of finding an approximate square root of a number which is not a perfect square.]

To compute (approximately) choose largest square number x. Then compute approximately abx and divide by bvx.

Comment:

Vabx

bVx

In the above Example 2, 169 4342 13 (L.414) = 13(0441.4) 400

476 approx_ 800 200

By Bhaskaracarya's method, choose X = 10000 = 100² and 169 169 x 8x10000 3677 477

8x8x 10000 800 800

Here we note that 36772 = 13520329 and 169x 8x 10000 = 13520000

Bhaskaracarya was aware of the fact that are irrational numbers and so gave a method to compute their values approximately. In his times, decimal fractions were not known.

Lilavati of Bhaskaracarya

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. Bhaskaracarya rationalizes the denominator.

Formula to find hypotenuse and altitude of a right-angled triangle given its base

base, b hypotenuse, altitude (see fig: of stanza CXLI)

Choose any convenient number y Then given to find b, c;

2ay b = ( +1) y

b=3 a2 (Ic-H('-))

OR: and

Comment: (1) Here b2 = ( +1)2 and ( -1)2 a2 + c2 = 0 4a2y = 2 + 4aly C+=b2

(2) In the second case +4 4aly + at 2aly + yt t)=b', a2 + c2 = a2 4y (a2 4y

Example

If the base of a right triangle is 12, find its altitude and hypotenuse in integers.

Mensuration

y _ 1 divides 2ay.

Comment: 1st Method. Choose y so that here a = 12 and we choose y = 2.

2ay = 2 x 12 x 2 = 48, b = a (y² + 1) = 12 (5²) = 60.

C = y² - 1 = 4 - 1 = 3.

2nd Method. Here y must divide a and we choose y = 6.

b = 12² = 144; c = 3 x 12² - 6 = 9.

6 15.

If we take y = 2 we get b = 37, c = 35.

NOTE: These are known as (Pythagorean) triplets. Several solutions 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 derived 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 Bhaskaracarya's times. There are well-known formulae to find the triplets:

[(2n + 1), (2n, (n² + 2n + 1)]

Zn),

and [2n, n, n + 1]; n².

Formula to find two sides when the hypotenuse is given:

Given hypotenuse b, choose a convenient number y. Then a = 2by + C = b (1 - y²) + 1.

Comment: a² + c² = b² [4y + 2y² + y] = b² + 1.

Lilavati of Bhaskaracarya

If the hypotenuse is 85 find the Other sides in integers

Comment: Here we choose y = 2, since y + 1 divides b.

a = 2 * 85 x 2 68. c = 85 x 3 = 5[ We could have chosen y = 4 and got a = 40, c = 75. By using (2n 1), (2n 2n), (2n? + 2n + 4) we find 2n 2n + [ = 85 n = 6 and we get (13, 84, 85).

Another formula when hypotenuse is given

srtiu 447 fju: tuse %:

a = b 2b	C = 2by
Y` + [

Comment: a = b(y2 _ V) C = 2by

y2 + ]

a? + c? = b2

These are the same (except for a sign) as in (CL), but a and c are interchanged.

Formula to construct right triangles with integral sides

Given any two numbers X, Y, there is a right triangle with a = 2xy, c = x_y (x > Y) and b = x? + y.

Comment: b2 = Y) = (-Y)? + 4xy = c + a

This formula can be used whether or not x, Y are integers.

Mensuration

Example

If all the three sides of a right triangle are unknown, friend, tell (mc) quickly their values:

Comment: Take any two unequal numbers, say x = 2, Y = 1. 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.

Suppose a bamboo BAC is broken at A and the part AC touches the ground at C. Given the length x of the bamboo and the distance BC = Y, find X₁, X₂, and X₃ by: X₁ = (1 X),

B	C	X = Z₀ + 2)

Comment: Here X₁ + X₂ = X and x₁ + y² = x². So x₁ - x₂ - y² = (-X₁)² - y² = x² - Z₁ X₁ + x₂ - y²

2X X₁ = X - y²

Then X₂ = X - X₁ + y

Example 1

424 tfag 2N: 49/1 CLVII I

Lilavati of Bhaskaracarya

A bamboo 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, O mathematician, tell (me) the height of the point where the bamboo broke.

Comment: Here X = 32, Y = 16.

32 16? (32 8) = 12.

32 2

Example 2

[Near the foot of the 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 BD = DC = S.

:{-Y)

T =

s = y - r

Comment: Here R = 8 _ x? = (y - r)? _ x?

Y 2yr + r SX

T = (y

Example

Ts9R

Mensuration

There was a snake at the foot of a pole, on the top of which was seated a domesticated peacock. The pole was 9 C (cubits) high. The peacock saw a snake crawl towards the post at a distance of 27 C from the post. The peacock pounced on the snake at the same speed as the snake's crawl and caught it at a certain distance from the pole. Find the distance quickly.

Here y = 27, x = 9

27 9² (27 3) = 12 C.

The peacock and the snake travelled equal distances of 15 C.

NOTE: Some modern 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 Bhaskaracuya was ignorant of gravity as the following verse (Goladhyaya Chapter Bhuvanakosa, Stanaza 6) shows:

That is, the earth attracts inert bodies in space towards 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 keeps down from palm; he purposefully pounces on the snake and his speed is 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 BD to be a straight line.

Formula to find height of lotus and depth of a pond

4a ada m&rq CLIX I

122 Lilavati of Bhaskaracwya

[Height of lotus flower is AC = AD When the lotus moves by wind and its top touches the water at D, BD = r: BC = y.]

Depth of water AB = Height of the lotus

base	AC = x = + Y	hypotenuse
0 friend, the distance between B, the point where the lotus stem meets the water surface and D where the tip touches the water level, is called the base. The height of the lotus above the water surface is AD	AB hypotenuse altitude. Altitude is like depth of water:
Comment:	Here (x + 7 = x ~ Y)?	2xy + y + r = x?
X = H{+)	#(+)	AB = X - y =

Example

In a lake there were a large number of ruddy geese and cranes. A lotus whose height above the water surface was One vita C (Cubit). and its tip, bent by a rustling wind, sank at a distance of 2 C. Mathematician! tell (me) quickly the depth of water:

Mensuration

Here y = I=2_

Depth of waler 45c.

Length of lotus stem = 4(1+4) - 4c.

A monkey jumps from point M on palmtree AM to D and then onto a well: B. To find MD.

MD + DB = r + s and MA + AB = c + x and r + s = c +

s² = x² + (e + r)²

M = x + 1 + 2cr and + s + 2rs = c² + x² + 2cx

Zcx 2cr - r = r + 2rs

Zcx 2cr = 2r 2r (c + x - r)

2r² + 2rc + Zrx 2r²

Zcx 4cr + 2rx

T = CX Height of Jump: 2c + x

Example

There was a palmtree 100 C (cubits) high and there was a well at a distance of 200 C from the tree. Two monkeys were on the top of the

Lilavati %f Bhaskaracarya

One of them came down the tree and walked to the well. The other one jumped up and then pounced on the well along the hypotenuse. If both covered equal distances, find the length of the second monkey's jump.

Comment: Jump 100 x 200 50 C.

2 x 100 + 200

An objection may be raised that the second monkey traverses a parabolic path: But as per the conditions on his movement, the path is to be taken as a straight line.

Formula: To solve a right triangle given its hypotenuse (b) and the sum of the other two (a + c).

To find the base and altitude when their sum and hypotenuse of a right triangle are given, subtract from twice the square of the hypotenuse, the square of the sum of the base and altitude. The square-root of this result subtracted from (resp. added to) the sum of the sides and divided by two happens to be the length of the base (resp. altitude).

Thus given b and a + c = $:

√(2b² - s²) = C = S + a and a = S =

Note: If the hypotenuse and the difference of two sides are given, then to find out the base (resp. altitude), this difference should be subtracted from (resp. added to) the square-root of (twice the square of the hypotenuse minus the square of the given difference) and divided by two; (see also the comment of stanza CLXVI).

Comment: b² = a² + c² = (a + c)² - 2ac = $ 2ac and (c - a) = (a + c)² - 4ac = $ + 2(b² - s²) = 2b² s².

C - a = √(2b²) and C + a = $.

Solving these equations, we have the formulae.

Mensuration

Bhaskaracarya calls the smaller a the base and larger c the altitude. This problem can also be solved by quadratic equations.

Example

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.

V2 x 172	232
Comment:	C = 23 + V2 x 172 = 15
and	a = 23 - V2 x 172 = 8

Example

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 = 7 and b = 13,
s + V2 b2	7 + V2 x 132 = 72 = 42
and	a = V2 b2 - s2 = 17 - 7 = 5.

NOTE: Khanapur Shastri used the same formula as in (CLXV) and got 5. We use a different formula: Although Bhaskaracarya knew about negative square root, he says that a base of a right triangle cannot be negative and this is accepted by modern mathematicians.

Lilavati of Bhaskaracarya

CLXVI

[There are (WO verlicalpolesAB, 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

Y = sr

S + r

BF = n FD = Sl.

S + r

Comment: ABEF, 4BCD arc similar X=BF

AEFD ad AABD are similar FD

Adding X X BF + FD=1 Y = sr Another proof can be given by co-ordinate geometry. Clearly is not needed to compute Y_

Example

Twopoleshave heights 15 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 = sr ISx10=6C

s + r 15+10

Mensuration

127 Existence of Triangles

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 thing is possible.

This is a well-known result in Euclidean Geometry. It is unfortunate that ancient Indian mathematicians did not perceive the need to offer proofs for their assertions.

Example

An idiot says that there is a triangle with sides 3, 6, 9, explain to him that they don't exist.

In triangle ABC, AD is perpendicular to BC. To find BD and DC:

In triangle (assuming side as base) form the product of the sum and difference of the remaining 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.

The 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,

Lilavati of Bhaskaracarya

BD = a + (c + b)(c - b)

DC = (c + b)(c - b)

p² = & = b² DC²

~BD²

Also Area AABC = p a:

Comment: These are given in school geometry. If C is obtuse, some modifications are necessary. See Example 2 below.

Example 1

If the base of a triangle is 4 and the other sides are 13, 15, find the length of the perpendicular on the base, the lengths of two parts of the base into which it is divided by the foot of the perpendicular and the area of the triangle.

Comment: BD = 1 14 + (28)

DC = 14 - (28)(22) = 5,

P = √(1s² - 9²) = 12

and AABC = 1/2 x 14 x 12 = 84 sq. units

Mensuration

Example 2

A triangle has sides 10, 17 and base 9. O mathematician, find its altitude, the two projections (on the base) and the area.

Comment: Here projection of AB BD = 9 + 27 * 7 | = 15

Projection of AC = CD

27 x 7 - 9 / - 6,

Vio? _6 = 8

p =

and area = V * 9 * 8 = 36 sq: units.

Here ACB is obtuse and so there is an appropriate modification.

Formulae

Area of (cyclic) quadrilateral and triangle

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 b, c, be the sides of the quadrilateral and a + b + c + d semi-perimeter: Imprecise area of a quadrilateral is

(s-a)(s-b)(s-c)(s-d) Area of triangle with sides 4, b, c is obtained by pulling d = 0, viz √(s-a)(s-b)(s-c)

Comment: The formula for the area of quadrilateral is exact for a cyclic quadrilateral. Formula for the area of a triangle was discovered by Heron (200 B.C.).

130 Lilavati of Bhaskaracarya

mer; as also pointed Out by Bhaskaracarya, is nOt correcl fr non- quadrilateral for which the correcl forula is

eyclie (s - a) (s - 6) (s - c) (s - d) abcd cos? P + F

This derivation can be found in a standard trigonometry book

Example 1

A quadrilateral has base 14, upper side 9 and (the other) (wo 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 rectangle and a right triangle.

So the area = 12 * 9 + 2 * 12 * 5 = 108 + 30 = 138.

Or using the primitive formula:

Area = √(s(s - b)(s - c)(s - d))

V(24 * 12) (24 - 9) (24 - 13) (24 - 14) = '19800 # 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.

Area = √(s(s - a)(s - b)(s - c)) = √(21 * 7 * 6 * 8) = 84

Mensuration

CLXXVII

The (primitive) formula for the area of a quadrilateral is not accurate. This is so because the lengths of its diagonals are indeterminate. 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.

CLXXVIII

Even if all the four sides of quadrilaterals are equal (i.e., a rhombus), diagonals can be different and consequently we get different areas. Comment: Nowadays 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.

I

If a perpendicular to 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.

CLXXXI

327 4|

CLXXXII

7s54d aqyufeum: | |

Lilavati %f Bhaskaracaya

Subtract the square of the 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 d1, d2]

(Given a and d1): d2 = √(4a2 - d12)
(Given d1, d2): Area = (d1 × d2) / 2

Area of a rectangle = product of its adjacent sides

Area of a trapezium = (sum of parallel sides) × height

Comment: These are standard formulae.

Example

In a rhombus of sides 25, find two diagonals and area:
If the diagonals are equal, what is its area?
Find the area of a rectangle if its base and vertical side are 6 and 8 respectively.

Comment: (1) Choose d1 = 30. Here a = 25.

d2 = √(4 × 252 - 302) = 40

Area = (30 × 40) / 2 = 600 sq. units.

Mensuration

If we choose d = 48 then dz = 14 and area is 336 sq units. Thus rhombus with given sides can have infinitely many pairs of diagonals and areas.

Area = 25 * 25 = 625 sq units
Area = 6 * 8 = 48 sq units

Example 1

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.

Area = (11 + 22) * 12 = 198 sq units

[Here Bhaskaracarya has given one more datum than is necessary. Given the four sides we can compute the altitude, as is clear from the figure.]

Example 2

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.

Lilavati of Bhaskaricarva

134

MC = 51

MD = 75

MC = 144

AM? 68" 1442 AM = 308

Area of the quadrilateral = AADC + AABC25

2310 + 924 = 3234 square units:

From the primitive formula 149 x 77 x 42 * 66 = 3234 which shows that the quadrilateral is cyclic. 40" = 7225

If we take C = 909, BD? 75? BD = 85.

Formula for a diagonal of a quadrilateral

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 = -AD? AM

CM = DC - DM

AC =

and Vmc? AM-

Formula to find the second diagonal

Mensuration

Sides of quadrilateral ABCD are given. Take some value for AC = 77.

Sides of AABC and AADC are known. Suppose BM = AC, DN = AC. Compute AM and MC. Similarly, altitudes can be computed. Thus we get AN, CN and DN in AACD. MN = DR.

BR = BM + MR = BM + DN.

Then BD² = BR² + DR².

Assume some convenient value for one diagonal. If we take this diagonal as the base, we get two triangles. Then we get perpendicular BM as well as AM, MC, ND, AN, NC.

BD² = BR² - RD² [BM + ND]² = MN².

Formula to find area of a quadrilateral

Area = ½ × (sum of the two smallest sides) × (perpendicular height)

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 perpendicular. So the length of the diagonal is chosen 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 51, 75, 40, 51, 68 but the fifth data is not given. Bhaskaracarya says that we can take the diagonal AC to have a length.

136 Lilavati of Bhaskaracarya

belween the base 75 and the altitude A'M of AA'B'C' where A'C' = 51 40 = 91, Obviously AC < 51 + 40 = 91. Now AM 66. So we can choose AC = 77 and then compule BD 85 and aea ABCD 3234 square units.

Formula to find the diagonals of a trapezium

Yon

Draw triangle with the difference of parallel sides of a trapezium as its base and the slanting sides as other two sides. Then find lengths of 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) 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 Il DC, BF L DA, BE = BC - AD_

Compute AM, BM, ME in AABE.

AC = VAM? MC?

BD = VBF? FD?

MC = ME + AD

B M E FD = BM + AD.

Example (Scalene Quadrilateral)

Mensuration

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 trapezium, find the lengths of the diagonals. This example was given by earlier teachers.

25	BM = 60	102 x 24	48
		60	5
AM = √(AB² + BM²) = 189	39	56	63	52
BN = 60 + 108 = 168
B M	N	DN = √(BD² + BN²) = 224
		60	AM ≠ DN

NOTE: If the reader knows Ptolemy's Theorem, it is easy to recognize ABCD as an acyclic quadrilateral.

AC	BD	AD	BC	AB	DC
63	56	25	60	39	52

i.e. 3528, 1500, 2028 which is true. In the above quadrilateral if BD = 32 (rather than 56), then what is AC?

BM = 32 + 64 x 14 = 30 (AM ⊥ BD)
				32	AM = 392	30²	AM ≠ 25
If CN ⊥ BD, then	BN = 1/2 (32 + 112 x 8) = 30	M = N

That is, diagonals of ABCD intersect at right angles. Many commentators on Lilavati have missed this point.

Mc?	60?	30?	CM	52
AC	25	52	77

Lilavati of Bhaskaracaya

Here the older writers should have taken approximate values rather than actual square roots.

Now consider the third example wherein AB II CD.

39 52 52 AN Il BC, AM L DC.

CN = 25, DN = 35

D 35 M R N 25 DM = 35 91x 13

AM = 9 1 V38o16

MC = 60 3 297

AC = VAM? MC? 71, BD = 25 +5 AM? 46.7

Area 2(60+ 25) x195 493 square units, slightly different from the exact value.

Current methods of approximation differ from those of Bhaskaracarya.

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. It 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.

(Diagonals of a cyclic quadrilateral). Find the sum of the products of sides holding a diagonal of quadrilateral, and obtain a similar sum for the other diagonal. Multiply them by the number obtained on adding the products of the opposite sides. Divide each of the multi-

Mensuration

plied sums by the other (unmultiplied) sum. Square-roots of these quotients are diagonals.

Comment: Let m and n be the diagonals of a cyclic quadrilateral (see the adjoining figure) Then

m = (ad + bc) (bd + ac)

ab + cd

and m = (ab + cd) (bd + ac)

ab + bc

Since & + 0 = %, cos & + cos 0 = 0

i.e. c2 + d2 - n2 = a2 + b2 - n2 = 0

i.e. ab(e2 + d2) + cd(a2 + b2) = n2

(ab + cd)

Simplyfying, we get Brahmagupta's formula.

Bhaskaracarya's Method

(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, the 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).

140 Lilavati of Bhaskaracarya

Comment: Take two right triangles of (generally integral) sides 1.b, C1, 42, bz, Cz, a; & b; < C; Form quadrilateral with sides C1az, Czal, C1bz, Czb1 (Products of hypotenuse with sides of the other triangle). One diagonal = a2 + b2. The other diagonal = b2 + a2.

NOTE: Bhaskaracarya claims that his method is much simpler than Brahmagupta's and wonders how it was not discovered earlier.

Example

Take a1 = 3, b1 = 4, C1 = 5, a2 = 5, b2 = 12, C2 = 13. Then the quadrilateral has sides 52, 39, 25, 60 which is the same one in sluza (CXCV). With sequence of sides (25, 39, 60, 52), the diagonals are:

5 * 3 + 12 * 4 = 63.
12 * 3 + 5 * 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 C1 C2 = 65.

There are no square roots nor big products using Brahmagupta. Bhaskaracarya's formula can be proved by substituting Cpaz, C2a/, Cbz, C2bp for the sides and nothing c1 = a1 + b1 and c2 = a2 + b2.

REMARK: It 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 * 8 + 4 * 15 = 84
3 * 15 + 4 * 8 = 77.

These are against the correct values 63, 56. Thus these right triangles are not compatible to 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 cyclic quadrilateral can be chosen in the following manner:

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 divide the remaining two sides of the quadrilateral by the hypotenuse of the first optional triangle. The quotients are the perpendicular sides of the optional right triangle.

Example

A cyclic quadrilateral 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:

GE, HF, DE, EC, DF, FC
PT, DT, TC
SR, DR, RC

DC = 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:

DE?	(195)?	(189)?	DE = 48	EC = 252
CF = √260?	224² = 132	FD = 168	EF = 120

Lilavati of Bhaskaracarya

From similar triangles AEC, YDC; CE DC DY = 225.

AE DY

Similarly, we get RC = 300 DR, 33 SR = 56 (300 DR), 16 SR 63 DR. Solving RC 3564 GE 64, GD = 80, HF 99_

HC = 165, RF = 7717 SA 195* 15 SB 2600

PT = 144, CT = 192.

I

(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 pitha or complements of the projections:

Then

GE DE x BF GD = DE x BD

DF DF

YD and LC are called 'poles'. They are at right angles to the base and their tops are on the diagonals YD = DC x AE. There we can get the height of the point of intersection and the segments of the base by the use of stanza CLXVII.

Mlensuration

143

DR = DC x Y2X* DC x J XXI

Y2* + YiN2 Yi * XL + X2

SR = DC x Yi Yi X2 J2lL

D x E R F x

Comment: Two equations (for DR and SR) derived in stanza CC are given here by Bhaskaracarya in different forms.

Formula to compute the circumference of a circle given its diameter

4frg 489: CCVII /

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 22/7 gives a rough approximation of the circumference.

Comment: Circumference 3927 Diameter 1250 22 Diameter.

It is known that circumference is a transcendental irrational number and is usually denoted by T. The diameter is not a root of any algebraic equation. For calculations, we use approximate values 22, 3927, 3.1416, etc. It shows that Bhaskara and his (Indian) foremathematicians were.

Lilavati of Bhaskaracarya

aware ol" the irralionality of circumference that is that of T. The diameler Vedic term for 7 is trita (Rgveda 1.52.5).

Example

Find the circumference of a circle of diameter 7 and the diameter of a circle whose circumference is 22.

Comment: Circumference = 22 x 7 = 22_

Diameter	22	22 = 7

This formula requires calculus for its proof:

Formulae: Area of a disk, surface area and volume of a sphere

Area of a disk (circumference) (diameter). Area of the surface of sphere = 4 x area of great circle. Volume of sphere = (surface area) * (diameter).

Comment: If the radius of the sphere is 5, Area of great circle = (2πr) (2r) = πr². Area of surface = 4πr².

Volume of the sphere = (4/3)πr³.

These formulae are proved in any calculus text.

Mensuration

Example

What is the area of a disk of diameter 72? What is the area of a net which just encloses a ball of diameter 72? What is its volume? O wise friend, answer if you know pellucid Lilavati well.

Comment:

Area of the disk	2 * (3) = 385 sq. units
Surface area	4 * 2 * 49 = 154 sq. units
Volume	22 * 3' * 532 = 1792 cu. units

For greater accuracy, Area of a disk = 3927 (Diameter)? , and 5000 for most practical purposes this may be 11 (Diameter)? 14

Comment:

Volume V = D^2 * 2 [ where D = Diameter: V = 22 D' = T 8r' = Ttr 2 x 21

Lilavati of Bhaskaracarya

Formula: Perpendicular to a chord

CCXI M

[These two stanzas give three relations between chord, its arrow and diameter of circle. Indeed, one of the three is computed when the other two are known, BC = chord, MB radius, AE Diameter 2r; AD Arrow.]

Subtract the (positive) 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 product 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 VAE+BC) (AE BC) BC = 2BD = 2VAD (AE AD) AE = AD + (BD)²]

AD Comment: FC? BF? _ BC? = AE? Bc? (AE + BC) (AE BC) TC = FC AD = AM MD

AE V(AE BC) (AE BC) BD² = AD * DE = AD (AE - AD) BC = 2BD = 2VAD (AE AD) ED x AD = BD²

Mensuration

147

ED = BD? AE = AD + BD?

AD AD

Example

0 friend, diameter 0l a cirele is 10 and its chord is 6. Find the length of the aTow. If the aTow is given, tell (me) the length of the chord: If the aTow and the chord are given, find the diameter;

Comment: AE = 10, BD = DC = 3.

MD = VMB? BD? Vs? _ 37 = 4 AD = 1.

In the second example, BD? BM? MD? = 9 BD = 3 and BC = 6.

In the third example, AE = AD + (BD)? = T + 10.

Formula: To find the sides of a regular inscribed polygon

No;of sides	Side
3	923a
4	853a
5	534a
6	000a
8	922a
9	034a

where a = Diameter

Lilavati of Bhaskaracarya

Comment: For regular polygon of n sides inscribed in circle of radius I, side = 2 r sin

Using Trigonometrical labels; we get different values for n = 3, 4, 5,

Formula for the length of a chord: rough approximation

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.

Length of the chord = 4d c (p - c)

p ~ c (p - c)

This formula 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 formula must be Xc (p - c).

Now we have two special cases:

p when c = p
P when c = R

Solving these two equations for X and Y we get the result.

Mensuration

Example

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.

Here d = 240, C = 18 P

4d	18 P	x	17 p	68d	17 x 240
18			42.06

Its exact value = d sin 10° = 240 * (0.1736) = 41.664, When € = 18 P; = 84.12.

In Bhaskaracarya's time (A.D. 1150), there were no tables of logarithms or trigonometric functions and so the students were given approximate formulae.

Formula to find the arc length, given the chord

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).

Comment: Arclength c = P t 4p² e

Lilavati of Bhaskaracarya

Solve the above formula (cf. stanza CCXX) for € and the two answers give the major arc and the minor arc respectively. (Bhaskaracarya intends to give only the minor arc.)

Example

Arithmetician! If you know the formula 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.

Comment: Here 42, d = 240, p = 754.16. Since p is large, to facilitate the calculations, we compute:

c =	5	+	4	17	or L for € = 42
p	2	√t	4d + €	18	18

If € = 82, € = 1 or 8

p = 9

CHAPTER 29 Volume

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

Measurements of an irregular solid were:

151

Lilavati of Bhaskaracarya

lengths 10, 11, 12 C (cubits), breadths 6, 5, 7 € and heights 2, 4, 3 C. Find its volume

Comment: Average length 10 + H+ 12=IC _

Average breadth 6+5+7=6€_

Average height 2 +3 +4 =3C _

Average volume H * 6 * 3 = 198 C.

It should be noted that the measurements should be carried out at equal intervals. It is clear that Bhaskaracarya gives practical methods.

Formula: Volume of a pyramid and its frustum

Find the sum of the areas of the mouth (top), bottom (base) and the rectangle whose (adjacent) 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) volume divided by three is volume of pyramid.

Comment: Volume of frustum MM' [pq + pq' + (p + p') (q + 4)]:

D Volume of Pyramid is that of prism N with the same base and same height.

Volume 153

These are standard formulae given in texts.

Here ABCD and A'B'CD' are similar rectangles.

Example 1

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 C? (cubits) and the base is half the size of the top, i.e. 5 x 6 C?. If its height is 7 C, find the volume of the well.

Volume = [120 + 30 + 15 + 18] = 490 C.

Examples 2, 3, 4, 5

(2) The top of a hole is 12 x 12 C? (cubits) and its depth is 9 C. Find its volume.

(3) The circular top of a hole is of diameter 10 C and its depth is 5 C. Find its volume.

Find the volume of a pyramid (cone) whose base and height are:

(4) as in (2): (5) as in (3).

(2) This is a box (rectangular parallelepiped) whose volume = [2 x 12 x 9 = 1296 C.

(3) This is a right circular cylinder whose volume = π (5)^2 * 5 = 392.86 C.

(4) Volume of the pyramid = 1296 / 3 = 432 C.

(5) Volume of the cone = 392.86 / 3 = 130.95 C.

Lilavati of Bhaskaricarya

Formula: Volume of a prism

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.

Comment: This is straightforward.

Example

Rectangular prism 8 * 5 x 3 (C = cubits) is to be constructed by using bricks of dimensions 1 x 12 x 3 A' (A = argulas). Find the volume of the prism, number of bricks required, and the number of layers. (1 C = 24 A).

Comment: Volume of the prism = 8 * 5 * 3 = 120 C

Volume of a brick = 1 x 12 x 3 = 36 A3

1 C = 24 A, thus:

Volume of a brick = 36 A = 36/24 C = 1.5 C

Number of bricks = 120 C / 1.5 C = 80

Number of layers = height of heap / height of brick = 3 C / (3 A / 24) = 24

CHAPTER 30 Wood Cutting

1 CCXXXI

[Log is a frustum of a cone AB = d1, DC = d2, MN = h.]

Area of a section (ABCD) = (d1 + d2)

Multiply this by the number of times the wood is cut to get the total area in A2 (A angular). To get the answer in C (cubits) divide by 576.

Comment: Section ABCD is a trapezium. If several sections are made, obviously they are not equal in area. So this gives an approximate answer. 1 C? = 24 x 24 A = 576 A?

Lilavati of Bhaskaracarya

Example

A log of wood has base 20 A (argulas) long; top 16 A long and height 100 A. If it is cut at four places; find the total surface area of the wooden pieces:

Comment: Area = 4 x 100 (20 + 16) = 576 c?

Formula: oblique cutting of a log

Example

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.

Example

A log is 32 A (angulas) long and 16 A thick: It was cut obliquely at 9 places. Find the total area of the sections:

Comment: Area = 9 x 16 x 32 = 576

CHAPTER 31 Volume of a Heap of Grain

If the grains are big and spherical, their heap (right circular cone) has height which is 10th of the circumference. If they are small and spherical, the proportion is 1/3 and if they are pointed, it is 1/4.

Volume can be computed by (height) × circumference. This gives the answer in C (cubits); it also represents the number of Magadha kharikas.

Lilavati of Bhiskaracarya

158

Comment: 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 nr h (r radius) as seen before. Bhaskaracarya's formula:

2πr h = nr². We take π = 3. This is for all good practical purposes for the village folk:

Example

32 CCXXXVI

On a flat ground there is a heap of big spherical grains of perimeter 60 C (cubits). O mathematician, find how many kharikas (K) are there? Find the same if the grains are small or pointed.

Comment: Big spherical: Circumference p = 60 C, h = 6 C; Volume = 600 K;

(ii) small spherical:	h = 60	Volume = 100 x 60 = 545.5 K;
(iii) pointed:	h = 60	Volume = 60 x 2 K.

Formula: Heap in a corner

Volume of a Heap of Grain

The heap is in an outer corner. The base is the area of the disk 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 wall, multiply by 2 and if it is in the interior it will be times. The actual volumes of the heaps of grain will be respectively of the volumes so obtained.

Example

Find the volumes of heaps of grain stored in 45 C (cubits).

(i) front corner with length of the base = 30 C
(ii) on wall = 30 C
(iii) inside corner = 15 C

Comment: In all cases circumference = 60 C. Let h be the height (determined according to stanza CCXXXVII).

Volume	Formula
(i)	Volume = 60 h = 3 100 h = 75 h c
(ii)	Volume = 100 x h = 50 h c
(iii)	Volume = 100 x h = 25 h c

NO_CONTENT_HERE

CHAPTER 32 Shadows

Formula: Length of a shadow

Obtain squares of the (given) difference of hypotenuses and that of shadows. Divide 576 by the difference of these (Wo 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 = Gits shadows are (t + q) and the corresponding hypotenuses are (c + p) P. We are given T, P, q and we have to find when q > p]

( = -q + p 576 q² - p)

The other shadow is (t + q).

161

Lilavati of Bhaskaracarya

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 4)² and c² = r + 1²

Zcp = 2qt + m where m = 4 - p

4c p² = (2qt + m)²

4p² (r + ()) = 447,? + 4qlm + m

41 (q² p²) + 4mqt + m 4pr = 0.

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 angulas.

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.

[Given AB, DC = 12, BD, find PD.]

PD	12	DB
AB - CD
Comment: By similar triangles, PCD, CAQ,	12	PD
CQ	PD	DB
CD	AQ	12
AB	BQ
PD	= 12 DB	AB CD

Shadows

If BD 3 C [Cubits 72 A (angulas)], AB 3 L C (844) and CD 12 A. Find the length of the shadow.

Comment: PD 12 x 72=12 A

84 - 12

Formula: to find the height of a lamp

AB = BD x CD / PD

Comment: From the similar triangles PCD, CAQ,

CQ PD AQ = CQ x CD / BD x CD

AQ = CD / PD

AB = BQ + AQ = CD + BD x CD / PD

Example

Given BD = 3 € (72A), PD = 16 A, find AB_

Comment: Take CD = 12 A_

AB = 72 x 12 + 12 = 66 A = C

Formula: Distance between the basepoints

BD PD (AB - CD)

164 Lilavati of Bhiskaricarya

1

[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

PD' _ QD =

QB = d x QD

PB = d x PD

AB = QB x CD

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

CD = 12 A, DQ = 8 A, DD' = 2C (= 48 A), PD' = 12 A. Find PB, QB and AB.

Comment: QD = &, DD' = 48, QD' 40 PD' = 12.

PQ = 52. BQ = 52 * 8 = 104 A

PB = 52 x 12 = 156 A

Shadows

Knowledgeable persons are fully aware that examples in Algebra and Arithmelie 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.

In this stanza, the importance of the Rule of Three is described. Nowadays, the Unitary System has replaced the Rule of Three with a consequent loss of speed.

NOTE: It appears that this stanza is not Bhaskaracarya's; it does not seem plausible that he considered himself as mediocre.

Lilavati of Bhaskaricarya

[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

PD' QD =

QB = d x QD

PB = d x PD'

AB = QB x CD

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.

Example

viaphfasfaiyarru 481faarreiyahl

CD = 12 A, DQ = 84, DD' = 2C (= 48 A);

PD' = 12 A. Find PB, QB and AB.

Comment: QD = 8, DD' = 48, QD' = 40, PD' = 12,

PQ = 52. BQ = 52 * 8 = 104 A

PB = 52 x 12 = 156 A

Shadows

165

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.

In this stanza, the importance of the Rule of Three is described. Nowadays, the Unitary System has replaced the Rule of Three with a consequent loss of speed.

NOTE: It appears that this stanza is not Bhaskaracarya's; it does not seem plausible that he considered himself as mediocre.

NO_CONTENT_HERE

CHAPTER 33 Pulverization (Kuttaka)

Before studying Bhaskaracarya's method, let us first see what is 'pulverization'. Take an equation ax + b = cy where a, b, c are integers, then to find y we have to choose those values of x that will make ax + b = cy. 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 kuttaka. Kutta means '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 + by = c stand for dividend, constant, divisor, multiplier and quotient respectively.)

For example, if one hundred is multiplied by an integer; 90 is added to or subtracted from the product; the results are exactly divisible by 63.

168 Lilavati of Bhaskaracwya

you are proficient in pulverization, tell me the multiplier correctly.

[That is if 100x * 90 is divisible by 63, find x.]

Let us solve 100x * 90 = 63y or

100x * 63y = -90

Let two consecutive solutions be (Xr, Yr), (xr+h, Yr+1)

Then 100xs+1 ~ 63yr+1 ~ 90 = [100xr 63yr]

100 (xr+1 - xr) = 63 (yr+1 - yr)

Since (100, 63) = 1, xr+1 - xr = 63 and yr+1 - yr = 100, i.e. (xr) is an A.P. with CD 63 and (yr) is an A.P. with CD 100. Now (T) is

10x Jx - 1 Clearly y and Jx - 1 must both be divisible by 10.

y = 10

30 satisfies this. So X = 10 (20) = 18. So the solutions are (18, 30), (81, 130), (144, 230).

Similarly, solutions of 100x * 63y = 90 are (45, 70), (108, 170), (171, 270).

Now let us consider Bhaskaracarya's method and compare it with current ones.

717y:

T: / CCLVII /

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:

Pulverization (Kuttaka)

First find the HCF of the dividend and the divisor which will be 1. For if it is not 1, it will divide the constant and after division, the dividend, divisor and the constant will be in Standard form. Go on dividing this standard dividend by the standard divisor till we get the remainder. Then write the quotients one below the other followed by the constant and 0. 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 = x) will be the desired y and x respectively. If the number of divisions is odd, just interchange.

Now we’ll solve the equations by Bhaskaracarya's method:

Dividend	Divisor	HCF
100	63	1

63 100 Divide bigger number by smaller; 26 37 write the quotients in the right and left columns and remainders below.

This way we get six quotients even number; 1, 1, 2 in the right column and 1, 2, 1 in the left one. They are, in the order; 1, 1, 1, 2, 2, 1.

Quotients	Remainders
1	90
1	270
2	90
2	90
1	90

Write these in a vertical column and write below the constant 90 and 90 1 + 0 is written in front of 90 2. 90 = 270 is written above 90.

Lilavati of Bhaskaracarya

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 [00 18 63 * 30 = 1800 _ 1890 = -90. So (18,30) is a solution.

Complete solution:

x	18	81	144	207
y	30	130	230	330

If the equation is [OOx 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 is odd.

Solve: 60x + 3 = Y in integers

We’ll first solve 60x + 3 = Y. The label with 60, 13 is as on the left side. The sequence of quotients is 4, 1, 1, 1, 1 odd in number. Here the constant is 3. We write 4, 1, 1, 1, 1, 3, 0 in the left column and proceed from the bottom upwards as in the 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 = 11. If the equation is 60x - 3 = y we get x = 2, y = 9.

Pulverization (Kuttaka)

Example

3oaTary 7I CCLVIII /

arithmetician! 221 is multiplied by an integer; 65 is added to the product; 195 divides the result exactly. Tell the multiplier quickly.

[Solve 2211 + 65 = y in integers]

195

Comment: Here we can divide by HCF (221, 195) 13 and get 17x + 5 = y. Here the quotients for 17, 15 are 1, 7. The top two numbers are 40, 35. 40 divided by 17 leaves remainder 6 and 35 leaves remainder 5. So X = 5, Y = 6,

Complete solution: (5, 6), (20, 23), (35, 40), (50, 57),

Example

Hafa myfariaa: 97: # 7 1 CCLIX

If in ax + b = Y, #, b have a common factor m, then

Xt 6

m m_Xey'

and solve this equation. Clearly m will be the same and values of y = my'.

Comment: We'll take 100x + 90 = y (stanza CCLI).

Divide by 10 to get

IOx + 9 X = Y_

63 10

172 Lilavati of Bhaskaracava

Sequence of quotients for [0,63 is 0,6,3 (odd).

27 27 + 10 leaves the remainder 7;

171 171 $ 63 leaves the remainder 45.

3 27 Since the number of quotients is odd,

X =63 45 = 18,y = 100 - 10 7 = 30_

The above methods are applicable when the constant is positive. If it is negative, take suitable complements (in a, b)

Example

4svr:

0 arithmetician! an integer is multiplied by 60; 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; 60x = 16=y in integers]

Comment: Here choosing +16 we get 60x + [6=y] Since 16 > 13

we consider 60x + 3=Y-1-y' As per (CCLVII), x = M,Y' =51.

X 11, y 52. To solve 60x 16=y consider

60x - 3=Y+I=y. Here x =2,Y' = 9,

the solution is (2, 8).

Example

47 Riyfum: 44Trifvifriym:

Pulverization (Kuttaka)

Five is multiplied by an integer; 23 is added (0 or subtracted from the product; divides the results. Tell the multiplier. [Solve: 51 + 23 = Y.]

Comment: We solve 51 + 2 = YFT = y.

A solution of 5x + 2 = y' is x = 2, y' = 4 and of 5x - 2 = y' is x = 1, y' = [_

So a solution of 5x + 23 = y is x = 2, y = H1 and of 5x - 23 = y is x = 7, y = 4.

Formula to solve the equation when the constant is zero:

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 divisor (that is one solution is x = 0 and y = ...

Example

The product of an integer and five is added to 0 or 65. The outcomes divided by 13 leave no residue. O arithmetician! tell me that number immediately. [Solve: 5x = Y and (ii) 5x + 65 / 13 = Y]

Comment: In (i), (0, 0) is one solution. Others are (13, 5), (26, 10).

In (ii), one solution is (0, 5). Others are (13, 10), (26, 15).

174 Lilavati of Bhaskaracarya

Formula: constant is [

First solve ax + ]=y ad gel solution (Xo, Yo) Then solution of ax *C=y is: x is the remainder in Co and y is the remainder in CV

Comment: Suppose we wish (0 solve 17x + 5 = y.

First solve [7x + ]=y and get Xo, 7, Yo = &.

5 * 7 + 15 leaves the remainder 5 = x and 8 * 5 + 17 leaves the remainder 6 = y: proof of Bhaskaracarya

Continued fractions provide s method, To solve ax - by = we for continued fraction for

e.g. 5

13 13/5 2 + 3 2 + 2 2 + 1 + 1 + 2

If we omit the last one, viz, and compute the remaining fraction we get and we note that 13 * 2 - 5 * 5 = ] and thus X = 2 + 2, y = S is a solution of 13x - 5y = 1. Similarly the penultimate convergent of provides a solution to ax - by = 1. Details can be found in an algebra text,

Formula: union of pulverizers with the same divisor

Fgcegewisr / / CCLXVI / /

Pulverization (Kuttaka)

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: Il axle = Y + blc, i.e. ax - b = cy and a'xlc = y b'/c, i.e. a'x - b' = cy' then (a + a')x (b + b') = c(y + y') is the union of two pulverizers and (ii).

Thus if ax ≡ 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.

Example

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. Sx - 7 = 21y. One solution is (14,3), and x = 14, 35, 56.

NO_CONTENT_HERE

CHAPTER 34 Concatenation (Permutations, Partitions etc.)

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 formed (by permuting the given n digits).

178 Lilavati of Bhaskarzcarya

Comment: The first part describes the familiar formula Pn = n! / (n - r)! It is known that nPr is the number of permutations from n different objects taken r at a time. This is proved in algebra texts. It is easy to derive the formula for the sum of the numbers formed:

it is (n - r)! (r!)(sum of the digits)

Example

Using (i) 2, 8, (ii) 3, 8, 9, (iii) 2, 3, 9 how many different numbers 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, P2 = 2! = 2 different numbers.

(ii) Here the answer is P3 = 3! = 6 different numbers. In, n = 3, sum of the digits = 20 and so answer = 2 x 999 x 20 = 39960.

(iii) n = 8, Sum of the digits = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 44. Answer = 8! (44) = 2463999975360.

Example

Lord Siva holds ten different weapons, namely a trap, a goad, a snake, a drum, a polsherd, a club, a spear, a missile, an arrow, and a bow.

Concatenation (Permutations, Partitions etc)

his hands. [Lord Siva has five heads and So it is presumed that he has ten arms.] Find the number of different Siva idols. Similarly, solve the problem for Visnu idols; Visnu has four objects: a mace, disc, lotus and a conch.

In the first example, there are 10! = 3628800 possible Siva idols. In the case of Lord Visnu, possible idols are 4! = 24 in number;

[NOTE: In the daily ritual (sandhvavandanam) there are 24 names of Lord Visnu]

Formula: Permutations with repetitions

(To find the total number of permutations of given n digits or objects) if certain digits are alike then form the product of the number of permutations of those places at which alike digits occur (assuming that each block of alike digits has different digits), and divide the number of permutations of all the given (n) digits (assuming them different) obtained by the previous method by this product. And sum of the numbers formed is obtained by the previous method.

Comment: Let a1 = a2 = ... = ar, b1 = b2 = ... = bs, C1 = C2 = ... = Ct be n objects. Then number of permutations n! = n = r + s + ... + t.

This is proved in standard texts.

Example

74 4aea

Lilavati of Bhaskaricarya

Find quickly the number of different numbers that can be formed with

(i) 2, 2, 1, 1
(ii) 4, 8, 5, 5, 5. Also find their sums

Comment: 4 24 = 6_

2121

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.

(ii) Number of permutations Si 20

General formula for the sum 34

n! (raj sbp tc+(10" 1)

r! sl ! n

In (ii), n = 5, r = 1, s = 1, = 3,

Sum S1 (ICS _

315 (4 + 8 + 15) - 1) = 1199988

Example

T #ifa: rgife af Atrii CCLXXH

Leaving aside 0, if the digits 0 to 9 are written six at a time, then find how many different numbers are formed

Comment: Pi sPo = 9 x 8 x 7 x 6 x 5 x 4 = 60480.

7TI

48a CCLXXVI /

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:

Concatenation (Permutations, Partitions etc )

181

(n - ) (n -r + 4) This method is applicable when n < r + 9. AlI (r _ 4) ! this has been given brielly because this ocean of mathematies is vast

Forula can be writlen as (n -)! (n-r)! (r -

NOTE: The great India mathematician (Lale) S. Ramanujan made outstanding discoveries in Ihe theory of partitions It is said that he got inspiration Irom the above stanza:

Example

1 CCLXXVI

Find the total number of different numbers of five digits whose sum is 13.

n = [3, r = 5 so the answer

12 !	495
8!	41
7

Aithough there is no multiplication division, squaring or cubing in this concatenation, yet it is asked to egolistical evil-minded lads of astronomers, their humbling is Certan.

This Lilavati clearly explains fractions, simple fractions, multiplication etc. It also beautifully describes problems in day-to-day transactions.

Lilavati of Bhaskaracarya

Those who master this Lilavati will be happy and prosperous.

Alternate meaning: (Lass) Lilavati is born in respectable family, stands out in any group of enlightened persons, has mastered idioms and proverbs. Whomsoever she embraces (marries) will be happy and prosperous.

3121 8947 (INTERPOLATED VERSE)

II CCLXXIX

Bhaskaracarya, 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 (Supreme) cannot be fathomed. [Obviously he had not mastered only Mathematics but was well versed in many branches of knowledge.]

Index of Verses

BEGINNING PART OF VERSE	STANZA NO.
XL
XXXVII
CLXXXII
CXCV
32	XCIX
37g3 #rrriri	CCXXXVII
CLXVI
XXXI
CXCVII
LV
XIV
LXXVII
CCXXXI
CCXX
183

Lilavati of Bhaskaracarya

184	a fut-	CCLXXIX
	LXX	Tg9f?
	CLVHII	CXXVII
	CXXXVII	XXIX
	CXXIX	TrfA:
	CXCH	CLXXXIX
	LXXX	LXXXIV
	LXV	CLII
grqiju	447	CLII
	LXVII
3	CLXXXI	CCXXX
327	CL	CLXXXVII
	XVI
rn	CXLI
71	CXLVIII	CXLVII
	CCXXIX	LII

Index of Verses

185

XII

CXIX

CXXII

CCCLVIII

CXXIV

CXVIII

CCCLXVI

Q

CCLVI

CCLXVII

CLXIV

CXCVI

CXC

CVII

#T:

XII

LXXXI

@ferrgr& 47

CXLIV

CLXXXIV

458 4 4

CLXXXV

437 T9371

CLXXII

87 77 YTT

CXCIX

CCLX

44 CCLXV

CCLXII

taynnafa @ &

XLVIII

CCXVHI

Lilavati of Bhaskaracarya

XXI

XXVII

CCXXVI

CXXXII

Trrd mn Ti Rn

CXXX

CCXXII

LXXI

Xv

Mprataai

CCXII

CLXI

CLXXVII

CLXX

CLXXXIII

CCXIX

CCXLI

H1

CCXLVII

CCXLIV

CCXXXV

XLIX

XLII

XXXV

LXXXV

CCXIV

CCXIII

CXLII

[ 1 H MHHi 8X ] 4 L 1 1 3V 1 3 V [ 4 ] 1 H 1 lix1JX 1 d L a 1 4 L 1 J 3 1 9 1 3 1 1 4 3

8 8306998 ? 19 8 & 8 8 8

Lilavati of Bhaskaracarya

CXXV

CCLXXII

RAErfxfaya 4

CLXII

3on

CXCIV

CCXXXIX

grlferyjvi 47

CLXIX

CCLXXVII

CVII

Tu FtT T7 772n

XXVIII

CCLXXV

CCLXXIV

XXX

XCVIII

LIII

CLXV

LXXXIX

CCLXXVI

XXXIX

LX

CII

CLXXXVI

CXXXI

CLI

93 XCIV

Cv

Index of Verses

TT 4fd4id:	CCLIII
4RffATT	CCXL
	LXIX
	CXXXIX
	IX
	LXXVI
	CCLXX
	CCXXXIII
fqoz	XCII
9Teagi 47 fyM	CXXXVII
	LXXXII
	CI
4 TAY:	CXXXIII
	CCXLV
	XCVII
	LXXIX
	CXXI
	LXXXVI
	CCXVII
aa	XVIII
	LXXIII
	CXCVIII
4a yu: yu 47 *7	XVI
Ifevrisr4 aff44	CII

+IH9 9 V H

4 4 4 4 4 1 2 { 4 1 ] 1 1 1 3

4 1 1 4 1 1 W 1 1 4 1 1

1 1 1 1j 4 L

11 8 2381 8 8 3 2 F

</p}>

Index of Verses

CLXXXVII

LI

4 d desf4	CCXXV
	LXXV
	CCLXXI
g2nef	LIV
47 4agfun: <gat:	CCLXIV
47 riyfora: 44	CCLXII
	CIV
4qi gonfaryoarifarm	CCLXXVII
	CLIV
4 & 49i Tfal & &nfral <f:	XLVI
as-R gauerirATi	XXXVI
	LXI
	LXXVI
	CXLIII
	LXVII
	LXIV
	LXVI
	XXXI
	CCI
	CLXXIX
	CCIV
F:	CCMI

1

WjV H

4 1 1 1 I{H } } 4 H H L 4 8

1 4 H { 1 li 1

1 1 1 1 ] 1 (i 4

4 4 4 4 1 3

1 4 2 4 4

4 8

[

2

0 3 1 2 I806 = 8 F 2 1 1

HHLHWH V 84U F

1	1j	L	4	3	3	1	1
1	1	1	1	1	1
1	{	4	7	3	1!	1	4	1
L	4	1	3	1	4	1	4	2
L	4j	1	]	3	1	1	1
H!	13	4	1
1	14	1	[	}	4
4	3	1
3

8 8 9 89148 2 9 1 * 4 1 X 1 2

194 Lilavati of Bhaskaracarya

CXXIII

CL VII

CCLXXII

CCLXVIII

RTTTTT7 XXVI

VI

LXXIV

CLXXII

CXII

CXIV

LVIII

CCLV

VII

CXVII

Agathale #farTari LVI

Subject Index

[Brackets containing number(s) stand for the verse number(s)]

Abja	(XII)
Acre	(VI)
Addition	(XII)
Addo	(XXXIV)
Adhaka	(VIII, LXXXVIII)
Adholi	(VIII)
Altitude	(CXLI-CLII, CLXIV)
Angula	(CV, CCXXXII-CCXXXIV, CCXXXVI; CCXLI, CCXLIII)
Anguli	(VI)
Antya	(XII)
Anustupa	(CXL)
Ardhasamavrtta	(CXXXIX)
Area	(CXLI, CLXXV, CCIX-CCXI)
Arithmetic progression (AP)	(CXXVI, CXXVIII-CXXXII, CXXXIV, CXXXV)
Arow	(CCXIV, CCXV)
Asu	(x)
Athave	(IX)
Bahu	(CXLI)
Bamboo	(VI)

195

Lilavati of Bhaskaracaya

Base	(CXLI-CLII)
Bhagiratha	(XXII)
Bhuja	(XLI)
Brahmagupta	(CXCV, CXCVI, CXCVIII)
Bhuvankosa	(CLVIII)
Champaka	(LVII)
Circumference of a circle	(CCVII, CCVIII)
Circumference of a disk	(CCIX)
Combination	(CXVII-CXXI)
Compound fraction	(XXXIII)
Concatenation	(CCLXVI, CCLXXVII)
Congruence	(CCLXVI)
Cowrie	(II, XXXIV, CXXXVII)
Cube	(XXVI-XXVIII)
Cube root	(XXX)
Dana	(II)
Danda	(v)
Datta	(XII)
Davadi	(I)
Dedo	(XXXIV)
Devanagari	(XII)
Dhabu Paisa	(II)
Dharana	(III)
Dhataka	(III)
Diagonal of a quadrilateral	(CLXXXVII-CXCVIII)
Diophantine	(CCXLIX)
Divakara	(XVIII)
Division	(XIX)
Doa cynosuroides	(XLII)
Dramma	(I, XXXIV, LXXXII, LXXXIII, XCIV, CVI, CVII, CXXIX, CXXXVII)
Drona	(VIII)
Dvisapta	(IX)
Dwadasasra	(VII)

Subject Index

Exchange of jewels (CVIII; CIX)

Existence of triangles (CLXIX)

Fractional residue (LVIII)

Fractions

addo and dedo (XXXv)
compound (XXXIII)
division (XLII)
multiplication (XL)
simple (XXXI)
squares, cubes, square roots and cube roots (XLIV, XLV)

Gadyana (II; IX)

Gadyanaka (IIL, LXXXVII)

Ganapati (XI)

Ganesha (I, XI)

Geometric progression (G.P.) (CXXXVI)

Ghanahasta (VII)

Ghatika (X)

Gnomon (CCXLI)

Goladhyaya (CLVIII)

Guija (III, IX, LXXXI, LXXXVII)

Hasta (V-VII)

Hectare (VI)

Hypotenuse (CXLI-CLII)

Iccha (LXXIX)

Infinite (XLVII)

Infinity (XLVII)

Interest (XCVII-C)

Inverse proportion (LXXXIV)

Jaggi (XII)

Jaladhi (XII)

Jnanaraja (XLVII)

Kadamba (LVII)

Kakini (I1, LXXXII, XCIV, CXXXVII)

Kala

Karna (LXXVI)

Karsa (Iv)

Lilavati of Bhaskaracarya

Kastha	Kavadi	(I, XXXIV)
Ketaki	(LX)
Khahara	(XLVI; XLVII)
Khandameru	(CXVIII-CXX, CXXII)
Khari	(VIL, VIII)
Kharika	(VII; LXXXIII, CCXXXVII, CCXXXVIII)
Kharva	(XII)
Khavuga	(IX)
Kosa	(V, VI; XCIV)
Koti	(XXXI)
Krosa
Ksna	(X)
Kudava	(VIII)
Kusa	(XLIII)
Kutta	(CCL)
Kuttaka	(CCL)
Madhya	(XII)
Mahapadma	(XII)
Malati	(LX)
Masa	(IV, LXXXI, CXI; CXIII, CXV, CXVII)
Maund	(IX)
Meru	(CXXII)
Method of transition	(LXI)
Metre	(CXXXVIII-CXL)
Miamsa	(CCLXXIX)
Mouth	(CCXXVI)
Naksatra	(X)
Newton	(CXXII)
Nikharva	(XII)
Nimisa	(I, LVIII, LXXXI, LXXXII, LXXXVI; XC, XCII, XCIII, XCVIII, C, CII, CVII, CXXXVII)

Subject Index

199

Nithave (IX)
Nivartana (VI)
Pai (I)
Pala (IV, X, LXXXI, LXXXII, CVII)
Pana (I, LXXXII, LXXXIII, CXXXVI)
Pankt (XXIII; XXIV)
Parardha (XII)
Pascal's triangle (CXX, CXXII)
Patali (LVII)
Pavah M
Pavali (VII)
Permutation (CCLXVIII)
Peshawa (II)
Phala (LXXIX)
Pirgalac rya (CXXII)
Pitha (CCII)
Pole (VI)
Pramana (LXXIX)
Prastha (VIII)
Progression (CXXIII)
- arithmetic (CXXIII see also A.P.)
geometrie (C; CXXXVI see also G.P.)

Projection (CLXXIV)

Prosody (CXXXVIII)
Ptolmey (CXCV)
Pulverize (CCL, CCLXV-CCLXVII)
Pythagoras (CXXIV)
Pythagorean triplet (CXLIX)
Quadratic equation (LXXI-LXXIII)
Ramanujan (CCLXXV)
Ratna (CCLXXIX)
Ratti (II)
Reverse process (XLIX, L)
Rgveda (CCVII)
Ruka (II)
Rule of
eleven (XCIV)
five (LXXXIX, XC)

Lilavati of Bhaskaracarya

nine (LXXXIx, XCII)
seven (LXXXIV, LXXXIX, XCII)
three (LXIX, LXXIX, LXXX, CCXLVIII, CCL)
Rules for barter (XCV)
Samavnta (CXXXIX)
Sandhi (CCII)
Szndiyavandanam (CCLXX)
Sarku (XII)
Sastrodita (VII)
Seer (VIII)
Series (CxXIII)
Shastri (XVIII, XXII, LXXXI, CLXVI)
Silindhra (LX)
Simple fraction (XXXI)
Singh (XII)
Siva (LV, CCLXX)
Square (XX)
rool (XXIII; XXIV)
transition (LXIII, LXIV)
Subtraction (XII)
Sulvasurakara (CXLII)
Suvarna (IV)
Tanka (IX)
Tatpara (IV)
Tola
Trita (CCVII)
Truti
Vala (Ii)
Valla (III)
Varalaka (I, LXXXII)
Visamavrta (CXXXIX)
Visnu (XLVII; LV, CCLXX)
Vita (CLXI)
Volume of a (CCXXVIII)
cone

Subject Index

ditch or solid (CCXXIII-CCXXV)
heap of grain (CCXXXVII-CCXL)
prism (CCXXIX-CCXXXII)
pyramid and frustum (CCXXV-CCXXVIII)
Yava (I;, V)
Yojana (VI; LXXXIII)
Zero
eight rules (XLVI; XLVII)

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Of Related Interest

THE SURYA SIDDHANTA

Phanindralal Gangooly

The oldest and most important treatise of the post-Christian period on Indian Astronomy that has come down to us is the Surya Siddhanta consisting of 14 chapters written in slokas. Alberuni says that Lata was the author of this work: According to the introductory verses Surya, the sun-god, revealed it to Asura Maya in the city of Romaka.

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Lilavati

Bhiskracarya

A Treatise of Mathematics of Vedic Tradition

LILAVATI OF BHASKARACARYA

LILAVATI OF BHASKARACARYA

Treatise of Mathematics of Vedic Tradition

with rationale in terms of modern mathematics

largely based on N. H. Phadke's Marathi translation of Lilavati

Translated by

Foreword

Foreword

Contents

Foreword by V.V Narlikar

Roman Transliteration of Devanagari

Translators' Preface

Bhaskaracarya: His Life and Work

Lilavati

1. Definitions and Tables

2. Place Values of Digits

3. Addition and Subtraction

4. Methods of Multiplication

5. Division

6. Methods of Finding Squares

7. Square Root

8. Method to Find the Cube

9. Cube Roots

10. Eight Operations on Fractions

11. Addition and Subtraction of Fractions

12. Multiplication of Fractions

13. Division of Fractions

Lilavati of Bhaskaracarya

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

15. Eight Rules Concerning Zero

16. Reverse Process

17. To Find an Unknown Quantity

18. Method of Transition

19. Square Transition

20. Quadratic Equation

21. The Rule of Three

22. Inverse Proportion

23. The Rule of Five

24. Rules for Barter

25. Simple Interest

26. Combinations

27. Progressions (Series)

28. Mensuration

29. Volume

30. Wood Cutting

31. Volume of Heap of Grain

32. Shadows

33. Pulverization

34. Concatenation (Permutations, Partitions etc.)

Index of Verses

Subject Index

Roman Transliteration of Devanagari

VOWELS

CONSONANTS

Lilavati

Bhaskaracarya

Bhaskaracarya: His Life and Work

Quatrains:

Lilavati of Bhaskaracarya

Location of Vijjalavida

Xviii

Lilavati of Bhaskaracarya

Bhaskaracarya's Education

Works

Valiant Bhaskaracarya

The Poet Bhaskaracarya

Valiant Bhaskaracarya

The Poet Bhaskaracarya

Teaching Ability

Social Conditions

Bhaskaracarya: His Life and Work

Praise of Lord Ganesa

CHAPTER

Definitions and Tables

Lilavati of Bhaskaracarya

Measure for gold

Units of Length