Contributed by:

I. Why Vedic Mathematics?

II. Vedic Mathematical Formulae

1. Ekadhikena Purvena

2. Nikhilam navatascaramam Dasatah

3. Urdhva - tiryagbhyam

4. Paravartya Yojayet

5. Sunyam Samya Samuccaye

6. Anurupye - Sunyamanyat

7. Sankalana - Vyavakalanabhyam

8. Puranapuranabhyam

9. Calana - Kalanabhyam

10. Ekanyunena Purvena

11. Anurupyena

12. Adyamadyenantya - mantyena

13. Yavadunam Tavadunikrtya Varganca Yojayet

14. Antyayor Dasakepi

15. Antyayoreva

16. Lopana Sthapanabhyam

17. Vilokanam

18. Gunita Samuccayah : Samuccaya Gunitah

III Vedic Mathematics - A briefing

1. Terms and Operations

2. Addition and Subtraction

3. Multiplication

4. Division

5. Miscellaneous Items

IV Conclusion

II. Vedic Mathematical Formulae

1. Ekadhikena Purvena

2. Nikhilam navatascaramam Dasatah

3. Urdhva - tiryagbhyam

4. Paravartya Yojayet

5. Sunyam Samya Samuccaye

6. Anurupye - Sunyamanyat

7. Sankalana - Vyavakalanabhyam

8. Puranapuranabhyam

9. Calana - Kalanabhyam

10. Ekanyunena Purvena

11. Anurupyena

12. Adyamadyenantya - mantyena

13. Yavadunam Tavadunikrtya Varganca Yojayet

14. Antyayor Dasakepi

15. Antyayoreva

16. Lopana Sthapanabhyam

17. Vilokanam

18. Gunita Samuccayah : Samuccaya Gunitah

III Vedic Mathematics - A briefing

1. Terms and Operations

2. Addition and Subtraction

3. Multiplication

4. Division

5. Miscellaneous Items

IV Conclusion

1.
Vedic Mathematics - Methods

Preface ------------------------------------------------------------------------------------------------ 1

I. Why Vedic Mathematics? --------------------------------------------------------------------------------- 3

II. Vedic Mathematical Formulae -------------------------------------------------------------------------- 5

1. Ekadhikena Purvena -------------------------------------------------------------------------------------- 7

2. Nikhilam navatascaramam Dasatah ----------------------------------------------------------------- 18

3. Urdhva - tiryagbhyam ---------------------------------------------------------------------------------- 31

4. Paravartya Yojayet -------------------------------------------------------------------------------------- 41

5. Sunyam Samya Samuccaye ---------------------------------------------------------------------------- 53

6. Anurupye - Sunyamanyat ------------------------------------------------------------------------------ 64

7. Sankalana - Vyavakalanabhyam ---------------------------------------------------------------------- 65

8. Puranapuranabhyam ------------------------------------------------------------------------------------ 67

9. Calana - Kalanabhyam -------------------------------------------------------- -------------------------- 68

10. Ekanyunena Purvena ---------------------------------------------------------------------------------- 69

11. Anurupyena ---------------------------------------------------------------------------------------------- 75

12. Adyamadyenantya - mantyena ---------------------------------------------------------------------- 82

13. Yavadunam Tavadunikrtya Varganca Yojayet --------------------------------------------------- 86

14. Antyayor Dasakepi ------------------------------------------------------------------------------------- 93

15. Antyayoreva --------------------------------------------------------------------------------------------- 96

16. Lopana Sthapanabhyam ---------------------------------------------------------------------------- 101

17. Vilokanam ----------------------------------------------------------------------------------------------- 106

18. Gunita Samuccayah : Samuccaya Gunitah ------------------------------------------------------ 113

III Vedic Mathematics - A briefing ---------------------------------------------------------------------- 115

1. Terms and Operations --------------------------------------------------------------------------------- 116

2. Addition and Subtraction ------------------------------------------------------------------------------ 130

Preface ------------------------------------------------------------------------------------------------ 1

I. Why Vedic Mathematics? --------------------------------------------------------------------------------- 3

II. Vedic Mathematical Formulae -------------------------------------------------------------------------- 5

1. Ekadhikena Purvena -------------------------------------------------------------------------------------- 7

2. Nikhilam navatascaramam Dasatah ----------------------------------------------------------------- 18

3. Urdhva - tiryagbhyam ---------------------------------------------------------------------------------- 31

4. Paravartya Yojayet -------------------------------------------------------------------------------------- 41

5. Sunyam Samya Samuccaye ---------------------------------------------------------------------------- 53

6. Anurupye - Sunyamanyat ------------------------------------------------------------------------------ 64

7. Sankalana - Vyavakalanabhyam ---------------------------------------------------------------------- 65

8. Puranapuranabhyam ------------------------------------------------------------------------------------ 67

9. Calana - Kalanabhyam -------------------------------------------------------- -------------------------- 68

10. Ekanyunena Purvena ---------------------------------------------------------------------------------- 69

11. Anurupyena ---------------------------------------------------------------------------------------------- 75

12. Adyamadyenantya - mantyena ---------------------------------------------------------------------- 82

13. Yavadunam Tavadunikrtya Varganca Yojayet --------------------------------------------------- 86

14. Antyayor Dasakepi ------------------------------------------------------------------------------------- 93

15. Antyayoreva --------------------------------------------------------------------------------------------- 96

16. Lopana Sthapanabhyam ---------------------------------------------------------------------------- 101

17. Vilokanam ----------------------------------------------------------------------------------------------- 106

18. Gunita Samuccayah : Samuccaya Gunitah ------------------------------------------------------ 113

III Vedic Mathematics - A briefing ---------------------------------------------------------------------- 115

1. Terms and Operations --------------------------------------------------------------------------------- 116

2. Addition and Subtraction ------------------------------------------------------------------------------ 130

2.
3. Multiplication ---------------------------------------------------------------------------------------- 139

4. Division ----------------------------------------------------------------------------------------------- 144

5. Miscellaneous Items ------------------------------------------------------------------------------- 151

IV Conclusion --------------------------------------------------------------------------------------------- 158

4. Division ----------------------------------------------------------------------------------------------- 144

5. Miscellaneous Items ------------------------------------------------------------------------------- 151

IV Conclusion --------------------------------------------------------------------------------------------- 158

3.
Preface

The Sanskrit word Veda is derived from the root Vid, meaning to know

without limit. The word Veda covers all Veda-sakhas known to humanity. The

Veda is a repository of all knowledge, fathomless, ever revealing as it is

delved deeper.

Swami Bharati Krishna Tirtha (1884-1960), former Jagadguru Sankaracharya

of Puri culled a set of 16 Sutras (aphorisms) and 13 Sub - Sutras (corollaries)

from the Atharva Veda. He developed methods and techniques for amplifying

the principles contained in the aphorisms and their corollaries, and called it

Vedic Mathematics.

According to him, there has been considerable literature on Mathematics in

the Veda-sakhas. Unfortunately most of it has been lost to humanity as of

now. This is evident from the fact that while, by the time of Patanjali, about

25 centuries ago, 1131 Veda-sakhas were known to the Vedic scholars, only

about ten Veda-sakhas are presently in the knowledge of the Vedic scholars

in the country.

The Sutras apply to and cover almost every branch of Mathematics. They

apply even to complex problems involving a large number of mathematical

operations. Application of the Sutras saves a lot of time and effort in solving

the problems, compared to the formal methods presently in vogue. Though

the solutions appear like magic, the application of the Sutras is perfectly

logical and rational. The computation made on the computers follows, in a

way, the principles underlying the Sutras. The Sutras provide not only

methods of calculation, but also ways of thinking for their application.

This book on Vedic Mathematics seeks to present an integrated approach to

learning Mathematics with keenness of observation and inquisitiveness,

avoiding the monotony of accepting theories and working from them

mechanically. The explanations offered make the processes clear to the

learners. The logical proof of the Sutras is detailed in algebra, which

eliminates the misconception that the Sutras are a jugglery.

Application of the Sutras improves the computational skills of the learners in

a wide area of problems, ensuring both speed and accuracy, strictly based on

rational and logical reasoning. The knowledge of such methods enables the

teachers to be more resourceful to mould the students and improve their

talent and creativity. Application of the Sutras to specific problems involves

rational thinking, which, in the process, helps improve intuition that is the

bottom - line of the mastery of the mathematical geniuses of the past and the

present such as Aryabhatta, Bhaskaracharya, Srinivasa Ramanujan, etc.

1

The Sanskrit word Veda is derived from the root Vid, meaning to know

without limit. The word Veda covers all Veda-sakhas known to humanity. The

Veda is a repository of all knowledge, fathomless, ever revealing as it is

delved deeper.

Swami Bharati Krishna Tirtha (1884-1960), former Jagadguru Sankaracharya

of Puri culled a set of 16 Sutras (aphorisms) and 13 Sub - Sutras (corollaries)

from the Atharva Veda. He developed methods and techniques for amplifying

the principles contained in the aphorisms and their corollaries, and called it

Vedic Mathematics.

According to him, there has been considerable literature on Mathematics in

the Veda-sakhas. Unfortunately most of it has been lost to humanity as of

now. This is evident from the fact that while, by the time of Patanjali, about

25 centuries ago, 1131 Veda-sakhas were known to the Vedic scholars, only

about ten Veda-sakhas are presently in the knowledge of the Vedic scholars

in the country.

The Sutras apply to and cover almost every branch of Mathematics. They

apply even to complex problems involving a large number of mathematical

operations. Application of the Sutras saves a lot of time and effort in solving

the problems, compared to the formal methods presently in vogue. Though

the solutions appear like magic, the application of the Sutras is perfectly

logical and rational. The computation made on the computers follows, in a

way, the principles underlying the Sutras. The Sutras provide not only

methods of calculation, but also ways of thinking for their application.

This book on Vedic Mathematics seeks to present an integrated approach to

learning Mathematics with keenness of observation and inquisitiveness,

avoiding the monotony of accepting theories and working from them

mechanically. The explanations offered make the processes clear to the

learners. The logical proof of the Sutras is detailed in algebra, which

eliminates the misconception that the Sutras are a jugglery.

Application of the Sutras improves the computational skills of the learners in

a wide area of problems, ensuring both speed and accuracy, strictly based on

rational and logical reasoning. The knowledge of such methods enables the

teachers to be more resourceful to mould the students and improve their

talent and creativity. Application of the Sutras to specific problems involves

rational thinking, which, in the process, helps improve intuition that is the

bottom - line of the mastery of the mathematical geniuses of the past and the

present such as Aryabhatta, Bhaskaracharya, Srinivasa Ramanujan, etc.

1

4.
This book makes use of the Sutras and Sub-Sutras stated above for

presentation of their application for learning Mathematics at the secondary

school level in a way different from what is taught at present, but strictly

embodying the principles of algebra for empirical accuracy. The innovation in

the presentation is the algebraic proof for every elucidation of the Sutra or

the Sub-Sutra concerned.

Sri Sathya Sai Veda Pratishtan

2

presentation of their application for learning Mathematics at the secondary

school level in a way different from what is taught at present, but strictly

embodying the principles of algebra for empirical accuracy. The innovation in

the presentation is the algebraic proof for every elucidation of the Sutra or

the Sub-Sutra concerned.

Sri Sathya Sai Veda Pratishtan

2

5.
I. Why Vedic Mathematics?

Many Indian Secondary School students consider Mathematics a very difficult

subject. Some students encounter difficulty with basic arithmetical

operations. Some students feel it difficult to manipulate symbols and balance

equations. In other words, abstract and logical reasoning is their hurdle.

Many such difficulties in learning Mathematics enter into a long list if prepared

by an experienced teacher of Mathematics. Volumes have been written on the

diagnosis of 'learning difficulties' related to Mathematics and remedial

techniques. Learning Mathematics is an unpleasant experience to some

students mainly because it involves mental exercise.

Of late, a few teachers and scholars have revived interest in Vedic

Mathematics which was developed, as a system derived from Vedic principles,

by Swami Bharati Krishna Tirthaji in the early decades of the 20th century.

Dr. Narinder Puri of the Roorke University prepared teaching materials based

on Vedic Mathematics during 1986 - 89. A few of his opinions are stated

i) Mathematics, derived from the Veda, provides one line, mental and super-

fast methods along with quick cross checking systems.

ii) Vedic Mathematics converts a tedious subject into a playful and blissful one

which students learn with smiles.

iii) Vedic Mathematics offers a new and entirely different approach to the

study of Mathematics based on pattern recognition. It allows for constant

expression of a student's creativity, and is found to be easier to learn.

iv) In this system, for any problem, there is always one general technique

applicable to all cases and also a number of special pattern problems. The

element of choice and flexibility at each stage keeps the mind lively and alert

to develop clarity of thought and intuition, and thereby a holistic development

of the human brain automatically takes place.

v) Vedic Mathematics with its special features has the inbuilt potential to

solve the psychological problem of Mathematics - anxiety.

J.T.Glover (London, 1995) says that the experience of teaching Vedic

Mathematics' methods to children has shown that a high degree of

mathematical ability can be attained from an early stage while the subject is

enjoyed for its own merits.

3

Many Indian Secondary School students consider Mathematics a very difficult

subject. Some students encounter difficulty with basic arithmetical

operations. Some students feel it difficult to manipulate symbols and balance

equations. In other words, abstract and logical reasoning is their hurdle.

Many such difficulties in learning Mathematics enter into a long list if prepared

by an experienced teacher of Mathematics. Volumes have been written on the

diagnosis of 'learning difficulties' related to Mathematics and remedial

techniques. Learning Mathematics is an unpleasant experience to some

students mainly because it involves mental exercise.

Of late, a few teachers and scholars have revived interest in Vedic

Mathematics which was developed, as a system derived from Vedic principles,

by Swami Bharati Krishna Tirthaji in the early decades of the 20th century.

Dr. Narinder Puri of the Roorke University prepared teaching materials based

on Vedic Mathematics during 1986 - 89. A few of his opinions are stated

i) Mathematics, derived from the Veda, provides one line, mental and super-

fast methods along with quick cross checking systems.

ii) Vedic Mathematics converts a tedious subject into a playful and blissful one

which students learn with smiles.

iii) Vedic Mathematics offers a new and entirely different approach to the

study of Mathematics based on pattern recognition. It allows for constant

expression of a student's creativity, and is found to be easier to learn.

iv) In this system, for any problem, there is always one general technique

applicable to all cases and also a number of special pattern problems. The

element of choice and flexibility at each stage keeps the mind lively and alert

to develop clarity of thought and intuition, and thereby a holistic development

of the human brain automatically takes place.

v) Vedic Mathematics with its special features has the inbuilt potential to

solve the psychological problem of Mathematics - anxiety.

J.T.Glover (London, 1995) says that the experience of teaching Vedic

Mathematics' methods to children has shown that a high degree of

mathematical ability can be attained from an early stage while the subject is

enjoyed for its own merits.

3

6.
A.P. Nicholas (1984) puts the Vedic Mathematics system as 'one of the most

delightful chapters of the 20th century mathematical history'.

Prof. R.C. Gupta (1994) says 'the system has great educational value because

the Sutras contain techniques for performing some elementary mathematical

operations in simple ways, and results are obtained quickly'.

Prof. J.N. Kapur says 'Vedic Mathematics can be used to remove math-

phobia, and can be taught to (school) children as enrichment material along

with other high speed methods'.

Dr. Michael Weinless, Chairman of the Department of Mathematics at the

M.I.U, Iowa says thus: 'Vedic Mathematics is easier to learn, faster to use and

less prone to error than conventional methods. Furthermore, the techniques

of Vedic Mathematics not only enable the students to solve specific

mathematical problems; they also develop creativity, logical thinking and

Keeping the above observations in view, let us enter Vedic Mathematics as

given by Sri Bharati Krishna Tirthaji (1884 - 1960), Sankaracharya of

Govardhana Math, Puri. Entering into the methods and procedures, one can

realize the importance and applicability of the different formulae (Sutras) and

4

delightful chapters of the 20th century mathematical history'.

Prof. R.C. Gupta (1994) says 'the system has great educational value because

the Sutras contain techniques for performing some elementary mathematical

operations in simple ways, and results are obtained quickly'.

Prof. J.N. Kapur says 'Vedic Mathematics can be used to remove math-

phobia, and can be taught to (school) children as enrichment material along

with other high speed methods'.

Dr. Michael Weinless, Chairman of the Department of Mathematics at the

M.I.U, Iowa says thus: 'Vedic Mathematics is easier to learn, faster to use and

less prone to error than conventional methods. Furthermore, the techniques

of Vedic Mathematics not only enable the students to solve specific

mathematical problems; they also develop creativity, logical thinking and

Keeping the above observations in view, let us enter Vedic Mathematics as

given by Sri Bharati Krishna Tirthaji (1884 - 1960), Sankaracharya of

Govardhana Math, Puri. Entering into the methods and procedures, one can

realize the importance and applicability of the different formulae (Sutras) and

4

7.
II. Vedic Mathematical Formulae

What we call VEDIC MATHEMATICS is a mathematical elaboration of 'Sixteen

Simple Mathematical formulae from theVedas' as brought out by Sri

Bharati Krishna Tirthaji. In the text authored by the Swamiji, nowhere has

the list of the Mathematical formulae (Sutras) been given. But the Editor of the

text has compiled the list of the formulae from stray references in the text. The

list so compiled contains Sixteen Sutras and Thirteen Sub - Sutras as stated

SIXTEEN SUTRAS

5

What we call VEDIC MATHEMATICS is a mathematical elaboration of 'Sixteen

Simple Mathematical formulae from theVedas' as brought out by Sri

Bharati Krishna Tirthaji. In the text authored by the Swamiji, nowhere has

the list of the Mathematical formulae (Sutras) been given. But the Editor of the

text has compiled the list of the formulae from stray references in the text. The

list so compiled contains Sixteen Sutras and Thirteen Sub - Sutras as stated

SIXTEEN SUTRAS

5

8.
THIRTEEN SUB – SUTRAS

In the text, the words Sutra, aphorism, formula are used synonymously. So are

also the words Upa-sutra, Sub-sutra, Sub-formula, corollary used.

Now we shall have the literal meaning, contextual meaning, process, different

methods of application along with examples for the Sutras. Explanation,

methods, further short-cuts, algebraic proof, etc follow. What follows relates to

a single formula or a group of formulae related to the methods of Vedic

6

In the text, the words Sutra, aphorism, formula are used synonymously. So are

also the words Upa-sutra, Sub-sutra, Sub-formula, corollary used.

Now we shall have the literal meaning, contextual meaning, process, different

methods of application along with examples for the Sutras. Explanation,

methods, further short-cuts, algebraic proof, etc follow. What follows relates to

a single formula or a group of formulae related to the methods of Vedic

6

9.
1. Ekadhikena Purvena

The Sutra (formula) Ekādhikena Pūrvena means: “By one more than the

previous one”.

i) Squares of numbers ending in 5 :

Now we relate the sutra to the ‘squaring of numbers ending in 5’. Consider the

example 252.

Here the number is 25. We have to find out the square of the number. For the

number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more

than the previous one', that is, 2+1=3. The Sutra, in this context, gives the

procedure'to multiply the previous digit 2 by one more than itself, that is, by 3'.

It becomes the L.H.S (left hand side) of the result, that is, 2 X 3 = 6. The R.H.S

(right hand side) of the result is52, that is, 25.

Thus 252 = 2 X 3 / 25 = 625.

In the same way,

352= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;

652= 6 X 7 / 25 = 4225;

1052= 10 X 11/25 = 11025;

1352= 13 X 14/25 = 18225;

Apply the formula to find the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.

Algebraic proof:

a) Consider (ax + b)2 Ξ a2. x2+ 2abx + b2.

Thisidentity for x = 10 and b = 5 becomes

(10a + 5) 2 = a2 . 102 + 2. 10a . 5 + 52

= a2 . 102+ a.102+ 52

= (a2+ a ) . 102 +52

7

The Sutra (formula) Ekādhikena Pūrvena means: “By one more than the

previous one”.

i) Squares of numbers ending in 5 :

Now we relate the sutra to the ‘squaring of numbers ending in 5’. Consider the

example 252.

Here the number is 25. We have to find out the square of the number. For the

number 25, the last digit is 5 and the 'previous' digit is 2. Hence, 'one more

than the previous one', that is, 2+1=3. The Sutra, in this context, gives the

procedure'to multiply the previous digit 2 by one more than itself, that is, by 3'.

It becomes the L.H.S (left hand side) of the result, that is, 2 X 3 = 6. The R.H.S

(right hand side) of the result is52, that is, 25.

Thus 252 = 2 X 3 / 25 = 625.

In the same way,

352= 3 X (3+1) /25 = 3 X 4/ 25 = 1225;

652= 6 X 7 / 25 = 4225;

1052= 10 X 11/25 = 11025;

1352= 13 X 14/25 = 18225;

Apply the formula to find the squares of the numbers 15, 45, 85, 125, 175 and verify the answers.

Algebraic proof:

a) Consider (ax + b)2 Ξ a2. x2+ 2abx + b2.

Thisidentity for x = 10 and b = 5 becomes

(10a + 5) 2 = a2 . 102 + 2. 10a . 5 + 52

= a2 . 102+ a.102+ 52

= (a2+ a ) . 102 +52

7

10.
= a (a + 1) . 102 + 25.

Clearly 10a + 5 represents two-digit numbers 15, 25, 35, -------,95 for the values a = 1, 2, 3, --

-----,9 respectively. In such a case the number (10a + 5)2 is of the form whose L.H.S is a (a + 1)

and R.H.S is 25, that is, a (a + 1) / 25.

Thus any such two digit number gives the result in the same fashion.

Example: 45 = (40 + 5)2, It is of the form (ax+b)2 for a = 4, x=10

and b = 5. giving the answer a (a+1) / 25

that is, 4 (4+1) / 25 + 4 X 5 / 25 = 2025.

b) Any three digit number is of the form ax2+bx+c for x =10, a ≠ 0, a, b, c Є

Now (ax2+bx+ c) 2 = a2 x4 + b2 x2 + c2 + 2abx3 + 2bcx + 2cax2

= a2 x4+2ab.x3+(b2+ 2ca)x2+2bc . x+ c2.

This identity for x = 10, c = 5becomes (a . 102 + b .10 + 5) 2

= a2.104+ 2.a.b.103 + (b2+ 2.5.a)102+ 2.b.5.10 + 52

= a2.104+ 2.a.b.103 + (b2 + 10 a)102 + b.102+ 52

= a2.104+ 2ab.103+ b2.102+ a . 103 + b 102+ 52

= a2.104 + (2ab + a).103 + (b2+ b)102 +52

= [ a2.102 +2ab.10 + a.10 + b2 + b] 102+ 52

= (10a + b) ( 10a+b+1).102 + 25

8

Clearly 10a + 5 represents two-digit numbers 15, 25, 35, -------,95 for the values a = 1, 2, 3, --

-----,9 respectively. In such a case the number (10a + 5)2 is of the form whose L.H.S is a (a + 1)

and R.H.S is 25, that is, a (a + 1) / 25.

Thus any such two digit number gives the result in the same fashion.

Example: 45 = (40 + 5)2, It is of the form (ax+b)2 for a = 4, x=10

and b = 5. giving the answer a (a+1) / 25

that is, 4 (4+1) / 25 + 4 X 5 / 25 = 2025.

b) Any three digit number is of the form ax2+bx+c for x =10, a ≠ 0, a, b, c Є

Now (ax2+bx+ c) 2 = a2 x4 + b2 x2 + c2 + 2abx3 + 2bcx + 2cax2

= a2 x4+2ab.x3+(b2+ 2ca)x2+2bc . x+ c2.

This identity for x = 10, c = 5becomes (a . 102 + b .10 + 5) 2

= a2.104+ 2.a.b.103 + (b2+ 2.5.a)102+ 2.b.5.10 + 52

= a2.104+ 2.a.b.103 + (b2 + 10 a)102 + b.102+ 52

= a2.104+ 2ab.103+ b2.102+ a . 103 + b 102+ 52

= a2.104 + (2ab + a).103 + (b2+ b)102 +52

= [ a2.102 +2ab.10 + a.10 + b2 + b] 102+ 52

= (10a + b) ( 10a+b+1).102 + 25

8

11.
= P (P+1) 102+ 25, where P = 10a+b.

Hence any three digit number whose last digit is 5 gives the same result as in

(a) for P=10a + b, the ‘previous’ of 5.

Example : 1652= (1 . 102 + 6 . 10 + 5) 2.

It is of the form (ax2+bx+c)2for a = 1, b = 6, c = 5 and x = 10. It gives the

answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ‘previous’. The

answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225.

Apply Ekadhikena purvena to find the squares of the numbers 95, 225,

375, 635, 745, 915, 1105, 2545.

ii) Vulgar fractions whose denominators are numbers ending in NINE :

We now take examples of 1 / a9, where a = 1, 2, -----, 9. In the conversion

of such vulgar fractions into recurring decimals, Ekadhika process can be

effectively used both in division and multiplication.

a) Division Method : Value of 1 / 19.

The numbers of decimal places before repetition is the difference of numerator

and denominator, i.e.,, 19 -1=18 places.

For the denominator 19, the purva (previous) is 1.

Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2.

The sutra is applied in a different context. Now the method of division is as

Step. 1 : Divide numerator 1 by 20.

i.e.,, 1 / 20 = 0.1 / 2 = .10 ( 0 times, 1 remainder)

Step. 2 : Divide 10 by 2

9

Hence any three digit number whose last digit is 5 gives the same result as in

(a) for P=10a + b, the ‘previous’ of 5.

Example : 1652= (1 . 102 + 6 . 10 + 5) 2.

It is of the form (ax2+bx+c)2for a = 1, b = 6, c = 5 and x = 10. It gives the

answer P(P+1) / 25, where P = 10a + b = 10 X 1 + 6 = 16, the ‘previous’. The

answer is 16 (16+1) / 25 = 16 X 17 / 25 = 27225.

Apply Ekadhikena purvena to find the squares of the numbers 95, 225,

375, 635, 745, 915, 1105, 2545.

ii) Vulgar fractions whose denominators are numbers ending in NINE :

We now take examples of 1 / a9, where a = 1, 2, -----, 9. In the conversion

of such vulgar fractions into recurring decimals, Ekadhika process can be

effectively used both in division and multiplication.

a) Division Method : Value of 1 / 19.

The numbers of decimal places before repetition is the difference of numerator

and denominator, i.e.,, 19 -1=18 places.

For the denominator 19, the purva (previous) is 1.

Hence Ekadhikena purva (one more than the previous) is 1 + 1 = 2.

The sutra is applied in a different context. Now the method of division is as

Step. 1 : Divide numerator 1 by 20.

i.e.,, 1 / 20 = 0.1 / 2 = .10 ( 0 times, 1 remainder)

Step. 2 : Divide 10 by 2

9

12.
i.e.,, 0.005( 5 times, 0 remainder )

Step. 3 : Divide 5 by 2

i.e.,, 0.0512 ( 2 times, 1 remainder )

Step. 4 : Divide 12 i.e.,, 12 by 2

i.e.,, 0.0526 ( 6 times, No remainder )

Step. 5 : Divide 6 by 2

i.e.,, 0.05263 ( 3 times, No remainder )

Step. 6 : Divide 3 by 2

i.e.,, 0.0526311(1 time, 1 remainder )

Step. 7 : Divide 11 i.e.,, 11 by 2

i.e.,, 0.05263115 (5 times, 1 remainder )

Step. 8 : Divide 15 i.e.,, 15 by 2

i.e.,, 0.052631517 ( 7 times, 1 remainder )

Step. 9 : Divide 17 i.e.,, 17 by 2

i.e.,, 0.05263157 18 (8 times, 1 remainder )

Step. 10 : Divide 18 i.e.,, 18 by 2

i.e.,, 0.0526315789 (9 times, No remainder )

Step. 11 : Divide 9 by 2

i.e.,, 0.0526315789 14 (4 times, 1 remainder )

Step. 12 : Divide 14 i.e.,, 14 by 2

i.e.,, 0.052631578947 ( 7 times, No remainder )

Step. 13 : Divide 7 by 2

10

Step. 3 : Divide 5 by 2

i.e.,, 0.0512 ( 2 times, 1 remainder )

Step. 4 : Divide 12 i.e.,, 12 by 2

i.e.,, 0.0526 ( 6 times, No remainder )

Step. 5 : Divide 6 by 2

i.e.,, 0.05263 ( 3 times, No remainder )

Step. 6 : Divide 3 by 2

i.e.,, 0.0526311(1 time, 1 remainder )

Step. 7 : Divide 11 i.e.,, 11 by 2

i.e.,, 0.05263115 (5 times, 1 remainder )

Step. 8 : Divide 15 i.e.,, 15 by 2

i.e.,, 0.052631517 ( 7 times, 1 remainder )

Step. 9 : Divide 17 i.e.,, 17 by 2

i.e.,, 0.05263157 18 (8 times, 1 remainder )

Step. 10 : Divide 18 i.e.,, 18 by 2

i.e.,, 0.0526315789 (9 times, No remainder )

Step. 11 : Divide 9 by 2

i.e.,, 0.0526315789 14 (4 times, 1 remainder )

Step. 12 : Divide 14 i.e.,, 14 by 2

i.e.,, 0.052631578947 ( 7 times, No remainder )

Step. 13 : Divide 7 by 2

10

13.
i.e.,, 0.05263157894713 ( 3 times, 1 remainder )

Step. 14 : Divide 13 i.e.,, 13 by 2

i.e.,, 0.052631578947316 ( 6 times, 1 remainder )

Step. 15 : Divide 16 i.e.,, 16 by 2

i.e.,, 0.052631578947368 (8 times, No remainder )

Step. 16 : Divide 8 by 2

i.e.,, 0.0526315789473684 ( 4 times, No remainder )

Step. 17 : Divide 4 by 2

i.e.,, 0.05263157894736842 ( 2 times, No remainder )

Step. 18 : Divide 2 by 2

i.e.,, 0.052631578947368421 ( 1 time, No remainder )

Now from step 19, i.e.,, dividing 1 by 2, Step 2 to Step. 18 repeats thus giving

0 __________________ . .

1 / 19 =0.052631578947368421 or 0.052631578947368421

Note that we have completed the process of division only by using ‘2’. Nowhere

the division by 19 occurs.

b) Multiplication Method: Value of 1 / 19

First we recognize the last digit of the denominator of the type 1 / a9. Here the

last digit is 9.

For a fraction of the form in whose denominator 9 is the last digit, we take the

case of 1 / 19 as follows:

For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2.

Therefore 2 is the multiplier for the conversion. We write the last digit in the

numerator as 1 and follow the steps leftwards.

Step. 1 : 1

11

Step. 14 : Divide 13 i.e.,, 13 by 2

i.e.,, 0.052631578947316 ( 6 times, 1 remainder )

Step. 15 : Divide 16 i.e.,, 16 by 2

i.e.,, 0.052631578947368 (8 times, No remainder )

Step. 16 : Divide 8 by 2

i.e.,, 0.0526315789473684 ( 4 times, No remainder )

Step. 17 : Divide 4 by 2

i.e.,, 0.05263157894736842 ( 2 times, No remainder )

Step. 18 : Divide 2 by 2

i.e.,, 0.052631578947368421 ( 1 time, No remainder )

Now from step 19, i.e.,, dividing 1 by 2, Step 2 to Step. 18 repeats thus giving

0 __________________ . .

1 / 19 =0.052631578947368421 or 0.052631578947368421

Note that we have completed the process of division only by using ‘2’. Nowhere

the division by 19 occurs.

b) Multiplication Method: Value of 1 / 19

First we recognize the last digit of the denominator of the type 1 / a9. Here the

last digit is 9.

For a fraction of the form in whose denominator 9 is the last digit, we take the

case of 1 / 19 as follows:

For 1 / 19, 'previous' of 19 is 1. And one more than of it is 1 + 1 = 2.

Therefore 2 is the multiplier for the conversion. We write the last digit in the

numerator as 1 and follow the steps leftwards.

Step. 1 : 1

11

14.
Step. 2 : 21(multiply 1 by 2, put to left)

Step. 3 : 421(multiply 2 by 2, put to left)

Step. 4 : 8421(multiply 4 by 2, put to left)

Step. 5 : 168421 (multiply 8 by 2 =16, 1 carried over, 6 put to left)

Step. 6 : 1368421 ( 6 X 2 =12,+1 [carry over]

= 13, 1 carried over, 3 put to left )

Step. 7 : 7368421 ( 3 X 2, = 6 +1 [Carryover]

= 7, put to left)

Step. 8 : 147368421 (as in the same process)

Step. 9 : 947368421 ( Do – continue to step 18)

Step. 10 : 18947368421

Step. 11 : 178947368421

Step. 12 : 1578947368421

Step. 13 : 11578947368421

Step. 14 : 31578947368421

Step. 15 : 631578947368421

Step. 16 : 12631578947368421

Step. 17 : 52631578947368421

Step. 18 : 1052631578947368421

Now from step 18 onwards the same numbers and order towards left continue.

Thus 1 / 19 = 0.052631578947368421

It is interesting to note that we have

i) not at all used division process

12

Step. 3 : 421(multiply 2 by 2, put to left)

Step. 4 : 8421(multiply 4 by 2, put to left)

Step. 5 : 168421 (multiply 8 by 2 =16, 1 carried over, 6 put to left)

Step. 6 : 1368421 ( 6 X 2 =12,+1 [carry over]

= 13, 1 carried over, 3 put to left )

Step. 7 : 7368421 ( 3 X 2, = 6 +1 [Carryover]

= 7, put to left)

Step. 8 : 147368421 (as in the same process)

Step. 9 : 947368421 ( Do – continue to step 18)

Step. 10 : 18947368421

Step. 11 : 178947368421

Step. 12 : 1578947368421

Step. 13 : 11578947368421

Step. 14 : 31578947368421

Step. 15 : 631578947368421

Step. 16 : 12631578947368421

Step. 17 : 52631578947368421

Step. 18 : 1052631578947368421

Now from step 18 onwards the same numbers and order towards left continue.

Thus 1 / 19 = 0.052631578947368421

It is interesting to note that we have

i) not at all used division process

12

15.
ii) instead of dividing 1 by 19 continuously, just multiplied 1 by 2 and continued to multiply

the resultant successively by 2.

Observations :

a) For any fraction of the form 1 / a9 i.e.,, in whose denominator 9 is the digit in the units place

and a is the set of remaining digits, the value of the fraction is in recurring decimal form and the

repeating block’s right most digit is 1.

b) Whatever may be a9, and the numerator, it is enough to follow the said process with (a+1)

either in division or in multiplication.

c) Starting from right most digit and counting from the right, we see ( in the given example 1 /

Sum of 1st digit + 10th digit = 1 + 8 = 9

Sum of 2nd digit + 11th digit = 2 + 7 = 9

- - - - - - - - -- - - - - - - - - - - - - - - - - - -

Sum of 9th digit + 18th digit = 9+ 0 = 9

From the above observations, we conclude that if we find first 9 digits, further digits can

be derived as complements of 9.

i) Thus at the step 8 in division process we have 0.052631517 and next step. 9 gives

Now the complements of the numbers

0, 5, 2, 6, 3, 1, 5, 7, 8 from 9

9, 4, 7, 3, 6, 8, 4, 2, 1 follow the right order

i.e.,, 0.052631578947368421

Now taking the multiplication process we have

Step. 8 : 147368421

13

the resultant successively by 2.

Observations :

a) For any fraction of the form 1 / a9 i.e.,, in whose denominator 9 is the digit in the units place

and a is the set of remaining digits, the value of the fraction is in recurring decimal form and the

repeating block’s right most digit is 1.

b) Whatever may be a9, and the numerator, it is enough to follow the said process with (a+1)

either in division or in multiplication.

c) Starting from right most digit and counting from the right, we see ( in the given example 1 /

Sum of 1st digit + 10th digit = 1 + 8 = 9

Sum of 2nd digit + 11th digit = 2 + 7 = 9

- - - - - - - - -- - - - - - - - - - - - - - - - - - -

Sum of 9th digit + 18th digit = 9+ 0 = 9

From the above observations, we conclude that if we find first 9 digits, further digits can

be derived as complements of 9.

i) Thus at the step 8 in division process we have 0.052631517 and next step. 9 gives

Now the complements of the numbers

0, 5, 2, 6, 3, 1, 5, 7, 8 from 9

9, 4, 7, 3, 6, 8, 4, 2, 1 follow the right order

i.e.,, 0.052631578947368421

Now taking the multiplication process we have

Step. 8 : 147368421

13

16.
Step. 9 : 947368421

Now the complements of 1, 2, 4, 8, 6, 3, 7, 4, 9 from 9

i.e.,, 8, 7, 5, 1, 3, 6, 2, 5, 0 precede in successive steps, giving the answer.

0.052631578947368421.

d) When we get (Denominator – Numerator) as the product in the multiplicative

process, half the work is done. We stop the multiplication there and

mechanically write the remaining half of the answer by merely taking down

complements from 9.

e) Either division or multiplication process of giving the answer can be put in a

single line form.

Algebraic proof :

Any vulgar fraction of the form 1 / a9 can be written as

1 / a9 = 1 / ( (a + 1 ) x - 1 ) where x = 10

1

= ________________________

(a+1) x [1 - 1/(a+1)x ]

1

= ___________

[1 - 1/(a+1)x]-1

(a+1)x

1

= __________ [1 + 1/(a+1)x + 1/(a+1)x2+ ----------]

(a+1)x

= 1/(a+1)x + 1/(a+1)2x2 +1/(a+1)3x3+ ----ad infinitum

= 10-1(1/(a+1))+10-2(1/(a+1)2)+10-3(1/(a+1)3) + ---ad infinitum

14

Now the complements of 1, 2, 4, 8, 6, 3, 7, 4, 9 from 9

i.e.,, 8, 7, 5, 1, 3, 6, 2, 5, 0 precede in successive steps, giving the answer.

0.052631578947368421.

d) When we get (Denominator – Numerator) as the product in the multiplicative

process, half the work is done. We stop the multiplication there and

mechanically write the remaining half of the answer by merely taking down

complements from 9.

e) Either division or multiplication process of giving the answer can be put in a

single line form.

Algebraic proof :

Any vulgar fraction of the form 1 / a9 can be written as

1 / a9 = 1 / ( (a + 1 ) x - 1 ) where x = 10

1

= ________________________

(a+1) x [1 - 1/(a+1)x ]

1

= ___________

[1 - 1/(a+1)x]-1

(a+1)x

1

= __________ [1 + 1/(a+1)x + 1/(a+1)x2+ ----------]

(a+1)x

= 1/(a+1)x + 1/(a+1)2x2 +1/(a+1)3x3+ ----ad infinitum

= 10-1(1/(a+1))+10-2(1/(a+1)2)+10-3(1/(a+1)3) + ---ad infinitum

14

17.
This series explains the process of ekadhik.

Now consider the problem of 1 / 19. From above we get

1 / 19 = 10-1 (1/(1+1)) + 10-2 (1/(1+1)2) + 10-3 (1/(1+1)3) + ----

( since a=1)

= 10-1 (1/2) + 10-2 (1/2)2 + 10-3 (1/3)3 + ----------

= 10-1 (0.5) + 10-2 (0.25) + 10-3 (0.125)+ ----------

= 0.05 + 0.0025 + 0.000125 + 0.00000625+ - - - - - - -

= 0.052631 - - - - - - -

Example1 :

1. Find 1 / 49 by ekadhikena process.

Now ‘previous’ is 4. ‘One more than the previous’ is 4 + 1 = 5.

Now by division right ward from the left by ‘5’.

1 / 49 = .10 - - - - - - - - - - - -(divide 1 by 50)

= .02 - - - - - - - - - (divide 2 by 5, 0 times, 2 remainder )

= .0220 - - - - - - --(divide 20 by 5, 4 times)

= .0204 - - - - - - -( divide 4 by 5, 0 times, 4 remainder )

= .020440 -- - -- - ( divide 40 by 5, 8 times )

= .020408 - - - - - (divide 8 by 5, 1 time, 3 remainder )

= .02040831 - - - -(divide 31 by 5, 6 times, 1 remainder )

= .02040811 6 - - - - - - - continue

15

Now consider the problem of 1 / 19. From above we get

1 / 19 = 10-1 (1/(1+1)) + 10-2 (1/(1+1)2) + 10-3 (1/(1+1)3) + ----

( since a=1)

= 10-1 (1/2) + 10-2 (1/2)2 + 10-3 (1/3)3 + ----------

= 10-1 (0.5) + 10-2 (0.25) + 10-3 (0.125)+ ----------

= 0.05 + 0.0025 + 0.000125 + 0.00000625+ - - - - - - -

= 0.052631 - - - - - - -

Example1 :

1. Find 1 / 49 by ekadhikena process.

Now ‘previous’ is 4. ‘One more than the previous’ is 4 + 1 = 5.

Now by division right ward from the left by ‘5’.

1 / 49 = .10 - - - - - - - - - - - -(divide 1 by 50)

= .02 - - - - - - - - - (divide 2 by 5, 0 times, 2 remainder )

= .0220 - - - - - - --(divide 20 by 5, 4 times)

= .0204 - - - - - - -( divide 4 by 5, 0 times, 4 remainder )

= .020440 -- - -- - ( divide 40 by 5, 8 times )

= .020408 - - - - - (divide 8 by 5, 1 time, 3 remainder )

= .02040831 - - - -(divide 31 by 5, 6 times, 1 remainder )

= .02040811 6 - - - - - - - continue

15

18.
= .0204081613322615306111222244448 - -- - - - -

On completing 21 digits, we get 48

i.e.,,Denominator - Numerator = 49 – 1 = 48 stands.

i.e, half of the process stops here. The remaining half can be obtained as

complements from 9.

.

Thus 1 / 49 = 0.020408163265306122448

.

979591836734693877551

Now finding 1 / 49 by process of multiplication left ward from right by 5, we

1 / 49 = ----------------------------------------------1

= ---------------------------------------------51

= -------------------------------------------2551

= ------------------------------------------27551

= ---- 483947294594118333617233446943383727551

i.e.,,Denominator – Numerator = 49 – 1 = 48 is obtained as 5X9+3

( Carry over ) = 45 + 3 = 48. Hence half of the process is over. The remaining

half is automatically obtained as complements of 9.

Thus 1 / 49 = ---------------979591836734693877551

.

= 0.020408163265306122448

.

979591836734693877551

16

On completing 21 digits, we get 48

i.e.,,Denominator - Numerator = 49 – 1 = 48 stands.

i.e, half of the process stops here. The remaining half can be obtained as

complements from 9.

.

Thus 1 / 49 = 0.020408163265306122448

.

979591836734693877551

Now finding 1 / 49 by process of multiplication left ward from right by 5, we

1 / 49 = ----------------------------------------------1

= ---------------------------------------------51

= -------------------------------------------2551

= ------------------------------------------27551

= ---- 483947294594118333617233446943383727551

i.e.,,Denominator – Numerator = 49 – 1 = 48 is obtained as 5X9+3

( Carry over ) = 45 + 3 = 48. Hence half of the process is over. The remaining

half is automatically obtained as complements of 9.

Thus 1 / 49 = ---------------979591836734693877551

.

= 0.020408163265306122448

.

979591836734693877551

16

19.
Example 2: Find 1 / 39 by Ekadhika process.

Now by multiplication method, Ekadhikena purva is 3 + 1 = 4

1 / 39 = -------------------------------------1

= -------------------------------------41

= ----------------------------------1641

= ---------------------------------25641

= --------------------------------225641

= -------------------------------1025641

Here the repeating block happens to be block of 6 digits. Now the rule

predicting the completion of half of the computation does not hold. The

complete block has to be computed by ekadhika process.

Now continue and obtain the result. Find reasons for the non–applicability of

the said ‘rule’.

Find the recurring decimal form of the fractions 1 / 29, 1 / 59,

1 / 69, 1 / 79, 1 / 89 using Ekadhika process if possible. Judge whether

the rule of completion of half the computation holds good in such cases.

Note : The Ekadhikena Purvena sutra can also be used for conversion of vulgar

fractions ending in 1, 3, 7 such as 1 / 11, 1 / 21, 1 / 31 - - -- ,1 / 13, 1 / 23, - -

- -, 1 / 7, 1 / 17, - - - - - by writing them in the following way and solving them.

17

Now by multiplication method, Ekadhikena purva is 3 + 1 = 4

1 / 39 = -------------------------------------1

= -------------------------------------41

= ----------------------------------1641

= ---------------------------------25641

= --------------------------------225641

= -------------------------------1025641

Here the repeating block happens to be block of 6 digits. Now the rule

predicting the completion of half of the computation does not hold. The

complete block has to be computed by ekadhika process.

Now continue and obtain the result. Find reasons for the non–applicability of

the said ‘rule’.

Find the recurring decimal form of the fractions 1 / 29, 1 / 59,

1 / 69, 1 / 79, 1 / 89 using Ekadhika process if possible. Judge whether

the rule of completion of half the computation holds good in such cases.

Note : The Ekadhikena Purvena sutra can also be used for conversion of vulgar

fractions ending in 1, 3, 7 such as 1 / 11, 1 / 21, 1 / 31 - - -- ,1 / 13, 1 / 23, - -

- -, 1 / 7, 1 / 17, - - - - - by writing them in the following way and solving them.

17

20.
2. Nikhilam navatascaramam Dasatah

The formula simply means : “all from 9 and the last from 10”

The formula can be very effectively applied in multiplication of numbers, which

are nearer to bases like 10, 100, 1000i.e., to the powers of 10 . The procedure

of multiplication using the Nikhilam involves minimum number of steps, space,

time saving and only mental calculation. The numbers taken can be either less

or more than the base considered.

The difference between the number and the base is termed as deviation.

Deviation may be positive or negative. Positive deviation is written without the

positive sign and the negative deviation, is written using Rekhank (a bar on the

number). Now observe the following table.

Number Base Number – Base Deviation

14 10 14 - 10 4

_

8 10 8 - 10 -2 or 2

__

97 100 97 - 100 -03 or 03

112 100 112 - 100 12

___

993 1000 993 - 1000 -007 or 007

1011 1000 1011 - 1000 011

Some rules of the method (near to the base) in Multiplication

a) Since deviation is obtained by Nikhilam sutra we call the method as Nikhilam

Eg : 94. Now deviation can be obtained by ‘all from 9 and the last from 10’

sutrai.e., the last digit 4 is from 10 and remaining digit 9 from 9 gives 06.

b) The two numbers under consideration are written one below the other. The

deviations are written on the right hand side.

Eg : Multiply 7 by 8.

Now the base is 10. Since it is near to both the numbers,

7

we write the numbers one below the other. 8

18

The formula simply means : “all from 9 and the last from 10”

The formula can be very effectively applied in multiplication of numbers, which

are nearer to bases like 10, 100, 1000i.e., to the powers of 10 . The procedure

of multiplication using the Nikhilam involves minimum number of steps, space,

time saving and only mental calculation. The numbers taken can be either less

or more than the base considered.

The difference between the number and the base is termed as deviation.

Deviation may be positive or negative. Positive deviation is written without the

positive sign and the negative deviation, is written using Rekhank (a bar on the

number). Now observe the following table.

Number Base Number – Base Deviation

14 10 14 - 10 4

_

8 10 8 - 10 -2 or 2

__

97 100 97 - 100 -03 or 03

112 100 112 - 100 12

___

993 1000 993 - 1000 -007 or 007

1011 1000 1011 - 1000 011

Some rules of the method (near to the base) in Multiplication

a) Since deviation is obtained by Nikhilam sutra we call the method as Nikhilam

Eg : 94. Now deviation can be obtained by ‘all from 9 and the last from 10’

sutrai.e., the last digit 4 is from 10 and remaining digit 9 from 9 gives 06.

b) The two numbers under consideration are written one below the other. The

deviations are written on the right hand side.

Eg : Multiply 7 by 8.

Now the base is 10. Since it is near to both the numbers,

7

we write the numbers one below the other. 8

18

21.
-----

Take the deviations of both the numbers from

the base and represent _

7 3

_

Rekhank or the minus sign before the deviations 8 2

------

------

or 7 -3

8 -2

-------

-------

or remainders 3 and 2 implies that the numbers to be multiplied are both less

than 10

c) The product or answer will have two parts, one on the left side and the other

on the right. A vertical or a slant linei.e., a slash may be drawn for the

demarcation of the two parts i.e.,

(or)

d) The R.H.S. of the answer is the product of the deviations of the numbers. It

shall contain the number of digits equal to number of zeroes in the base.

_

i.e., 7 3

_

8 2

_____________

/ (3x2) = 6

Since base is 10, 6 can be taken as it is.

e) L.H.S of the answer is the sum of one number with the deviation of the

other. It can be arrived at in any one of the four ways.

i) Cross-subtract deviation 2 on the second row from the original number7 in

the first row i.e., 7-2 = 5.

ii) Cross–subtract deviation 3 on the first row from the original number8 in the

19

Take the deviations of both the numbers from

the base and represent _

7 3

_

Rekhank or the minus sign before the deviations 8 2

------

------

or 7 -3

8 -2

-------

-------

or remainders 3 and 2 implies that the numbers to be multiplied are both less

than 10

c) The product or answer will have two parts, one on the left side and the other

on the right. A vertical or a slant linei.e., a slash may be drawn for the

demarcation of the two parts i.e.,

(or)

d) The R.H.S. of the answer is the product of the deviations of the numbers. It

shall contain the number of digits equal to number of zeroes in the base.

_

i.e., 7 3

_

8 2

_____________

/ (3x2) = 6

Since base is 10, 6 can be taken as it is.

e) L.H.S of the answer is the sum of one number with the deviation of the

other. It can be arrived at in any one of the four ways.

i) Cross-subtract deviation 2 on the second row from the original number7 in

the first row i.e., 7-2 = 5.

ii) Cross–subtract deviation 3 on the first row from the original number8 in the

19

22.
second row (converse way of(i))

i.e., 8 - 3 = 5

iii) Subtract the base 10 from the sum of the given numbers.

i.e., (7 + 8) – 10 = 5

iv) Subtract the sum of the two deviations from the base.

i.e., 10 – ( 3 + 2) = 5

Hence 5 is left hand side of the answer.

_

Thus 7 3

_

8 2

‾‾‾‾‾‾‾‾‾‾‾‾

5/

Now (d) and (e) together give the solution

_

7 3 7

_

8 2 i.e., X 8

‾‾‾‾‾‾‾ ‾‾‾‾‾‾

5/ 6 56

f) If R.H.S. contains less number of digits than the number of zeros in the base,

the remaining digits are filled up by giving zero or zeroes on the left side of the

R.H.S. If the number of digits are more than the number of zeroes in the base,

the excess digit or digits are to be added to L.H.S of the answer.

The general form of the multiplication under Nikhilam can be shown as follows :

Let N1 and N2 be two numbers near to a given base in powers of 10, andD1 and

D2 are their respective deviations from the base. ThenN1 X N2 can be

represented as

Case (i) : Both the numbers are lower than the base.We have already

considered the example 7 x 8 , with base 10.

20

i.e., 8 - 3 = 5

iii) Subtract the base 10 from the sum of the given numbers.

i.e., (7 + 8) – 10 = 5

iv) Subtract the sum of the two deviations from the base.

i.e., 10 – ( 3 + 2) = 5

Hence 5 is left hand side of the answer.

_

Thus 7 3

_

8 2

‾‾‾‾‾‾‾‾‾‾‾‾

5/

Now (d) and (e) together give the solution

_

7 3 7

_

8 2 i.e., X 8

‾‾‾‾‾‾‾ ‾‾‾‾‾‾

5/ 6 56

f) If R.H.S. contains less number of digits than the number of zeros in the base,

the remaining digits are filled up by giving zero or zeroes on the left side of the

R.H.S. If the number of digits are more than the number of zeroes in the base,

the excess digit or digits are to be added to L.H.S of the answer.

The general form of the multiplication under Nikhilam can be shown as follows :

Let N1 and N2 be two numbers near to a given base in powers of 10, andD1 and

D2 are their respective deviations from the base. ThenN1 X N2 can be

represented as

Case (i) : Both the numbers are lower than the base.We have already

considered the example 7 x 8 , with base 10.

20

23.
Now let us solve some more examples by taking bases 100 and 1000

Ex. 1: Find 97 X 94. Here base is 100. Now following the rules, the working is

as follows:

Ex. 2: 98 X 97 Base is 100.

Ex. 3: 75X95. Base is 100.

Ex. 4: 986 X 989. Base is 1000.

Ex. 5: 994X988. Base is 1000.

21

Ex. 1: Find 97 X 94. Here base is 100. Now following the rules, the working is

as follows:

Ex. 2: 98 X 97 Base is 100.

Ex. 3: 75X95. Base is 100.

Ex. 4: 986 X 989. Base is 1000.

Ex. 5: 994X988. Base is 1000.

21

24.
Ex. 6: 750X995.

Case ( ii) : Both the numbers are higher than the base.

The method and rules follow as they are. The only difference is the positive

deviation. Instead of cross – subtract, we follow cross – add.

Ex. 7: 13X12. Base is 10

Ex. 8: 18X14. Base is 10.

Ex. 9: 104X102. Base is 100.

104 04

102 02

‾‾‾‾‾‾‾‾‾‾‾‾

106 / 4x2 = 10608 ( rule -f )

‾‾‾‾‾‾‾‾‾‾‾‾

Ex. 10: 1275X1004. Base is 1000.

22

Case ( ii) : Both the numbers are higher than the base.

The method and rules follow as they are. The only difference is the positive

deviation. Instead of cross – subtract, we follow cross – add.

Ex. 7: 13X12. Base is 10

Ex. 8: 18X14. Base is 10.

Ex. 9: 104X102. Base is 100.

104 04

102 02

‾‾‾‾‾‾‾‾‾‾‾‾

106 / 4x2 = 10608 ( rule -f )

‾‾‾‾‾‾‾‾‾‾‾‾

Ex. 10: 1275X1004. Base is 1000.

22

25.
1275 275

1004 004

‾‾‾‾‾‾‾‾‾‾‾‾

1279/ 275x4 = 1279 / 1100 ( rule -f )

____________ = 1280100

Case ( iii ): One number is more and the other is less than the base.

In this situation one deviation is positive and the other is negative. So the

product of deviations becomes negative. So the right hand side of the answer

obtained will therefore have to be subtracted. To have a clear representation

and understanding a vinculum is used. It proceeds into normalization.

Ex.11: 13X7. Base is 10

Note : Conversion of common number into vinculum number and vice versa.

Eg :

__

9 = 10 –1 = 11

_

98 = 100 – 2 = 102

_

196 = 200 – 4 = 204

_

32 = 30 – 2 = 28

_

145 = 140 – 5 = 135

_

322 = 300 – 22 = 278. etc

The procedure can be explained in detail using Nikhilam Navatascaram Dasatah,

Ekadhikenapurvena, Ekanyunena purvena in the foregoing pages of this book.]

Ex. 12: 108 X 94. Base is 100.

23

1004 004

‾‾‾‾‾‾‾‾‾‾‾‾

1279/ 275x4 = 1279 / 1100 ( rule -f )

____________ = 1280100

Case ( iii ): One number is more and the other is less than the base.

In this situation one deviation is positive and the other is negative. So the

product of deviations becomes negative. So the right hand side of the answer

obtained will therefore have to be subtracted. To have a clear representation

and understanding a vinculum is used. It proceeds into normalization.

Ex.11: 13X7. Base is 10

Note : Conversion of common number into vinculum number and vice versa.

Eg :

__

9 = 10 –1 = 11

_

98 = 100 – 2 = 102

_

196 = 200 – 4 = 204

_

32 = 30 – 2 = 28

_

145 = 140 – 5 = 135

_

322 = 300 – 22 = 278. etc

The procedure can be explained in detail using Nikhilam Navatascaram Dasatah,

Ekadhikenapurvena, Ekanyunena purvena in the foregoing pages of this book.]

Ex. 12: 108 X 94. Base is 100.

23

26.
Ex. 13: 998 X 1025. Base is 1000.

Algebraic Proof:

Case ( i ):

Let the two numbers N1 and N2 be less than the selected base say x.

N1 = (x-a), N2 = (x-b). Here a and b are the corresponding deviations of the

numbersN1 and N2 from the base x. Observe that x is a multiple of 10.

Now N1 X N2 = (x-a) (x-b) = x.x – x.b – a.x + ab

= x (x – a – b ) + ab. [rule – e(iv), d ]

= x [(x – a) – b] + ab = x (N1–b) + ab[rule–e(i),d]

or = x [(x – b) – a] = x (N2 – a) + ab. [rule –e (ii),d]

x (x – a – b) + ab can also be written as

x[(x – a) + (x – b) – x] + ab = x[N1+N2 – x] + ab [rule –e(iii),d].

A difficult can be faced, if the vertical multiplication of the deficit digits or

deviationsi.e., a.b yields a product consisting of more than the required digits.

Then rule-f will enable us to surmount the difficulty.

Case ( ii ) :

When both the numbers exceed the selected base, we have N1 = x + a,N2 = x +

b, x being the base. Now the identity (x+a) (x+b) = x(x+a+b) + a.b holds

good, of course with relevant details mentioned in case(i).

24

Algebraic Proof:

Case ( i ):

Let the two numbers N1 and N2 be less than the selected base say x.

N1 = (x-a), N2 = (x-b). Here a and b are the corresponding deviations of the

numbersN1 and N2 from the base x. Observe that x is a multiple of 10.

Now N1 X N2 = (x-a) (x-b) = x.x – x.b – a.x + ab

= x (x – a – b ) + ab. [rule – e(iv), d ]

= x [(x – a) – b] + ab = x (N1–b) + ab[rule–e(i),d]

or = x [(x – b) – a] = x (N2 – a) + ab. [rule –e (ii),d]

x (x – a – b) + ab can also be written as

x[(x – a) + (x – b) – x] + ab = x[N1+N2 – x] + ab [rule –e(iii),d].

A difficult can be faced, if the vertical multiplication of the deficit digits or

deviationsi.e., a.b yields a product consisting of more than the required digits.

Then rule-f will enable us to surmount the difficulty.

Case ( ii ) :

When both the numbers exceed the selected base, we have N1 = x + a,N2 = x +

b, x being the base. Now the identity (x+a) (x+b) = x(x+a+b) + a.b holds

good, of course with relevant details mentioned in case(i).

24

27.
Case ( iii ) :

When one number is less and another is more than the base, we can use (x-

a)(x+b) = x(x–a+ b)–ab. and the procedure is evident from the examples

Find the following products by Nikhilam formula.

1) 7 X 4 2) 93 X 85 3) 875 X 994

4) 1234 X 1002 5) 1003 X 997 6) 11112 X 9998

7) 1234 X 1002 8) 118 X 105

Nikhilam in Division

Consider some two digit numbers (dividends) and same divisor 9. Observe the

following example.

i) 13 ÷ 9 The quotient (Q) is 1, Remainder (R) is 4.

since 9 ) 13 ( 1

9

____

4

ii) 34 ÷ 9, Q is 3, R is 7.

iii) 60 ÷ 9, Q is 6, R is 6.

iv) 80 ÷ 9, Q is 8, R is 8.

Now we have another type of representation for the above examples as given

i) Split each dividend into a left hand part for the Quotient and right - hand part

for the remainder by a slant line or slash.

Eg. 13 as 1 / 3, 34 as 3 / 4 , 80 as 8 / 0.

ii) Leave some space below such representation, draw a horizontal line.

25

When one number is less and another is more than the base, we can use (x-

a)(x+b) = x(x–a+ b)–ab. and the procedure is evident from the examples

Find the following products by Nikhilam formula.

1) 7 X 4 2) 93 X 85 3) 875 X 994

4) 1234 X 1002 5) 1003 X 997 6) 11112 X 9998

7) 1234 X 1002 8) 118 X 105

Nikhilam in Division

Consider some two digit numbers (dividends) and same divisor 9. Observe the

following example.

i) 13 ÷ 9 The quotient (Q) is 1, Remainder (R) is 4.

since 9 ) 13 ( 1

9

____

4

ii) 34 ÷ 9, Q is 3, R is 7.

iii) 60 ÷ 9, Q is 6, R is 6.

iv) 80 ÷ 9, Q is 8, R is 8.

Now we have another type of representation for the above examples as given

i) Split each dividend into a left hand part for the Quotient and right - hand part

for the remainder by a slant line or slash.

Eg. 13 as 1 / 3, 34 as 3 / 4 , 80 as 8 / 0.

ii) Leave some space below such representation, draw a horizontal line.

25

28.
Eg. 1/3 3/4 8/0

______ , ______ , ______

iii) Put the first digit of the dividend as it is under the horizontal line. Put the

same digit under the right hand part for the remainder, add the two and place

the sumi.e., sum of the digits of the numbers as the remainder.

1/3 3/4 8/0

1 3 8

______ , ______ , ______

1/4 3/7 8/8

Now the problem is over. i.e.,

13 ÷ 9 gives Q = 1, R = 4

34 ÷ 9 gives Q = 3, R = 7

80 ÷ 9 gives Q = 8, R = 8

Proceeding for some more of the two digit number division by 9, we get

a) 21 ÷ 9 as

9) 2 / 1 i.e Q=2, R=3

2

‾‾‾‾‾‾

2 / 3

b) 43 ÷ 9 as

9) 4 / 3 i.e Q = 4, R = 7.

4

‾‾‾‾‾‾

4 / 7

The examples given so far convey that in the division of two digit numbers by 9,

we canmechanically take the first digit down for the quotient – column and that,

by adding the quotient to the second digit, we can get the remainder.

Now in the case of 3 digit numbers, let us proceed as follows.

9 ) 104 ( 11 9 ) 10 / 4

99 1 / 1

26

______ , ______ , ______

iii) Put the first digit of the dividend as it is under the horizontal line. Put the

same digit under the right hand part for the remainder, add the two and place

the sumi.e., sum of the digits of the numbers as the remainder.

1/3 3/4 8/0

1 3 8

______ , ______ , ______

1/4 3/7 8/8

Now the problem is over. i.e.,

13 ÷ 9 gives Q = 1, R = 4

34 ÷ 9 gives Q = 3, R = 7

80 ÷ 9 gives Q = 8, R = 8

Proceeding for some more of the two digit number division by 9, we get

a) 21 ÷ 9 as

9) 2 / 1 i.e Q=2, R=3

2

‾‾‾‾‾‾

2 / 3

b) 43 ÷ 9 as

9) 4 / 3 i.e Q = 4, R = 7.

4

‾‾‾‾‾‾

4 / 7

The examples given so far convey that in the division of two digit numbers by 9,

we canmechanically take the first digit down for the quotient – column and that,

by adding the quotient to the second digit, we can get the remainder.

Now in the case of 3 digit numbers, let us proceed as follows.

9 ) 104 ( 11 9 ) 10 / 4

99 1 / 1

26

29.
‾‾‾‾‾‾ as ‾‾‾‾‾‾‾

5 11 / 5

9 ) 212 ( 23 9 ) 21 / 2

207 2 / 3

‾‾‾‾‾ as ‾‾‾‾‾‾‾

5 23 / 5

9 ) 401 (44 9 ) 40 / 1

396 4 / 4

‾‾‾‾‾ as ‾‾‾‾‾‾‾‾

5 44 / 5

Note that the remainder is the sum of the digits of the dividend. The first digit

of the dividend from left is added mechanically to the second digit of the

dividend to obtain the second digit of the quotient. This digit added to the third

digit sets the remainder. The first digit of the dividend remains as the first digit

of the quotient.

Consider 511 ÷ 9

Add the first digit 5 to second digit 1 getting 5 + 1 = 6. Hence Quotient is 56.

Now second digit of 56 i.e., 6 is added to third digit 1 of dividend to get the

remainder i.e., 1 + 6 = 7

9) 51 / 1

5/ 6

‾‾‾‾‾‾‾

56 / 7

Q is 56, R is 7.

Extending the same principle even to bigger numbers of still more digits, we can

get the results.

Eg : 1204 ÷ 9

i) Add first digit 1 to the second digit 2. 1 + 2 = 3

ii) Add the second digit of quotient 13. i.e., 3 to third digit ‘0’ and obtain the

Quotient. 3 + 0 = 3, 133

27

5 11 / 5

9 ) 212 ( 23 9 ) 21 / 2

207 2 / 3

‾‾‾‾‾ as ‾‾‾‾‾‾‾

5 23 / 5

9 ) 401 (44 9 ) 40 / 1

396 4 / 4

‾‾‾‾‾ as ‾‾‾‾‾‾‾‾

5 44 / 5

Note that the remainder is the sum of the digits of the dividend. The first digit

of the dividend from left is added mechanically to the second digit of the

dividend to obtain the second digit of the quotient. This digit added to the third

digit sets the remainder. The first digit of the dividend remains as the first digit

of the quotient.

Consider 511 ÷ 9

Add the first digit 5 to second digit 1 getting 5 + 1 = 6. Hence Quotient is 56.

Now second digit of 56 i.e., 6 is added to third digit 1 of dividend to get the

remainder i.e., 1 + 6 = 7

9) 51 / 1

5/ 6

‾‾‾‾‾‾‾

56 / 7

Q is 56, R is 7.

Extending the same principle even to bigger numbers of still more digits, we can

get the results.

Eg : 1204 ÷ 9

i) Add first digit 1 to the second digit 2. 1 + 2 = 3

ii) Add the second digit of quotient 13. i.e., 3 to third digit ‘0’ and obtain the

Quotient. 3 + 0 = 3, 133

27

30.
iii) Add the third digit of Quotient 133 i.e.,3 to last digit ‘4’ of the dividend and

write the final Quotient and Remainder. R = 3 + 4 = 7, Q = 133

In symbolic form 9 ) 120 / 4

13 / 3

‾‾‾‾‾‾‾‾

133 / 7

Another example.

9 ) 13210 /1 132101 ÷ 9

gives

1467 / 7 Q = 14677, R = 8

‾‾‾‾‾‾‾‾‾‾

14677 / 8

In all the cases mentioned above, the remainder is less than the divisor. What

about the case when the remainder is equal or greater than the divisor?

9) 3 / 6 9) 24 / 6

3 2 / 6

‾‾‾‾‾‾ or ‾‾‾‾‾‾‾‾

3 / 9 (equal) 26 / 12 (greater).

We proceed by re-dividing the remainder by 9, carrying over this Quotient to

the quotient side and retaining the final remainder in the remainder side.

9) 3 / 6 9 ) 24 / 6

/ 3 2 / 6

‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾

3 / 9 26 / 12

‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾

4 / 0 27 / 3

Q = 4, R=0 Q = 27, R = 3.

When the remainder is greater than divisor, it can also be represented as

9 ) 24 / 6

2 / 6

‾‾‾‾‾‾‾‾

26 /1 / 2

/1

‾‾‾‾‾‾‾‾

1 /3

28

write the final Quotient and Remainder. R = 3 + 4 = 7, Q = 133

In symbolic form 9 ) 120 / 4

13 / 3

‾‾‾‾‾‾‾‾

133 / 7

Another example.

9 ) 13210 /1 132101 ÷ 9

gives

1467 / 7 Q = 14677, R = 8

‾‾‾‾‾‾‾‾‾‾

14677 / 8

In all the cases mentioned above, the remainder is less than the divisor. What

about the case when the remainder is equal or greater than the divisor?

9) 3 / 6 9) 24 / 6

3 2 / 6

‾‾‾‾‾‾ or ‾‾‾‾‾‾‾‾

3 / 9 (equal) 26 / 12 (greater).

We proceed by re-dividing the remainder by 9, carrying over this Quotient to

the quotient side and retaining the final remainder in the remainder side.

9) 3 / 6 9 ) 24 / 6

/ 3 2 / 6

‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾

3 / 9 26 / 12

‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾

4 / 0 27 / 3

Q = 4, R=0 Q = 27, R = 3.

When the remainder is greater than divisor, it can also be represented as

9 ) 24 / 6

2 / 6

‾‾‾‾‾‾‾‾

26 /1 / 2

/1

‾‾‾‾‾‾‾‾

1 /3

28

31.
‾‾‾‾‾‾‾‾

27 / 3

Now consider the divisors of two or more digits whose last digit is 9,when

divisor is 89.

We Know 113 = 1 X 89 + 24, Q =1, R = 24

10015 = 112 X 89 + 47, Q = 112, R = 47.

Representing in the previous form of procedure, we have

89 ) 1 / 13 89 ) 100 / 15

/ 11 12 / 32

‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾

1 / 24 112 / 47

But how to get these? What is the procedure?

Now Nikhilam rule comes to rescue us. The nikhilam states “all from 9 and the

last from 10”. Now if you want to find 113 ÷ 89, 10015 ÷ 89, you have to apply

nikhilam formula on 89 and get the complement 11.Further while carrying the

added numbers to the place below the next digit, we have to multiply by this

89 ) 1 / 13 89 ) 100 /15

‾‾

/ 11 11 11 / first digit 1 x 11

‾‾‾‾‾‾‾‾

1 / 24 1/ 1 total second is 1x11

22 total of 3rd digit is 2 x 11

‾‾‾‾‾‾‾‾‾‾

112 / 47

What is 10015 ÷ 98 ? Apply Nikhilam and get 100 – 98 = 02. Set off the 2

digits from the right as the remainder consists of 2 digits. While carrying the

added numbers to the place below the next digit, multiply by 02.

98 ) 100 / 15

‾‾

02 02 / i.e., 10015 ÷ 98 gives

0 / 0 Q = 102, R = 19

/ 04

‾‾‾‾‾‾‾‾‾‾

102 / 19

29

27 / 3

Now consider the divisors of two or more digits whose last digit is 9,when

divisor is 89.

We Know 113 = 1 X 89 + 24, Q =1, R = 24

10015 = 112 X 89 + 47, Q = 112, R = 47.

Representing in the previous form of procedure, we have

89 ) 1 / 13 89 ) 100 / 15

/ 11 12 / 32

‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾

1 / 24 112 / 47

But how to get these? What is the procedure?

Now Nikhilam rule comes to rescue us. The nikhilam states “all from 9 and the

last from 10”. Now if you want to find 113 ÷ 89, 10015 ÷ 89, you have to apply

nikhilam formula on 89 and get the complement 11.Further while carrying the

added numbers to the place below the next digit, we have to multiply by this

89 ) 1 / 13 89 ) 100 /15

‾‾

/ 11 11 11 / first digit 1 x 11

‾‾‾‾‾‾‾‾

1 / 24 1/ 1 total second is 1x11

22 total of 3rd digit is 2 x 11

‾‾‾‾‾‾‾‾‾‾

112 / 47

What is 10015 ÷ 98 ? Apply Nikhilam and get 100 – 98 = 02. Set off the 2

digits from the right as the remainder consists of 2 digits. While carrying the

added numbers to the place below the next digit, multiply by 02.

98 ) 100 / 15

‾‾

02 02 / i.e., 10015 ÷ 98 gives

0 / 0 Q = 102, R = 19

/ 04

‾‾‾‾‾‾‾‾‾‾

102 / 19

29

32.
In the same way

897 ) 11 / 422

‾‾‾

103 1 / 03

/ 206

‾‾‾‾‾‾‾‾‾

12 / 658

gives 11,422 ÷ 897, Q = 12, R=658.

In this way we have to multiply the quotient by 2 in the case of 8, by 3 in the

case of 7, by 4 in the case of 6 and so on. i.e., multiply the Quotient digit by the

divisors complement from 10. In case of more digited numbers we apply

Nikhilam and proceed. Any how, this method is highly useful and effective for

division when the numbers are near to bases of 10.

* Guess the logic in the process of division by 9.

* Obtain the Quotient and Remainder for the following problems.

1) 311 ÷ 9 2) 120012 ÷ 9 3) 1135 ÷ 97

4) 2342 ÷ 98 5) 113401 ÷ 997

6) 11199171 ÷ 99979

Observe that by nikhilam process of division, even lengthier divisions involve no

division or no subtraction but only a few multiplications of single digits with

small numbers and a simple addition. But we know fairly well that only a special

type of cases are being dealt and hence many questions about various other

types of problems arise. The answer lies in Vedic Methods.

30

897 ) 11 / 422

‾‾‾

103 1 / 03

/ 206

‾‾‾‾‾‾‾‾‾

12 / 658

gives 11,422 ÷ 897, Q = 12, R=658.

In this way we have to multiply the quotient by 2 in the case of 8, by 3 in the

case of 7, by 4 in the case of 6 and so on. i.e., multiply the Quotient digit by the

divisors complement from 10. In case of more digited numbers we apply

Nikhilam and proceed. Any how, this method is highly useful and effective for

division when the numbers are near to bases of 10.

* Guess the logic in the process of division by 9.

* Obtain the Quotient and Remainder for the following problems.

1) 311 ÷ 9 2) 120012 ÷ 9 3) 1135 ÷ 97

4) 2342 ÷ 98 5) 113401 ÷ 997

6) 11199171 ÷ 99979

Observe that by nikhilam process of division, even lengthier divisions involve no

division or no subtraction but only a few multiplications of single digits with

small numbers and a simple addition. But we know fairly well that only a special

type of cases are being dealt and hence many questions about various other

types of problems arise. The answer lies in Vedic Methods.

30

33.
3. Urdhva - tiryagbhyam

Urdhva – tiryagbhyam is the general formula applicable to all cases of

multiplication and also in the division of a large number by another large

number. It means

(a) Multiplication of two 2 digit numbers.

Ex.1: Find the product 14 X 12

i) The right hand most digit of the multiplicand, the first number (14) i.e.,4 is

multiplied by the right hand most digit of the multiplier, the second number

(12)i.e., 2. The product 4 X 2 = 8 forms the right hand most part of the answer.

ii) Now, diagonally multiply the first digit of the multiplicand (14) i.e., 4 and

second digit of the multiplier (12)i.e., 1 (answer 4 X 1=4); then multiply the

second digit of the multiplicand i.e.,1 and first digit of the multiplier i.e., 2

(answer 1 X 2 = 2); add these two i.e.,4 + 2 = 6. It gives the next, i.e., second

digit of the answer. Hence second digit of the answer is 6.

iii) Now, multiply the second digit of the multiplicand i.e., 1 and second digit of

the multiplieri.e., 1 vertically, i.e., 1 X 1 = 1. It gives the left hand most part of

the answer.

Thus the answer is 16 8.

Symbolically we can represent the process as follows :

The symbols are operated from right to left .

Step i) :

31

Urdhva – tiryagbhyam is the general formula applicable to all cases of

multiplication and also in the division of a large number by another large

number. It means

(a) Multiplication of two 2 digit numbers.

Ex.1: Find the product 14 X 12

i) The right hand most digit of the multiplicand, the first number (14) i.e.,4 is

multiplied by the right hand most digit of the multiplier, the second number

(12)i.e., 2. The product 4 X 2 = 8 forms the right hand most part of the answer.

ii) Now, diagonally multiply the first digit of the multiplicand (14) i.e., 4 and

second digit of the multiplier (12)i.e., 1 (answer 4 X 1=4); then multiply the

second digit of the multiplicand i.e.,1 and first digit of the multiplier i.e., 2

(answer 1 X 2 = 2); add these two i.e.,4 + 2 = 6. It gives the next, i.e., second

digit of the answer. Hence second digit of the answer is 6.

iii) Now, multiply the second digit of the multiplicand i.e., 1 and second digit of

the multiplieri.e., 1 vertically, i.e., 1 X 1 = 1. It gives the left hand most part of

the answer.

Thus the answer is 16 8.

Symbolically we can represent the process as follows :

The symbols are operated from right to left .

Step i) :

31

34.
Step ii) :

Step iii) :

Now in the same process, answer can be written as

23

13

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

2 : 6 + 3 : 9 = 299 (Recall the 3 steps)

41

X 41

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

16 : 4 + 4 : 1 = 1681.

32

Step iii) :

Now in the same process, answer can be written as

23

13

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

2 : 6 + 3 : 9 = 299 (Recall the 3 steps)

41

X 41

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

16 : 4 + 4 : 1 = 1681.

32

35.
What happens when one of the results i.e., either in the last digit or in the

middle digit of the result, contains more than 1 digit ? Answer is simple. The

right – hand – most digit there of is to be put down there and the

preceding,i.e., left –hand –side digit or digits should be carried over to the left

and placed under the previous digit or digits of the upper row. The digits carried

over may be written as in Ex. 4.

Ex.4: 32 X 24

Step (i) : 2X4=8

Step (ii) : 3 X 4 = 12; 2 X 2 = 4; 12 + 4 = 16.

Here 6 is to be retained. 1 is to be carried out to left side.

Step (iii) : 3 X 2 = 6. Now the carried over digit 1 of 16 is to be added.

i.e., 6 + 1 = 7.

Thus 32 X 24 = 768

We can write it as follows

32

24

‾‾‾‾

668

1

‾‾‾‾

768.

33

middle digit of the result, contains more than 1 digit ? Answer is simple. The

right – hand – most digit there of is to be put down there and the

preceding,i.e., left –hand –side digit or digits should be carried over to the left

and placed under the previous digit or digits of the upper row. The digits carried

over may be written as in Ex. 4.

Ex.4: 32 X 24

Step (i) : 2X4=8

Step (ii) : 3 X 4 = 12; 2 X 2 = 4; 12 + 4 = 16.

Here 6 is to be retained. 1 is to be carried out to left side.

Step (iii) : 3 X 2 = 6. Now the carried over digit 1 of 16 is to be added.

i.e., 6 + 1 = 7.

Thus 32 X 24 = 768

We can write it as follows

32

24

‾‾‾‾

668

1

‾‾‾‾

768.

33

36.
Note that the carried over digit from the result (3X4) + (2X2) = 12+4 = 16

i.e.,1 is placed under the previous digit 3 X 2 = 6 and added.

After sufficient practice, you feel no necessity of writing in this way and simply

operate or perform mentally.

Ex.5 28 X 35.

Step (i) : 8 X 5 = 40. 0 is retained as the first digit of the answer and 4 is

carried over.

Step (ii) : 2 X 5 = 10; 8 X 3 = 24; 10 + 24 = 34; add the carried over 4 to

34. Now the result is 34 + 4 = 38. Now 8 is retained as the second digit of the

answer and3 is carried over.

Step (iii) : 2 X 3 = 6; add the carried over 3 to 6. The result 6 + 3 = 9 is the

third or final digit from right to left of the answer.

Thus 28 X 35 = 980.

48

47

‾‾‾‾‾‾

1606

65

‾‾‾‾‾‾‾

2256

Step (i): 8 X 7 = 56; 5, the carried over digit is placed below the second

Step (ii): ( 4 X 7) + (8 X 4) = 28 + 32 = 60; 6, the carried over digit is

placed below the third digit.

Step (iii): Respective digits are added.

Algebraic proof :

a) Let the two 2 digit numbers be (ax+b) and (cx+d). Note that x = 10. Now

consider the product

34

i.e.,1 is placed under the previous digit 3 X 2 = 6 and added.

After sufficient practice, you feel no necessity of writing in this way and simply

operate or perform mentally.

Ex.5 28 X 35.

Step (i) : 8 X 5 = 40. 0 is retained as the first digit of the answer and 4 is

carried over.

Step (ii) : 2 X 5 = 10; 8 X 3 = 24; 10 + 24 = 34; add the carried over 4 to

34. Now the result is 34 + 4 = 38. Now 8 is retained as the second digit of the

answer and3 is carried over.

Step (iii) : 2 X 3 = 6; add the carried over 3 to 6. The result 6 + 3 = 9 is the

third or final digit from right to left of the answer.

Thus 28 X 35 = 980.

48

47

‾‾‾‾‾‾

1606

65

‾‾‾‾‾‾‾

2256

Step (i): 8 X 7 = 56; 5, the carried over digit is placed below the second

Step (ii): ( 4 X 7) + (8 X 4) = 28 + 32 = 60; 6, the carried over digit is

placed below the third digit.

Step (iii): Respective digits are added.

Algebraic proof :

a) Let the two 2 digit numbers be (ax+b) and (cx+d). Note that x = 10. Now

consider the product

34

37.
(ax + b) (cx + d) = ac.x2 + adx + bcx + b.d

= ac.x2 + (ad + bc)x + b.d

Observe that

i) The first term i.e., the coefficient of x2 (i.e., 100, hence the digit in the100th

place) is obtained by vertical multiplication of a and c i.e.,the digits in10th place

(coefficient of x) of both the numbers;

ii) The middle term, i.e., the coefficient of x (i.e., digit in the 10th place) is

obtained by cross wise multiplication of a and d; and of b and c; and the

addition of the two products;

iii) The last (independent of x) term is obtained by vertical multiplication of the

independent terms b and d.

b) Consider the multiplication of two 3 digit numbers.

Let the two numbers be(ax2 + bx + c) and (dx2 + ex + f). Note that x=10

Now the product is

ax2 + bx + c

x dx2 + ex + f

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

ad.x4+bd.x3+cd.x2+ae.x3+be.x2+ce.x+af.x2+bf.x+cf

= ad.x4 + (bd + ae). x3 + (cd + be + af).x2 + (ce + bf)x + cf

35

= ac.x2 + (ad + bc)x + b.d

Observe that

i) The first term i.e., the coefficient of x2 (i.e., 100, hence the digit in the100th

place) is obtained by vertical multiplication of a and c i.e.,the digits in10th place

(coefficient of x) of both the numbers;

ii) The middle term, i.e., the coefficient of x (i.e., digit in the 10th place) is

obtained by cross wise multiplication of a and d; and of b and c; and the

addition of the two products;

iii) The last (independent of x) term is obtained by vertical multiplication of the

independent terms b and d.

b) Consider the multiplication of two 3 digit numbers.

Let the two numbers be(ax2 + bx + c) and (dx2 + ex + f). Note that x=10

Now the product is

ax2 + bx + c

x dx2 + ex + f

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

ad.x4+bd.x3+cd.x2+ae.x3+be.x2+ce.x+af.x2+bf.x+cf

= ad.x4 + (bd + ae). x3 + (cd + be + af).x2 + (ce + bf)x + cf

35

38.
Note the following points :

i) The coefficient of x4 , i.e., ad is obtained by the vertical multiplication of the

firstcoefficient from the left side :

ii)The coefficient of x3 , i.e., (ae + bd) is obtained by the cross –wise

multiplication of the first two coefficients and by the addition of the two

iii) The coefficient of x2 is obtained by the multiplication of the first coefficient

of the multiplicand(ax2+bx +c) i.e., a; by the last coefficient of the multiplier

(dx2 +ex +f)i.e.,f ; of the middle one i.e., b of the multiplicand by the middle

one i.e., e of the multiplier and of the last onei.e., c of the multiplicand by the

first one i.e., d of the multiplier and by the addition of all the three productsi.e.,

af + be +cd :

iv) The coefficient of x is obtained by the cross wise multiplication of the second

coefficienti.e., b of the multiplicand by the third one i.e., f of the multiplier, and

conversely the third coefficienti.e., c of the multiplicand by the second

coefficient i.e., e of the multiplier and by addition of the two products,i.e., bf +

ce ;

36

i) The coefficient of x4 , i.e., ad is obtained by the vertical multiplication of the

firstcoefficient from the left side :

ii)The coefficient of x3 , i.e., (ae + bd) is obtained by the cross –wise

multiplication of the first two coefficients and by the addition of the two

iii) The coefficient of x2 is obtained by the multiplication of the first coefficient

of the multiplicand(ax2+bx +c) i.e., a; by the last coefficient of the multiplier

(dx2 +ex +f)i.e.,f ; of the middle one i.e., b of the multiplicand by the middle

one i.e., e of the multiplier and of the last onei.e., c of the multiplicand by the

first one i.e., d of the multiplier and by the addition of all the three productsi.e.,

af + be +cd :

iv) The coefficient of x is obtained by the cross wise multiplication of the second

coefficienti.e., b of the multiplicand by the third one i.e., f of the multiplier, and

conversely the third coefficienti.e., c of the multiplicand by the second

coefficient i.e., e of the multiplier and by addition of the two products,i.e., bf +

ce ;

36

39.
v) And finally the last (independent of x) term is obtained by the vertical

multiplication of the last coefficients c and f i.e., cf

Thus the process can be put symbolically as (from left to right)

Consider the following example

124 X 132.

Proceeding from right to left

i) 4 X 2 = 8. First digit = 8

ii) (2 X 2) + (3 X 4) = 4 + 12 = 16. The digit 6 is retained and 1 is carried over

to left side. Second digit = 6.

iii) (1 X 2) + (2 X 3) + (1 X 4) = 2 + 6 + 4 =12. The carried over 1 of above

step is added i.e., 12 + 1 = 13. Now 3 is retained and 1 is carried over to left

37

multiplication of the last coefficients c and f i.e., cf

Thus the process can be put symbolically as (from left to right)

Consider the following example

124 X 132.

Proceeding from right to left

i) 4 X 2 = 8. First digit = 8

ii) (2 X 2) + (3 X 4) = 4 + 12 = 16. The digit 6 is retained and 1 is carried over

to left side. Second digit = 6.

iii) (1 X 2) + (2 X 3) + (1 X 4) = 2 + 6 + 4 =12. The carried over 1 of above

step is added i.e., 12 + 1 = 13. Now 3 is retained and 1 is carried over to left

37

40.
side. Thus third digit = 3.

iv) ( 1X 3 ) + ( 2 X 1 ) = 3 + 2 = 5. the carried over 1 of above step is added

i.e., 5 + 1 = 6 . It is retained. Thus fourth digit = 6

v) ( 1 X 1 ) = 1. As there is no carried over number from the previous step it is

retained. Thus fifth digit = 1

124 X 132 = 16368.

Let us work another problem by placing the carried over digits under the first

row and proceed.

234

x 316

‾‾‾‾‾‾‾

61724

1222

‾‾‾‾‾‾‾

73944

i) 4 X 6 = 24 : 2, the carried over digit is placed below the second digit.

ii) (3 X 6) + (4 x 1) = 18 + 4 = 22 ; 2, the carried over digit is placed below

third digit.

iii) (2 X 6) + (3 X 1) + (4 X 3) = 12 + 3 + 12 = 27 ; 2, the carried over digit is

placed below fourth digit.

iv) (2 X 1) + ( 3 X 3) = 2 + 9 = 11; 1, the carried over digit is placed below

fifth digit.

v) ( 2 X 3 ) = 6.

vi) Respective digits are added.

Note :

1. We can carry out the multiplication in urdhva - tiryak process from left to

right or right to left.

2. The same process can be applied even for numbers having more digits.

3. urdhva –tiryak process of multiplication can be effectively used in

multiplication regarding algebraic expressions.

38

iv) ( 1X 3 ) + ( 2 X 1 ) = 3 + 2 = 5. the carried over 1 of above step is added

i.e., 5 + 1 = 6 . It is retained. Thus fourth digit = 6

v) ( 1 X 1 ) = 1. As there is no carried over number from the previous step it is

retained. Thus fifth digit = 1

124 X 132 = 16368.

Let us work another problem by placing the carried over digits under the first

row and proceed.

234

x 316

‾‾‾‾‾‾‾

61724

1222

‾‾‾‾‾‾‾

73944

i) 4 X 6 = 24 : 2, the carried over digit is placed below the second digit.

ii) (3 X 6) + (4 x 1) = 18 + 4 = 22 ; 2, the carried over digit is placed below

third digit.

iii) (2 X 6) + (3 X 1) + (4 X 3) = 12 + 3 + 12 = 27 ; 2, the carried over digit is

placed below fourth digit.

iv) (2 X 1) + ( 3 X 3) = 2 + 9 = 11; 1, the carried over digit is placed below

fifth digit.

v) ( 2 X 3 ) = 6.

vi) Respective digits are added.

Note :

1. We can carry out the multiplication in urdhva - tiryak process from left to

right or right to left.

2. The same process can be applied even for numbers having more digits.

3. urdhva –tiryak process of multiplication can be effectively used in

multiplication regarding algebraic expressions.

38

41.
Example 1 : Find the product of (a+2b) and (3a+b).

Example 2 :

3a2 + 2a + 4

x 2a2 + 5a + 3

‾‾‾‾‾‾‾‾‾‾‾‾‾‾

i) 4X3 = 12

ii) (2 X 3) + ( 4 X 5 ) = 6 + 20 =26 i.e., 26a

iii) (3 X 3) + ( 2 X 5 ) + ( 4 X 2 ) = 9 + 10 + 8 = 27 i.e., 27a2

iv) (3 X 5) + ( 2 X 2 ) = 15 + 4 = 19 i.e., 19 a3

v) 3X2 = 6 i.e.,6a4

Hence the product is 6a4 +19a3 + 27a2 + 26a + 12

Example 3 : Find (3x2 + 4x + 7) (5x +6)

Now 3.x2 + 4x + 7

0.x2 + 5x + 6

‾‾‾‾‾‾‾‾‾‾‾‾

i) 7 X 6 = 42

ii) (4 X 6) + (7 X 5) = 24 + 35 = 59 i.e., 59x

iii) (3 X 6) + (4 X 5) + (7 X 0) = 18 + 20 + 0 = 38 i.e., 38x2

iv) (3 X 5) + (0 X 4) = 15 + 0 = 15 i.e., 15x3

v) 3 X 0 = 0

39

Example 2 :

3a2 + 2a + 4

x 2a2 + 5a + 3

‾‾‾‾‾‾‾‾‾‾‾‾‾‾

i) 4X3 = 12

ii) (2 X 3) + ( 4 X 5 ) = 6 + 20 =26 i.e., 26a

iii) (3 X 3) + ( 2 X 5 ) + ( 4 X 2 ) = 9 + 10 + 8 = 27 i.e., 27a2

iv) (3 X 5) + ( 2 X 2 ) = 15 + 4 = 19 i.e., 19 a3

v) 3X2 = 6 i.e.,6a4

Hence the product is 6a4 +19a3 + 27a2 + 26a + 12

Example 3 : Find (3x2 + 4x + 7) (5x +6)

Now 3.x2 + 4x + 7

0.x2 + 5x + 6

‾‾‾‾‾‾‾‾‾‾‾‾

i) 7 X 6 = 42

ii) (4 X 6) + (7 X 5) = 24 + 35 = 59 i.e., 59x

iii) (3 X 6) + (4 X 5) + (7 X 0) = 18 + 20 + 0 = 38 i.e., 38x2

iv) (3 X 5) + (0 X 4) = 15 + 0 = 15 i.e., 15x3

v) 3 X 0 = 0

39

42.
Hence the product is 15x3 +38x2 + 59x + 42

Find the products using urdhva tiryagbhyam process.

1) 25 X 16 2) 32 X 48 3) 56 X 56

4) 137 X 214 5) 321 X 213 6) 452 X 348

7) (2x + 3y) (4x + 5y) 8) (5a2 + 1) (3a2 + 4)

9) (6x2 + 5x + 2 ) (3x2 + 4x +7) 10) (4x2 + 3) (5x + 6)

Urdhva – tiryak in converse for division process:

As per the statement it an used as a simple argumentation for division process

particularly in algebra.

Consider the division of (x3 +5x2 + 3x + 7) by (x – 2)process by converse of

urdhva – tiryak :

i) x3 divided by x gives x2 . x3 + 5x2 + 3x + 7

It is the first term of the Quotient. ___________________

x–2

2

Q=x +-----------

ii) x2 X – 2 = - 2x2 . But 5x2 in the dividend hints7x2 more since 7x2 – 2x2 =

5x2 . This ‘more’ can be obtained from the multiplication of x by 7x. Hence

second term of Q is 7x.

x3 + 5x2 + 3x + 7

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ gives Q = x2 + 7x + - - - - - - - -

x–2

iii)We now have – 2 X 7x = -14x. But the 3rd term in the dividend is 3x for

which ‘17x more’ is required since 17x – 14x =3x.

Now multiplication of x by 17 gives 17x. Hence third term of quotient is 17

40

Find the products using urdhva tiryagbhyam process.

1) 25 X 16 2) 32 X 48 3) 56 X 56

4) 137 X 214 5) 321 X 213 6) 452 X 348

7) (2x + 3y) (4x + 5y) 8) (5a2 + 1) (3a2 + 4)

9) (6x2 + 5x + 2 ) (3x2 + 4x +7) 10) (4x2 + 3) (5x + 6)

Urdhva – tiryak in converse for division process:

As per the statement it an used as a simple argumentation for division process

particularly in algebra.

Consider the division of (x3 +5x2 + 3x + 7) by (x – 2)process by converse of

urdhva – tiryak :

i) x3 divided by x gives x2 . x3 + 5x2 + 3x + 7

It is the first term of the Quotient. ___________________

x–2

2

Q=x +-----------

ii) x2 X – 2 = - 2x2 . But 5x2 in the dividend hints7x2 more since 7x2 – 2x2 =

5x2 . This ‘more’ can be obtained from the multiplication of x by 7x. Hence

second term of Q is 7x.

x3 + 5x2 + 3x + 7

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ gives Q = x2 + 7x + - - - - - - - -

x–2

iii)We now have – 2 X 7x = -14x. But the 3rd term in the dividend is 3x for

which ‘17x more’ is required since 17x – 14x =3x.

Now multiplication of x by 17 gives 17x. Hence third term of quotient is 17

40

43.
Thus

x3 + 5x2 + 3x + 7

_________________ gives Q= x2 + 7x +17

x–2

iv) Now last term of Q, i.e., 17 multiplied by –2 gives 17X–2 = -34 but the

relevant term in dividend is 7. So 7 + 34 = 41 ‘more’ is required. As there no

more terms left in dividend, 41 remains as the remainder.

x3 + 5x2 + 3x + 7

________________ gives Q=x2 + 7x +17 and R = 41.

x–2

Find the Q and R in the following divisions by using the converse

process of urdhva – tiryagbhyam method :

1) 3x2 – x – 6 2) 16x2 + 24x +9

‾‾‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾‾‾

3x – 7 4x+3

3) x3+ 2x2 +3x + 5 4) 12x4 – 3x2 – 3x + 12

‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

x-3 x2 + 1

4. Paravartya Yojayet

'Paravartya – Yojayet' means 'transpose and apply'

(i) Consider the division by divisors of more than one digit, and when the

divisors are slightly greater than powers of 10.

Example 1 : Divide 1225 by 12.

Step 1 : (From left to right ) write the Divisor leaving the first digit, write the

other digit or digits using negative (-) sign and place them below the divisor

as shown.

12

-2

‾‾‾‾

Step 2 : Write down the dividend to the right. Set apart the last digit for the

remainder.

41

x3 + 5x2 + 3x + 7

_________________ gives Q= x2 + 7x +17

x–2

iv) Now last term of Q, i.e., 17 multiplied by –2 gives 17X–2 = -34 but the

relevant term in dividend is 7. So 7 + 34 = 41 ‘more’ is required. As there no

more terms left in dividend, 41 remains as the remainder.

x3 + 5x2 + 3x + 7

________________ gives Q=x2 + 7x +17 and R = 41.

x–2

Find the Q and R in the following divisions by using the converse

process of urdhva – tiryagbhyam method :

1) 3x2 – x – 6 2) 16x2 + 24x +9

‾‾‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾‾‾

3x – 7 4x+3

3) x3+ 2x2 +3x + 5 4) 12x4 – 3x2 – 3x + 12

‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

x-3 x2 + 1

4. Paravartya Yojayet

'Paravartya – Yojayet' means 'transpose and apply'

(i) Consider the division by divisors of more than one digit, and when the

divisors are slightly greater than powers of 10.

Example 1 : Divide 1225 by 12.

Step 1 : (From left to right ) write the Divisor leaving the first digit, write the

other digit or digits using negative (-) sign and place them below the divisor

as shown.

12

-2

‾‾‾‾

Step 2 : Write down the dividend to the right. Set apart the last digit for the

remainder.

41

44.
i.e.,, 12 122 5

-2

Step 3 : Write the 1st digit below the horizontal line drawn under

thedividend. Multiply the digit by –2, write the product below the 2nd digit

and add.

i.e.,, 12 122 5

-2 -2

‾‾‾‾‾ ‾‾‾‾

10

Since 1 x –2 = -2and 2 + (-2) = 0

Step 4 : We get second digits’ sum as ‘0’. Multiply the second digits’ sum

thus obtained by –2 and writes the product under 3rd digit and add.

12 122 5

-2 -20

‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾

102 5

Step 5 : Continue the process to the last digit.

i.e., 12 122 5

-2 -20 -4

‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾

102 1

Step 6: The sum of the last digit is the Remainder and the result to its left is

Thus Q = 102 andR = 1

Example 2 : Divide 1697 by 14.

14 169 7

-4 -4–8–4

‾‾‾‾ ‾‾‾‾‾‾‾

121 3

Q = 121, R = 3.

Example 3 : Divide 2598 by 123.

Note that the divisor has 3 digits. So we have to set up the last two

42

-2

Step 3 : Write the 1st digit below the horizontal line drawn under

thedividend. Multiply the digit by –2, write the product below the 2nd digit

and add.

i.e.,, 12 122 5

-2 -2

‾‾‾‾‾ ‾‾‾‾

10

Since 1 x –2 = -2and 2 + (-2) = 0

Step 4 : We get second digits’ sum as ‘0’. Multiply the second digits’ sum

thus obtained by –2 and writes the product under 3rd digit and add.

12 122 5

-2 -20

‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾

102 5

Step 5 : Continue the process to the last digit.

i.e., 12 122 5

-2 -20 -4

‾‾‾‾‾ ‾‾‾‾‾‾‾‾‾‾

102 1

Step 6: The sum of the last digit is the Remainder and the result to its left is

Thus Q = 102 andR = 1

Example 2 : Divide 1697 by 14.

14 169 7

-4 -4–8–4

‾‾‾‾ ‾‾‾‾‾‾‾

121 3

Q = 121, R = 3.

Example 3 : Divide 2598 by 123.

Note that the divisor has 3 digits. So we have to set up the last two

42

45.
digits of the dividend for the remainder.

123 25 98 Step ( 1 ) & Step ( 2 )

-2-3

‾‾‾‾‾ ‾‾‾‾‾‾‾‾

Now proceed the sequence of steps write –2 and –3 as follows :

123 25 98

-2-3 -4 -6

‾‾‾‾‾ -2–3

‾‾‾‾‾‾‾‾‾‾

21 15

Since 2 X (-2, -3)= -4 , -6;5 – 4 = 1

and (1 X (-2,-3); 9 – 6 – 2 = 1; 8 – 3 = 5.

Hence Q = 21 and R = 15.

Example 4 : Divide 239479 by 11213. The divisor has 5 digits. So the last

4digits of the dividend are to be set up for Remainder.

1 1213 23 9 479

-1-2-1-3 -2 -4-2-6 with 2

‾‾‾‾‾‾‾‾ -1-2-1-3 with 1

‾‾‾‾‾‾‾‾‾‾‾‾‾

21 40 0 6

Hence Q = 21, R = 4006.

Example 5 : Divide 13456 by 1123

1 12 3 134 5 6

-1–2–3 -1-2-3

‾‾‾‾‾‾‾ -2-4 –6

‾‾‾‾‾‾‾‾‾‾‾‾‾

1 2 0–2 0

Note that the remainder portion contains –20, i.e.,, a negative quantity. To

over come this situation, take 1 over from the quotient column, i.e.,, 1123

over to the right side, subtract the remainder portion 20 to get the actual

Thus Q = 12 – 1 = 11, and R = 1123 - 20 = 1103.

43

123 25 98 Step ( 1 ) & Step ( 2 )

-2-3

‾‾‾‾‾ ‾‾‾‾‾‾‾‾

Now proceed the sequence of steps write –2 and –3 as follows :

123 25 98

-2-3 -4 -6

‾‾‾‾‾ -2–3

‾‾‾‾‾‾‾‾‾‾

21 15

Since 2 X (-2, -3)= -4 , -6;5 – 4 = 1

and (1 X (-2,-3); 9 – 6 – 2 = 1; 8 – 3 = 5.

Hence Q = 21 and R = 15.

Example 4 : Divide 239479 by 11213. The divisor has 5 digits. So the last

4digits of the dividend are to be set up for Remainder.

1 1213 23 9 479

-1-2-1-3 -2 -4-2-6 with 2

‾‾‾‾‾‾‾‾ -1-2-1-3 with 1

‾‾‾‾‾‾‾‾‾‾‾‾‾

21 40 0 6

Hence Q = 21, R = 4006.

Example 5 : Divide 13456 by 1123

1 12 3 134 5 6

-1–2–3 -1-2-3

‾‾‾‾‾‾‾ -2-4 –6

‾‾‾‾‾‾‾‾‾‾‾‾‾

1 2 0–2 0

Note that the remainder portion contains –20, i.e.,, a negative quantity. To

over come this situation, take 1 over from the quotient column, i.e.,, 1123

over to the right side, subtract the remainder portion 20 to get the actual

Thus Q = 12 – 1 = 11, and R = 1123 - 20 = 1103.

43

46.
Find the Quotient and Remainder for the problems using

paravartya – yojayet method.

1) 1234 ÷ 112 2) 11329 ÷ 1132

3) 12349÷ 133 4) 239479÷1203

Now let us consider the application of paravartya – yojayet in algebra.

Example 1 : Divide 6x2 + 5x + 4 by x – 1

X- 1 6x2 + 5x + 4

‾‾‾‾‾‾

1 6+ 11

‾‾‾‾‾‾‾‾‾‾‾‾

6x + 11 + 15 ThusQ = 6x+11,R=15.

Example 2 : Divide x3 –3x2 + 10x – 4 by x - 5

X- 5 x3 – 3x2 + 10x – 4

‾‾‾‾‾

5 5 + 10 100

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

x2 + 2x + 20, + 96

Thus Q= x2 + 2x + 20, R = 96.

The procedure as a mental exercise comes as follows :

i) x3 / xgives x2 i.e.,, 1 the first coefficient in the Quotient.

ii) Multiply 1 by + 5,(obtained after reversing the sign of second term in the

Quotient) and add to the next coefficient in the dividend. It gives 1 X( +5) =

+5, adding to the next coefficient, i.e.,, –3 + 5 = 2. This is next coefficient in

iii) Continue the process : multiply 2 by +5, i.e.,, 2 X +5 =10, add to the

next coefficient 10 + 10 = 20. This is next coefficient in Quotient. Thus

Quotient is x2 + 2x + 20

iv) Nowmultiply 20 by + 5i.e.,, 20 x 5 = 100. Add to the next (last) term,

100 + (-4) = 96, which becomesR,i.e.,, R =9.

44

paravartya – yojayet method.

1) 1234 ÷ 112 2) 11329 ÷ 1132

3) 12349÷ 133 4) 239479÷1203

Now let us consider the application of paravartya – yojayet in algebra.

Example 1 : Divide 6x2 + 5x + 4 by x – 1

X- 1 6x2 + 5x + 4

‾‾‾‾‾‾

1 6+ 11

‾‾‾‾‾‾‾‾‾‾‾‾

6x + 11 + 15 ThusQ = 6x+11,R=15.

Example 2 : Divide x3 –3x2 + 10x – 4 by x - 5

X- 5 x3 – 3x2 + 10x – 4

‾‾‾‾‾

5 5 + 10 100

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

x2 + 2x + 20, + 96

Thus Q= x2 + 2x + 20, R = 96.

The procedure as a mental exercise comes as follows :

i) x3 / xgives x2 i.e.,, 1 the first coefficient in the Quotient.

ii) Multiply 1 by + 5,(obtained after reversing the sign of second term in the

Quotient) and add to the next coefficient in the dividend. It gives 1 X( +5) =

+5, adding to the next coefficient, i.e.,, –3 + 5 = 2. This is next coefficient in

iii) Continue the process : multiply 2 by +5, i.e.,, 2 X +5 =10, add to the

next coefficient 10 + 10 = 20. This is next coefficient in Quotient. Thus

Quotient is x2 + 2x + 20

iv) Nowmultiply 20 by + 5i.e.,, 20 x 5 = 100. Add to the next (last) term,

100 + (-4) = 96, which becomesR,i.e.,, R =9.

44

47.
Example 3:

x4 –3x3 + 7x2 + 5x + 7

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

x+4

Now thinking the method as in example ( 1 ), we proceed as follows.

x+4 x4 - 3x3 +7x2 + 5x + 7

‾‾‾‾‾

-4 - 4 + 28 - 140 + 540

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

x3 - 7x2 + 35x - 135 547

Thus Q = x3 – 7x2 + 35x – 135 and R = 547.

or we proceed orally as follows:

x4 / x gives 1 as first coefficient.

i) -4 X 1 = - 4 : add to next coefficient – 4 + (-3) = - 7 which gives next

coefficient in Q.

ii) – 7 X - 4 = 28 : then 28 + 7 = 35, the next coefficient in Q.

iii) 35 X - 4 = - 140 : then – 140 + 5 = - 135, the next coefficient in Q.

iv) - 135 X - 4 = 540 : then 540 + 7 = 547 becomes R.

Thus Q = x3 – 7x2 + 35x – 135 , R = 547.

Note :

1. We can follow the same procedure even the number of terms is more.

2. If any term is missing, we have to take the coefficient of the term as zero

and proceed.

Now consider the divisors of second degree or more as in the following

Example :4 2x4 – 3x3 – 3x + 2 by x2 + 1.

Here x2 term is missing in the dividend. Hence treat it as 0 .x2 or 0 .

And the x term in divisor is also absent we treat it as 0 . x. Now

45

x4 –3x3 + 7x2 + 5x + 7

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

x+4

Now thinking the method as in example ( 1 ), we proceed as follows.

x+4 x4 - 3x3 +7x2 + 5x + 7

‾‾‾‾‾

-4 - 4 + 28 - 140 + 540

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

x3 - 7x2 + 35x - 135 547

Thus Q = x3 – 7x2 + 35x – 135 and R = 547.

or we proceed orally as follows:

x4 / x gives 1 as first coefficient.

i) -4 X 1 = - 4 : add to next coefficient – 4 + (-3) = - 7 which gives next

coefficient in Q.

ii) – 7 X - 4 = 28 : then 28 + 7 = 35, the next coefficient in Q.

iii) 35 X - 4 = - 140 : then – 140 + 5 = - 135, the next coefficient in Q.

iv) - 135 X - 4 = 540 : then 540 + 7 = 547 becomes R.

Thus Q = x3 – 7x2 + 35x – 135 , R = 547.

Note :

1. We can follow the same procedure even the number of terms is more.

2. If any term is missing, we have to take the coefficient of the term as zero

and proceed.

Now consider the divisors of second degree or more as in the following

Example :4 2x4 – 3x3 – 3x + 2 by x2 + 1.

Here x2 term is missing in the dividend. Hence treat it as 0 .x2 or 0 .

And the x term in divisor is also absent we treat it as 0 . x. Now

45

48.
x2 +1 2x4 -3x3 + 0 . x2 - 3x + 2

x2+ 0 . x + 1 0 -2

‾‾‾‾‾‾‾‾‾‾‾‾

0 -1 0 +3

0 +2

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

2 -3 -2 0 4

Thus Q = 2x2 - 3x - 2andR = 0 . x + 4 = 4.

Example 5 : 2x5 – 5x4 + 3x2 – 4x + 7by x3 – 2x2 + 3.

We treat the dividend as 2x5 –5x4 + 0. x3 + 3x2 – 4x + 7 and divisor as

x3 –2x2 + 0 . x + 3 and proceed as follows :

x3 – 2x2 + 0 . x + 3 2x5 – 5x4 + 0.x3 + 3x2 – 4x + 7

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

2 0 -3 4 -60

-2

0 + 3

-4 0 + 6

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

2 - 1 -2 - 7 - 1 +13

Thus Q = 2x2 – x – 2, R = - 7x2 – x + 13.

You may observe a very close relation of the method paravartya in this

aspect with regard to REMAINDER THEOREM and HORNER PROCESS of

Synthetic division. And yet paravartya goes much farther and is capable of

numerous applications in other directions also.

Apply paravartya – yojayet to find out the Quotient and Remainder in

each of the following problems.

1) (4x2 + 3x + 5)÷(x+1)

2) (x3 – 4x2 + 7x + 6) ÷ (x – 2)

3) (x4 – x3 + x2 + 2x + 4) ÷(x2 - x – 1)

4) (2x5 +x3 – 3x + 7) ÷ (x3 + 2x – 3)

5) (7x6 + 6x5 – 5x4 + 4x3 –3x2 + 2x – 1)÷ (x-1)

46

x2+ 0 . x + 1 0 -2

‾‾‾‾‾‾‾‾‾‾‾‾

0 -1 0 +3

0 +2

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

2 -3 -2 0 4

Thus Q = 2x2 - 3x - 2andR = 0 . x + 4 = 4.

Example 5 : 2x5 – 5x4 + 3x2 – 4x + 7by x3 – 2x2 + 3.

We treat the dividend as 2x5 –5x4 + 0. x3 + 3x2 – 4x + 7 and divisor as

x3 –2x2 + 0 . x + 3 and proceed as follows :

x3 – 2x2 + 0 . x + 3 2x5 – 5x4 + 0.x3 + 3x2 – 4x + 7

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

2 0 -3 4 -60

-2

0 + 3

-4 0 + 6

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

2 - 1 -2 - 7 - 1 +13

Thus Q = 2x2 – x – 2, R = - 7x2 – x + 13.

You may observe a very close relation of the method paravartya in this

aspect with regard to REMAINDER THEOREM and HORNER PROCESS of

Synthetic division. And yet paravartya goes much farther and is capable of

numerous applications in other directions also.

Apply paravartya – yojayet to find out the Quotient and Remainder in

each of the following problems.

1) (4x2 + 3x + 5)÷(x+1)

2) (x3 – 4x2 + 7x + 6) ÷ (x – 2)

3) (x4 – x3 + x2 + 2x + 4) ÷(x2 - x – 1)

4) (2x5 +x3 – 3x + 7) ÷ (x3 + 2x – 3)

5) (7x6 + 6x5 – 5x4 + 4x3 –3x2 + 2x – 1)÷ (x-1)

46

49.
Paravartya in solving simple equations :

Recall that 'paravartya yojayet' means 'transpose and apply'. The rule

relating to transposition enjoins invariable change of sign with every change

of side. i.e., + becomes - and conversely ; and X becomes ÷ and conversely.

Further it can be extended to the transposition of terms from left to right and

conversely and from numerator to denominator and conversely in the

concerned problems.

Type ( i ) :

Consider the problem 7x – 5 = 5x + 1

7x – 5x = 1 + 5

i.e.,, 2x = 6 x = 6 ÷ 2 = 3.

Observe that the problem is of the type ax + b = cx + d from which we get

by ‘transpose’ (d – b), (a – c) and

d - b.

x= ‾‾‾‾‾‾‾‾

a-c

In this example a = 7, b = - 5, c = 5, d = 1

Hence 1 – (- 5) 1+5 6

x = _______ = ____ = __ = 3

7–5 7-5 2

Example 2: Solve for x,3x + 4 = 2x + 6

d-b 6-4 2

x = _____ = _____ = __ = 2

a-c 3-2 1

Type ( ii ) :Consider problems of the type (x + a) (x+b) = (x+c) (x+d). By

paravartya, we get

cd - ab

x = ______________

(a + b) – (c + d)

47

Recall that 'paravartya yojayet' means 'transpose and apply'. The rule

relating to transposition enjoins invariable change of sign with every change

of side. i.e., + becomes - and conversely ; and X becomes ÷ and conversely.

Further it can be extended to the transposition of terms from left to right and

conversely and from numerator to denominator and conversely in the

concerned problems.

Type ( i ) :

Consider the problem 7x – 5 = 5x + 1

7x – 5x = 1 + 5

i.e.,, 2x = 6 x = 6 ÷ 2 = 3.

Observe that the problem is of the type ax + b = cx + d from which we get

by ‘transpose’ (d – b), (a – c) and

d - b.

x= ‾‾‾‾‾‾‾‾

a-c

In this example a = 7, b = - 5, c = 5, d = 1

Hence 1 – (- 5) 1+5 6

x = _______ = ____ = __ = 3

7–5 7-5 2

Example 2: Solve for x,3x + 4 = 2x + 6

d-b 6-4 2

x = _____ = _____ = __ = 2

a-c 3-2 1

Type ( ii ) :Consider problems of the type (x + a) (x+b) = (x+c) (x+d). By

paravartya, we get

cd - ab

x = ______________

(a + b) – (c + d)

47

50.
It is trivial form the following steps

(x + a) (x + b) = (x + c) (x + d)

x2 + bx + ax + ab = x2 + dx + cx + cd

bx + ax – dx – cx = cd – ab

x( a + b – c – d) = cd – ab

cd – ab cd - ab

x = ____________ x = _________________

a+b–c–d ( a + b ) – (c + d.)

Example 1 : (x – 3) (x – 2 ) = (x + 1 ) (x + 2 ).

By paravartya

cd – ab 1 (2) – (-3) (-2)

x= __________ = ______________

a + b – c –d -3–2–1–2

2-6 -4 1

= _______ = ___ = __

-8 -8 2

Example 2 : (x + 7) (x – 6) = (x +3) (x – 4).

Now cd - ab (3) (-4) – (7) (-6)

x= ___________ = ________________

a+b–c–d 7 + (-6) – 3 - (-4)

- 12 + 42 30

= ____________ = ___ = 15

7–6–3+4 2

Note that if cd - ab = 0 i.e.,, cd = ab, i.e.,, if the product of the absolute

terms be the same on both sides, the numerator becomes zero giving x = 0.

For the problem (x + 4) (x + 3) = (x – 2 ) ( x – 6 )

Solution is x = 0 since 4 X 3 = - 2 X - 6. = 12

Type ( iii) :

Consider the problems of the type ax + b m

______ = __

cx + d n

48

(x + a) (x + b) = (x + c) (x + d)

x2 + bx + ax + ab = x2 + dx + cx + cd

bx + ax – dx – cx = cd – ab

x( a + b – c – d) = cd – ab

cd – ab cd - ab

x = ____________ x = _________________

a+b–c–d ( a + b ) – (c + d.)

Example 1 : (x – 3) (x – 2 ) = (x + 1 ) (x + 2 ).

By paravartya

cd – ab 1 (2) – (-3) (-2)

x= __________ = ______________

a + b – c –d -3–2–1–2

2-6 -4 1

= _______ = ___ = __

-8 -8 2

Example 2 : (x + 7) (x – 6) = (x +3) (x – 4).

Now cd - ab (3) (-4) – (7) (-6)

x= ___________ = ________________

a+b–c–d 7 + (-6) – 3 - (-4)

- 12 + 42 30

= ____________ = ___ = 15

7–6–3+4 2

Note that if cd - ab = 0 i.e.,, cd = ab, i.e.,, if the product of the absolute

terms be the same on both sides, the numerator becomes zero giving x = 0.

For the problem (x + 4) (x + 3) = (x – 2 ) ( x – 6 )

Solution is x = 0 since 4 X 3 = - 2 X - 6. = 12

Type ( iii) :

Consider the problems of the type ax + b m

______ = __

cx + d n

48

51.
By cross – multiplication,

n ( ax + b) = m (cx + d)

nax + nb = mcx + md

nax - mcx = md – nb

x( na – mc ) = md – nb

md - nb

x = ________

na - mc.

Now look at the problem once again

ax + b m

_____ = __

cx+ d n

paravartya gives md - nb, na - mc and

md - nb

x = _______

na - mc

Example 1: 3x + 1 13

_______ = ___

4x + 3 19

md - nb 13 (3) - 19(1) 39 - 19 20

x = ______ = ____________ = _______ = __

na- mc 19 (3) - 13(4) 57 - 52 5

= 4

Example 2: 4x + 5 7

________ = __

3x + 13/2 8

(7) (13/2) -(8)(5)

x = _______________

(8)(4) - (7)(3)

(91/2) - 40 (91 - 80)/2 11 1

= __________ = _________ = ______ = __

32 – 21 32 – 21 2 X 11 2

49

n ( ax + b) = m (cx + d)

nax + nb = mcx + md

nax - mcx = md – nb

x( na – mc ) = md – nb

md - nb

x = ________

na - mc.

Now look at the problem once again

ax + b m

_____ = __

cx+ d n

paravartya gives md - nb, na - mc and

md - nb

x = _______

na - mc

Example 1: 3x + 1 13

_______ = ___

4x + 3 19

md - nb 13 (3) - 19(1) 39 - 19 20

x = ______ = ____________ = _______ = __

na- mc 19 (3) - 13(4) 57 - 52 5

= 4

Example 2: 4x + 5 7

________ = __

3x + 13/2 8

(7) (13/2) -(8)(5)

x = _______________

(8)(4) - (7)(3)

(91/2) - 40 (91 - 80)/2 11 1

= __________ = _________ = ______ = __

32 – 21 32 – 21 2 X 11 2

49

52.
Type (iv) :Consider the problems of the type m n

_____ + ____ = 0

x+a x+b

Take L.C.M and proceed.

m(x+b) + n (x+a)

______________ = 0

(x + a) (x +b)

mx + mb + nx + na

________________ = 0

x + a)(x + b)

(m + n)x + mb + na = 0 (m + n)x = - mb - na

-mb - na

x = ________

(m + n)

Thus the problem m n

____ + ____ = 0, by paravartya process

x+a x+b

gives directly

-mb - na

x = ________

(m + n)

Example 1 : 3 4

____ + ____ = 0

x+4 x–6

gives -mb - na

x = ________ Note that m = 3, n = 4, a = 4, b = - 6

(m + n)

-(3)(-6) – (4) (4) 18 - 16 2

= _______________ = ______ = __

( 3 + 4) 7 7

50

_____ + ____ = 0

x+a x+b

Take L.C.M and proceed.

m(x+b) + n (x+a)

______________ = 0

(x + a) (x +b)

mx + mb + nx + na

________________ = 0

x + a)(x + b)

(m + n)x + mb + na = 0 (m + n)x = - mb - na

-mb - na

x = ________

(m + n)

Thus the problem m n

____ + ____ = 0, by paravartya process

x+a x+b

gives directly

-mb - na

x = ________

(m + n)

Example 1 : 3 4

____ + ____ = 0

x+4 x–6

gives -mb - na

x = ________ Note that m = 3, n = 4, a = 4, b = - 6

(m + n)

-(3)(-6) – (4) (4) 18 - 16 2

= _______________ = ______ = __

( 3 + 4) 7 7

50

53.
Example 2 :

5 6

____ + _____ = 0

x+1 x – 21

gives -(5) (-21) - (6) (1) 105 - 6 99

x = ________________ = ______ = __ = 9

5+6 11 11

I . Solve the following problems using the sutra Paravartya – yojayet.

1) 3x + 5 = 5x – 3 6) (x + 1) ( x + 2) = ( x – 3) (x – 4)

2) (2x/3) + 1=x - 1 7) (x – 7) (x – 9)= (x – 3) (x – 22)

3) 7x + 2 5 8) (x + 7) (x + 9)= (x + 3 ) (x + 21)

______ = __

3x- 5 8

4) x + 1 / 3

_______ = 1

3x - 1

5) 5 2

____ + ____ = 0

x+3 x–4

1.Show that for the type of equations

m n p

____ + ____ + ____ = 0, the solution is

x+a x+b x+c

- mbc – nca – pab

x = ________________________ , if m + n + p =0.

m(b + c) + n(c+a) + p(a + b)

51

5 6

____ + _____ = 0

x+1 x – 21

gives -(5) (-21) - (6) (1) 105 - 6 99

x = ________________ = ______ = __ = 9

5+6 11 11

I . Solve the following problems using the sutra Paravartya – yojayet.

1) 3x + 5 = 5x – 3 6) (x + 1) ( x + 2) = ( x – 3) (x – 4)

2) (2x/3) + 1=x - 1 7) (x – 7) (x – 9)= (x – 3) (x – 22)

3) 7x + 2 5 8) (x + 7) (x + 9)= (x + 3 ) (x + 21)

______ = __

3x- 5 8

4) x + 1 / 3

_______ = 1

3x - 1

5) 5 2

____ + ____ = 0

x+3 x–4

1.Show that for the type of equations

m n p

____ + ____ + ____ = 0, the solution is

x+a x+b x+c

- mbc – nca – pab

x = ________________________ , if m + n + p =0.

m(b + c) + n(c+a) + p(a + b)

51

54.
2. Apply the above formula to set the solution for the problem

Problem 3 2 5

____ + ____ - ____ = 0

x+4 x+6 x+5

some more simple solutions :

m n m+n

____ + ____ = _____

x+a x+b x+c

Now this can be written as,

m n m n

____ + ____ = _____ + _____

x+a x+b x+c x+c

m m n n

____ - ____ = _____ - _____

x+a x+c x+c x+b

m(x +c) – m(x + a) n(x + b) – n(x + c)

________________ = ________________

(x + a) (x + c) (x + c) (x + b)

mx + mc – mx – ma nx + nb – nx – nc

________________ = _______________

(x + a) (x + c) (x +c ) (x + b)

m (c – a) n (b –c)

____________ = ___________

x +a x+b

m (c - a).x + m (c - a).b = n (b - c). x + n(b - c).a

x [ m(c - a)- n(b - c) ] = na(b - c) – mb (c - a)

or x [ m(c - a) + n(c - b) ] = na(b - c) + mb (a - c)

52

Problem 3 2 5

____ + ____ - ____ = 0

x+4 x+6 x+5

some more simple solutions :

m n m+n

____ + ____ = _____

x+a x+b x+c

Now this can be written as,

m n m n

____ + ____ = _____ + _____

x+a x+b x+c x+c

m m n n

____ - ____ = _____ - _____

x+a x+c x+c x+b

m(x +c) – m(x + a) n(x + b) – n(x + c)

________________ = ________________

(x + a) (x + c) (x + c) (x + b)

mx + mc – mx – ma nx + nb – nx – nc

________________ = _______________

(x + a) (x + c) (x +c ) (x + b)

m (c – a) n (b –c)

____________ = ___________

x +a x+b

m (c - a).x + m (c - a).b = n (b - c). x + n(b - c).a

x [ m(c - a)- n(b - c) ] = na(b - c) – mb (c - a)

or x [ m(c - a) + n(c - b) ] = na(b - c) + mb (a - c)

52

55.
Thus mb(a - c) + na (b - c)

x = ___________________

m(c-a) + n(c-b).

By paravartya rule we can easily remember the formula.

Example 1 : solve 3

5. Sunyam Samya Samuccaye

The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that

Samuccaya is Zero.' i.e., it should be equated to zero. The term 'Samuccaya'

has several meanings under different contexts.

i) We interpret, 'Samuccaya' as a term which occurs as a common factor in all

the terms concerned and proceed as follows.

Example 1: The equation 7x + 3x = 4x + 5x has the same factor ‘ x ‘ in all

its terms. Hence by the sutra it is zero,i.e., x = 0.

Otherwise we have to work like this:

7x + 3x = 4x + 5x

10x = 9x

10x – 9x = 0

x=0

This is applicable not only for ‘x’ but also any such unknown quantity as

Example 2: 5(x+1) = 3(x+1)

No need to proceed in the usual procedure like

5x + 5 = 3x + 3

5x – 3x = 3 – 5

2x = -2 or x = -2 ÷ 2 = -1

Simply think of the contextual meaning of 'Samuccaya'

53

x = ___________________

m(c-a) + n(c-b).

By paravartya rule we can easily remember the formula.

Example 1 : solve 3

5. Sunyam Samya Samuccaye

The Sutra 'Sunyam Samyasamuccaye' says the 'Samuccaya is the same, that

Samuccaya is Zero.' i.e., it should be equated to zero. The term 'Samuccaya'

has several meanings under different contexts.

i) We interpret, 'Samuccaya' as a term which occurs as a common factor in all

the terms concerned and proceed as follows.

Example 1: The equation 7x + 3x = 4x + 5x has the same factor ‘ x ‘ in all

its terms. Hence by the sutra it is zero,i.e., x = 0.

Otherwise we have to work like this:

7x + 3x = 4x + 5x

10x = 9x

10x – 9x = 0

x=0

This is applicable not only for ‘x’ but also any such unknown quantity as

Example 2: 5(x+1) = 3(x+1)

No need to proceed in the usual procedure like

5x + 5 = 3x + 3

5x – 3x = 3 – 5

2x = -2 or x = -2 ÷ 2 = -1

Simply think of the contextual meaning of 'Samuccaya'

53

56.
Now Samuccaya is ( x + 1)

x+1=0 gives x = -1

ii) Now we interpret 'Samuccaya' as product of independent terms in

expressions like (x+a) (x+b)

Example 3: ( x + 3 ) ( x + 4) = ( x – 2) ( x – 6 )

Here Samuccaya is 3 x 4 = 12 = -2 x -6

Since it is same , we derive x = 0

This example, we have already dealt in type ( ii ) of Paravartya in solving

simple equations.

iii) We interpret ' Samuccaya 'as the sum of the denominators of two fractions

having the same numerical numerator.

Consider the example.

1 1

____ + ____ = 0

3x-2 2x-1

for this we proceed by takingL.C.M.

(2x-1)+(3x–2)

____________ = 0

(3x–2)(2x–1)

5x–3

__________ = 0

(3x–2)(2x–1)

5x – 3 = 0 5x = 3

3

x = __

5

Instead of this, we can directly put the Samuccaya i.e., sum of the

i.e., 3x – 2 + 2x - 1 = 5x - 3 = 0

54

x+1=0 gives x = -1

ii) Now we interpret 'Samuccaya' as product of independent terms in

expressions like (x+a) (x+b)

Example 3: ( x + 3 ) ( x + 4) = ( x – 2) ( x – 6 )

Here Samuccaya is 3 x 4 = 12 = -2 x -6

Since it is same , we derive x = 0

This example, we have already dealt in type ( ii ) of Paravartya in solving

simple equations.

iii) We interpret ' Samuccaya 'as the sum of the denominators of two fractions

having the same numerical numerator.

Consider the example.

1 1

____ + ____ = 0

3x-2 2x-1

for this we proceed by takingL.C.M.

(2x-1)+(3x–2)

____________ = 0

(3x–2)(2x–1)

5x–3

__________ = 0

(3x–2)(2x–1)

5x – 3 = 0 5x = 3

3

x = __

5

Instead of this, we can directly put the Samuccaya i.e., sum of the

i.e., 3x – 2 + 2x - 1 = 5x - 3 = 0

54

57.
giving 5x = 3 x=3/5

It is true and applicable for all problems of the type

m m

____ + _____ = 0

ax+b cx+d

Samuccaya is ax+b+cx+d and solution is ( m ≠ 0 )

-(b+d)

x = _________

(a+c)

iii) We now interpret 'Samuccaya' as combination or total.

If the sum of the numerators and the sum of the denominators be the

same, then that sum = 0.

Consider examples of type

ax+ b ax + c

_____ = ______

ax+ c ax + b

In this case, (ax+b) (ax+b) =(ax+c) (ax+c)

a2x2 + 2abx + b2 = a2x2 + 2acx +c2

2abx – 2acx = c2 – b2

x ( 2ab – 2ac ) = c2 – b2

c2–b2 (c+b)(c-b) -(c+b)

x= ______ = _________ = _____

2a(b-c) 2a(b-c) 2a

As per Samuccaya (ax+b) + (ax+c) = 0

2ax+b+c = 0

2ax = -b-c

-(c+b)

x = ______

2a Hence the statement.

55

It is true and applicable for all problems of the type

m m

____ + _____ = 0

ax+b cx+d

Samuccaya is ax+b+cx+d and solution is ( m ≠ 0 )

-(b+d)

x = _________

(a+c)

iii) We now interpret 'Samuccaya' as combination or total.

If the sum of the numerators and the sum of the denominators be the

same, then that sum = 0.

Consider examples of type

ax+ b ax + c

_____ = ______

ax+ c ax + b

In this case, (ax+b) (ax+b) =(ax+c) (ax+c)

a2x2 + 2abx + b2 = a2x2 + 2acx +c2

2abx – 2acx = c2 – b2

x ( 2ab – 2ac ) = c2 – b2

c2–b2 (c+b)(c-b) -(c+b)

x= ______ = _________ = _____

2a(b-c) 2a(b-c) 2a

As per Samuccaya (ax+b) + (ax+c) = 0

2ax+b+c = 0

2ax = -b-c

-(c+b)

x = ______

2a Hence the statement.

55

58.
Example 4:

3x+ 4 3x + 5

______ = ______

3x+ 5 3x + 4

Since N1 + N2 = 3x + 4 + 3x + 5 = 6x + 9 ,

And D1 + D2 = 3x + 4 + 3x + 5 = 6x + 9

We haveN1 + N2 = D1 + D2 = 6x + 9

Hence from Sunya Samuccaya we get 6x + 9 = 0

6x = -9

-9 -3

x = __ = __

6 2

Example 5:

5x +7 5x + 12

_____ = _______

5x+12 5x + 7

Hence N1 + N2 = 5x + 7 + 5x + 12 = 10x + 19

And D1 + D2 = 5x + 12 + 5x + 7 = 10x + 19

N1 + N2 = D1 + D2 gives 10x + 19 = 0

10x = -19

-19

x = ____

10

Consider the examples of the type, where N1 + N2 = K (D1 + D2 ), where K is

a numerical constant, then also by removing the numerical constant K, we

can proceed as above.

Example 6:

2x + 3 x +1

_____ = ______

4x + 5 2x + 3

Here N1 + N2 = 2x + 3 + x + 1 = 3x + 4

56

3x+ 4 3x + 5

______ = ______

3x+ 5 3x + 4

Since N1 + N2 = 3x + 4 + 3x + 5 = 6x + 9 ,

And D1 + D2 = 3x + 4 + 3x + 5 = 6x + 9

We haveN1 + N2 = D1 + D2 = 6x + 9

Hence from Sunya Samuccaya we get 6x + 9 = 0

6x = -9

-9 -3

x = __ = __

6 2

Example 5:

5x +7 5x + 12

_____ = _______

5x+12 5x + 7

Hence N1 + N2 = 5x + 7 + 5x + 12 = 10x + 19

And D1 + D2 = 5x + 12 + 5x + 7 = 10x + 19

N1 + N2 = D1 + D2 gives 10x + 19 = 0

10x = -19

-19

x = ____

10

Consider the examples of the type, where N1 + N2 = K (D1 + D2 ), where K is

a numerical constant, then also by removing the numerical constant K, we

can proceed as above.

Example 6:

2x + 3 x +1

_____ = ______

4x + 5 2x + 3

Here N1 + N2 = 2x + 3 + x + 1 = 3x + 4

56

59.
D1 + D2 = 4x + 5 + 2x + 3 = 6x + 8

= 2 ( 3x + 4 )

Removing the numerical factor 2, we get 3x + 4 on both sides.

3x + 4 = 0 3x = -4 x = - 4 / 3.

v) 'Samuccaya' with the same meaning as above, i.e., case (iv), we solve the

problems leading to quadratic equations. In this context, we take the

problems as follows;

If N1 + N2 = D1 + D2 and also the differences

N1 ~ D1 = N2 ~ D2 then both the things are equated to zero, the

solution gives the two values for x.

Example 7:

3x + 2 2x + 5

_____ = ______

2x + 5 3x + 2

In the conventional text book method, we work as follows :

3x + 2 2x + 5

_____ = ______

2x + 5 3x + 2

( 3x + 2 ) ( 3x + 2 ) = ( 2x + 5 ) ( 2x + 5 )

9x2 + 12x + 4 = 4x2 + 20x + 25

9x2 + 12x + 4 - 4x2 - 20x – 25 = 0

5x2 – 8x – 21 = 0

5x2 – 15x + 7x – 21 = 0

5x ( x – 3 ) + 7 ( x – 3 ) = 0

(x – 3 ) ( 5x + 7 ) = 0

x – 3 = 0 or 5x + 7 = 0

x = 3 or - 7 / 5

Now ‘Samuccaya’ sutra comes to help us in a beautiful way as follows :

Observe N1 + N2 = 3x + 2 + 2x + 5 = 5x + 7

D1 + D2 = 2x + 5 + 3x + 2 = 5x + 7

Further N1 ~ D1 = ( 3x + 2 ) – ( 2x + 5 ) = x – 3

N2 ~ D2 = ( 2x + 5) – ( 3x + 2 ) = - x + 3 = - ( x – 3 )

57

= 2 ( 3x + 4 )

Removing the numerical factor 2, we get 3x + 4 on both sides.

3x + 4 = 0 3x = -4 x = - 4 / 3.

v) 'Samuccaya' with the same meaning as above, i.e., case (iv), we solve the

problems leading to quadratic equations. In this context, we take the

problems as follows;

If N1 + N2 = D1 + D2 and also the differences

N1 ~ D1 = N2 ~ D2 then both the things are equated to zero, the

solution gives the two values for x.

Example 7:

3x + 2 2x + 5

_____ = ______

2x + 5 3x + 2

In the conventional text book method, we work as follows :

3x + 2 2x + 5

_____ = ______

2x + 5 3x + 2

( 3x + 2 ) ( 3x + 2 ) = ( 2x + 5 ) ( 2x + 5 )

9x2 + 12x + 4 = 4x2 + 20x + 25

9x2 + 12x + 4 - 4x2 - 20x – 25 = 0

5x2 – 8x – 21 = 0

5x2 – 15x + 7x – 21 = 0

5x ( x – 3 ) + 7 ( x – 3 ) = 0

(x – 3 ) ( 5x + 7 ) = 0

x – 3 = 0 or 5x + 7 = 0

x = 3 or - 7 / 5

Now ‘Samuccaya’ sutra comes to help us in a beautiful way as follows :

Observe N1 + N2 = 3x + 2 + 2x + 5 = 5x + 7

D1 + D2 = 2x + 5 + 3x + 2 = 5x + 7

Further N1 ~ D1 = ( 3x + 2 ) – ( 2x + 5 ) = x – 3

N2 ~ D2 = ( 2x + 5) – ( 3x + 2 ) = - x + 3 = - ( x – 3 )

57

60.
Hence 5x + 7 = 0 , x – 3 = 0

5x = -7 , x = 3

i.e., x = -7 / 5 , x = 3

Note that all these can be easily calculated by mere observation.

Example 8:

3x + 4 5x + 6

______ = _____

6x + 7 2x + 3

Observe that

N1 + N2 = 3x + 4 + 5x + 6 = 8x + 10

andD1 + D2 = 6x + 7 + 2x + 3 = 8x + 10

Further N1 ~D1 = (3x + 4) – (6x + 7)

= 3x + 4 – 6x – 7

= -3x – 3 = -3 ( x + 1 )

N2 ~ D2 = (5x + 6) – (2x + 3) = 3x + 3 = 3( x + 1)

By ‘Sunyam Samuccaye’ we have

8x + 10 = 0 3( x + 1 ) = 0

8x = -10 x+1=0

x = - 10 / 8 x = -1

=-5/4

vi)‘Samuccaya’ with the same sense but with a different context and

application .

Example 9:

1 1 1 1

____ + _____ = ____ + ____

x-4 x–6 x-2 x-8

Usually we proceed as follows.

x–6+x-4 x–8+x-2

___________ = ___________

(x–4) (x–6) (x–2) (x-8)

58

5x = -7 , x = 3

i.e., x = -7 / 5 , x = 3

Note that all these can be easily calculated by mere observation.

Example 8:

3x + 4 5x + 6

______ = _____

6x + 7 2x + 3

Observe that

N1 + N2 = 3x + 4 + 5x + 6 = 8x + 10

andD1 + D2 = 6x + 7 + 2x + 3 = 8x + 10

Further N1 ~D1 = (3x + 4) – (6x + 7)

= 3x + 4 – 6x – 7

= -3x – 3 = -3 ( x + 1 )

N2 ~ D2 = (5x + 6) – (2x + 3) = 3x + 3 = 3( x + 1)

By ‘Sunyam Samuccaye’ we have

8x + 10 = 0 3( x + 1 ) = 0

8x = -10 x+1=0

x = - 10 / 8 x = -1

=-5/4

vi)‘Samuccaya’ with the same sense but with a different context and

application .

Example 9:

1 1 1 1

____ + _____ = ____ + ____

x-4 x–6 x-2 x-8

Usually we proceed as follows.

x–6+x-4 x–8+x-2

___________ = ___________

(x–4) (x–6) (x–2) (x-8)

58

61.
2x-10 2x-10

_________ = _________

x2–10x+24 x2–10x+16

( 2x – 10 ) ( x2 – 10x + 16 ) = ( 2x – 10 ) ( x2 – 10x + 24)

2x3–20x2+32x–10x2+100x–160 = 2x3–20x2+48x–10x2+100x-240

2x3 – 30x2 + 132x – 160 = 2x3 – 30x2 + 148x – 240

132x – 160 = 148x – 240

132x – 148x = 160 – 240

– 16x = - 80

x = - 80 / - 16 = 5

Now ‘Samuccaya’ sutra, tell us that, if other elements being equal, the sum-

total of the denominators on the L.H.S. and their total on the R.H.S. be the

same, that total is zero.

Now D1 + D2 = x – 4 + x – 6 = 2x – 10, and

D3 + D4 = x – 2 + x – 8 = 2x – 10

By Samuccaya, 2x – 10 gives 2x = 10

10

x = __ = 5

2

Example 10:

1 1 1 1

____ + ____ = ____ + _____

x -8 x–9 x -5 x – 12

D1 +D2 = x – 8 + x – 9 = 2x – 17, and

D3 +D4 = x – 5 + x –12 = 2x – 17

Now 2x – 17 = 0 gives 2x = 17

17

x = __ = 8½

2

Example 11:

1 1 1 1

____ - _____ = ____ - _____

x +7 x + 10 x +6 x+9

59

_________ = _________

x2–10x+24 x2–10x+16

( 2x – 10 ) ( x2 – 10x + 16 ) = ( 2x – 10 ) ( x2 – 10x + 24)

2x3–20x2+32x–10x2+100x–160 = 2x3–20x2+48x–10x2+100x-240

2x3 – 30x2 + 132x – 160 = 2x3 – 30x2 + 148x – 240

132x – 160 = 148x – 240

132x – 148x = 160 – 240

– 16x = - 80

x = - 80 / - 16 = 5

Now ‘Samuccaya’ sutra, tell us that, if other elements being equal, the sum-

total of the denominators on the L.H.S. and their total on the R.H.S. be the

same, that total is zero.

Now D1 + D2 = x – 4 + x – 6 = 2x – 10, and

D3 + D4 = x – 2 + x – 8 = 2x – 10

By Samuccaya, 2x – 10 gives 2x = 10

10

x = __ = 5

2

Example 10:

1 1 1 1

____ + ____ = ____ + _____

x -8 x–9 x -5 x – 12

D1 +D2 = x – 8 + x – 9 = 2x – 17, and

D3 +D4 = x – 5 + x –12 = 2x – 17

Now 2x – 17 = 0 gives 2x = 17

17

x = __ = 8½

2

Example 11:

1 1 1 1

____ - _____ = ____ - _____

x +7 x + 10 x +6 x+9

59

62.
This is not in the expected form. But a little work regarding transposition

makes the above as follows.

1 1 1 1

____ + ____ = ____ + _____

x +7 x+9 x +6 x + 10

Now ‘Samuccaya’ sutra applies

D1 +D2 = x + 7 + x + 9 = 2x + 16, and

D3 +D4 = x + 6 + x + 10 = 2x + 16

Solution is given by 2x + 16 = 0 i.e., 2 x = - 16.

x = - 16 / 2 = - 8.

Solve the following problems using Sunyam Samya-Samuccaye

1. 7(x+2)+3(x+2)=6(x+2)+5(x+2)

2. ( x + 6 ) ( x + 3 ) = ( x – 9 ) ( x – 2)

3. ( x - 1 ) ( x + 14 ) = ( x + 2 ) ( x – 7)

1 1

4. ______ + ____ = 0

4x-3 x–2

4 4

5. _____ + _____ = 0

3x + 1 5x + 7

2x + 11 2x+5

6. ______ = _____

2x+ 5 2x+11

3x + 4 x+1

7. ______ = _____

6x + 7 2x + 3

60

makes the above as follows.

1 1 1 1

____ + ____ = ____ + _____

x +7 x+9 x +6 x + 10

Now ‘Samuccaya’ sutra applies

D1 +D2 = x + 7 + x + 9 = 2x + 16, and

D3 +D4 = x + 6 + x + 10 = 2x + 16

Solution is given by 2x + 16 = 0 i.e., 2 x = - 16.

x = - 16 / 2 = - 8.

Solve the following problems using Sunyam Samya-Samuccaye

1. 7(x+2)+3(x+2)=6(x+2)+5(x+2)

2. ( x + 6 ) ( x + 3 ) = ( x – 9 ) ( x – 2)

3. ( x - 1 ) ( x + 14 ) = ( x + 2 ) ( x – 7)

1 1

4. ______ + ____ = 0

4x-3 x–2

4 4

5. _____ + _____ = 0

3x + 1 5x + 7

2x + 11 2x+5

6. ______ = _____

2x+ 5 2x+11

3x + 4 x+1

7. ______ = _____

6x + 7 2x + 3

60

63.
4x - 3 x+ 4

8. ______ = _____

2x+ 3 3x - 2

1 1 1 1

9. ____ + ____ = ____ + _____

x-2 x-5 x-3 x-4

1 1 1 1

10. ____ - ____ = _____ - _____

x-7 x-6 x - 10 x-9

Sunyam Samya Samuccaye in Certain Cubes:

Consider the problem ( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5)3. For the solution

by the traditional method we follow the steps as given below:

( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3

x3 – 12x2 + 48x – 64 + x3 – 18x2 + 108x – 216

= 2 ( x3 – 15x2 + 75x – 125 )

2x3 – 30x2 + 156x – 280 = 2x3 – 30x2 + 150x – 250

156x – 280 = 150x – 250

156x – 150x = 280 – 250

6x = 30

x = 30 / 6 = 5

But once again observe the problem in the vedic sense

We have ( x – 4 ) + ( x – 6 ) = 2x – 10. Taking out the numerical factor 2,

we have ( x – 5 ) = 0, which is the factor under the cube on R.H.S. In such a

case “Sunyam samya Samuccaye” formula gives that x – 5 = 0. Hence x = 5

Think of solving the problem (x–249)3 + (x+247)3 = 2(x–1)3

The traditional method will be horrible even to think of.

But ( x – 249 ) + ( x + 247 ) = 2x – 2 = 2 ( x – 1 ). And x – 1. on R.H.S.

61

8. ______ = _____

2x+ 3 3x - 2

1 1 1 1

9. ____ + ____ = ____ + _____

x-2 x-5 x-3 x-4

1 1 1 1

10. ____ - ____ = _____ - _____

x-7 x-6 x - 10 x-9

Sunyam Samya Samuccaye in Certain Cubes:

Consider the problem ( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5)3. For the solution

by the traditional method we follow the steps as given below:

( x – 4 )3 + ( x – 6 )3 = 2 ( x – 5 )3

x3 – 12x2 + 48x – 64 + x3 – 18x2 + 108x – 216

= 2 ( x3 – 15x2 + 75x – 125 )

2x3 – 30x2 + 156x – 280 = 2x3 – 30x2 + 150x – 250

156x – 280 = 150x – 250

156x – 150x = 280 – 250

6x = 30

x = 30 / 6 = 5

But once again observe the problem in the vedic sense

We have ( x – 4 ) + ( x – 6 ) = 2x – 10. Taking out the numerical factor 2,

we have ( x – 5 ) = 0, which is the factor under the cube on R.H.S. In such a

case “Sunyam samya Samuccaye” formula gives that x – 5 = 0. Hence x = 5

Think of solving the problem (x–249)3 + (x+247)3 = 2(x–1)3

The traditional method will be horrible even to think of.

But ( x – 249 ) + ( x + 247 ) = 2x – 2 = 2 ( x – 1 ). And x – 1. on R.H.S.

61

64.
cube, it is enough to state that x – 1 = 0 by the ‘sutra’.

x = 1 is the solution. No cubing or any other mathematical operations.

Algebraic Proof :

Consider ( x – 2a)3 + ( x – 2b )3 = 2 ( x – a – b )3 it is clear that

x – 2a + x – 2b = 2x – 2a – 2b

=2(x–a–b)

Now the expression,

x3 -6x2a + 12xa2 – 8a3 + x3 – 6x2b +12xb2 – 8b3 =

2(x3–3x2a–3x2b+3xa2+3xb2+6axb–a3–3a2b–3ab2–b3)

= 2x3–6x2a–6x2b+6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3

cancel the common terms on both sides

12xa2+12xb2–8a3–8b3 = 6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3

6xa2 + 6xb2 – 12xab = 6a3 + 6b3 –6a2b – 6ab2

6x ( a2 + b2 – 2ab ) = 6 [ a3 + b3 – ab ( a + b )]

x ( a – b )2 = [ ( a + b ) ( a2 + b2 –ab ) – ( a + b )ab]

= ( a + b ) ( a2 + b2 – 2ab )

= ( a + b ) ( a – b )2

x=a+b

Solve the following using “Sunyam Samuccaye” process :

1. ( x – 3 )3 + ( x – 9 )3 = 2 ( x – 6 )3

2. ( x + 4 )3 + ( x – 10 )3 = 2 ( x – 3 )3

3. ( x + a + b – c )3 + ( x + b + c – a )3 = 2 ( x + b )3

62

x = 1 is the solution. No cubing or any other mathematical operations.

Algebraic Proof :

Consider ( x – 2a)3 + ( x – 2b )3 = 2 ( x – a – b )3 it is clear that

x – 2a + x – 2b = 2x – 2a – 2b

=2(x–a–b)

Now the expression,

x3 -6x2a + 12xa2 – 8a3 + x3 – 6x2b +12xb2 – 8b3 =

2(x3–3x2a–3x2b+3xa2+3xb2+6axb–a3–3a2b–3ab2–b3)

= 2x3–6x2a–6x2b+6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3

cancel the common terms on both sides

12xa2+12xb2–8a3–8b3 = 6xa2+6xb2+12xab–2a3–6a2b–6ab2–2b3

6xa2 + 6xb2 – 12xab = 6a3 + 6b3 –6a2b – 6ab2

6x ( a2 + b2 – 2ab ) = 6 [ a3 + b3 – ab ( a + b )]

x ( a – b )2 = [ ( a + b ) ( a2 + b2 –ab ) – ( a + b )ab]

= ( a + b ) ( a2 + b2 – 2ab )

= ( a + b ) ( a – b )2

x=a+b

Solve the following using “Sunyam Samuccaye” process :

1. ( x – 3 )3 + ( x – 9 )3 = 2 ( x – 6 )3

2. ( x + 4 )3 + ( x – 10 )3 = 2 ( x – 3 )3

3. ( x + a + b – c )3 + ( x + b + c – a )3 = 2 ( x + b )3

62

65.
Example :

(x + 2)3 x+1

______ = _____

(x + 3)3 x+4

with the text book procedures we proceed as follows

x3 + 6x2 + 12x +8 x+1

_______________ = _____

x3 + 9x2 + 27x +27 x+4

Now by cross multiplication,

( x + 4 ) ( x3 +6x2 + 12x + 8 ) = ( x + 1 ) ( x3 + 9x2 + 27x + 27 )

x4 + 6x3 + 12x2+ 8x + 4x3 + 24x2 + 48x + 32 =

x4 + 9x3 + 27x2 + 27x + x3 + 9x2 + 27x + 27

x4 +10x3 + 36x2 + 56x + 32 = x4 + 10x3 +36x2 + 54x + 27

56x + 32 = 54x + 27

56x – 54x = 27 – 32

2x = - 5

x=-5/2

Observe that ( N1 + D1 ) with in the cubes on

L.H.S. is x + 2 + x + 3 = 2x + 5 and

N2 + D2 on the right hand side

is x + 1 + x + 4 = 2x + 5.

By vedic formula we have 2x + 5 = 0 x = - 5 / 2.

Solve the following by using vedic method :

1.

(x + 3)3 x+1

______ = ____

(x + 5)3 x+7

63

(x + 2)3 x+1

______ = _____

(x + 3)3 x+4

with the text book procedures we proceed as follows

x3 + 6x2 + 12x +8 x+1

_______________ = _____

x3 + 9x2 + 27x +27 x+4

Now by cross multiplication,

( x + 4 ) ( x3 +6x2 + 12x + 8 ) = ( x + 1 ) ( x3 + 9x2 + 27x + 27 )

x4 + 6x3 + 12x2+ 8x + 4x3 + 24x2 + 48x + 32 =

x4 + 9x3 + 27x2 + 27x + x3 + 9x2 + 27x + 27

x4 +10x3 + 36x2 + 56x + 32 = x4 + 10x3 +36x2 + 54x + 27

56x + 32 = 54x + 27

56x – 54x = 27 – 32

2x = - 5

x=-5/2

Observe that ( N1 + D1 ) with in the cubes on

L.H.S. is x + 2 + x + 3 = 2x + 5 and

N2 + D2 on the right hand side

is x + 1 + x + 4 = 2x + 5.

By vedic formula we have 2x + 5 = 0 x = - 5 / 2.

Solve the following by using vedic method :

1.

(x + 3)3 x+1

______ = ____

(x + 5)3 x+7

63

66.
2.

(x - 5)3 x-3

______ = ____

(x - 7)3 x-9

6. Anurupye - Sunyamanyat

The Sutra Anurupye Sunyamanyat says : 'If one is in ratio, the other one is

We use this Sutra in solving a special type of simultaneous simple equations

in which the coefficients of 'one' variable are in the same ratio to each other

as the independent terms are to each other. In such a context the Sutra says

the 'other' variable is zero from which we get two simple equations in the first

variable (already considered) and of course give the same value for the

Example 1:

3x + 7y = 2

4x + 21y = 6

Observe that the y-coefficients are in the ratio 7 : 21 i.e., 1 : 3, which is

same as the ratio of independent terms i.e., 2 : 6 i.e., 1 : 3. Hence the other

variable x = 0 and 7y = 2 or 21y = 6 gives y = 2 / 7

Example 2:

323x + 147y = 1615

969x + 321y = 4845

The very appearance of the problem is frightening. But just an observation

and anurupye sunyamanyat give the solution x = 5, because coefficient of x

ratio is

323 : 969 = 1 : 3 and constant terms ratio is 1615 : 4845 = 1 : 3.

y = 0 and 323 x = 1615 or 969 x = 4845 gives x = 5.

Solve the following by anurupye sunyamanyat.

1. 12x + 78y = 12 2. 3x + 7y = 24

64

(x - 5)3 x-3

______ = ____

(x - 7)3 x-9

6. Anurupye - Sunyamanyat

The Sutra Anurupye Sunyamanyat says : 'If one is in ratio, the other one is

We use this Sutra in solving a special type of simultaneous simple equations

in which the coefficients of 'one' variable are in the same ratio to each other

as the independent terms are to each other. In such a context the Sutra says

the 'other' variable is zero from which we get two simple equations in the first

variable (already considered) and of course give the same value for the

Example 1:

3x + 7y = 2

4x + 21y = 6

Observe that the y-coefficients are in the ratio 7 : 21 i.e., 1 : 3, which is

same as the ratio of independent terms i.e., 2 : 6 i.e., 1 : 3. Hence the other

variable x = 0 and 7y = 2 or 21y = 6 gives y = 2 / 7

Example 2:

323x + 147y = 1615

969x + 321y = 4845

The very appearance of the problem is frightening. But just an observation

and anurupye sunyamanyat give the solution x = 5, because coefficient of x

ratio is

323 : 969 = 1 : 3 and constant terms ratio is 1615 : 4845 = 1 : 3.

y = 0 and 323 x = 1615 or 969 x = 4845 gives x = 5.

Solve the following by anurupye sunyamanyat.

1. 12x + 78y = 12 2. 3x + 7y = 24

64

67.
16x + 96y =16 12x + 5y = 96

3. 4x – 6y = 24 4. ax + by = bm

7x – 9y = 36 cx + dy = dm

In solving simultaneous quadratic equations, also we can take the help of the

‘sutra’ in the following way:

Example 3 :

Solve for x and y

x + 4y = 10

2 2

x + 5xy + 4y + 4x - 2y = 20

x2 + 5xy + 4y2 + 4x - 2y = 20 can be written as

( x + y ) ( x + 4y ) + 4x – 2y = 20

10 ( x + y ) + 4x – 2y = 20 ( Since x + 4y = 10 )

10x + 10y + 4x – 2y = 20

14x + 8y = 20

Now x + 4y = 10

14x + 8y = 20 and 4 : 8 :: 10 : 20

from the Sutra, x = 0 and 4y = 10, i.e.,, 8y= 20 y = 10/4 = 2½

Thus x = 0 and y = 2½ is the solution.

7. Sankalana - Vyavakalanabhyam

This Sutra means 'by addition and by subtraction'. It can be applied in solving

a special type of simultaneous equations where the x - coefficients and the y

- coefficients are found interchanged.

Example 1:

45x – 23y = 113

23x – 45y = 91

In the conventional method we have to make equal either the coefficient of x

or coefficient of y in both the equations. For that we have to multiply equation

( 1 ) by 45 and equation ( 2 ) by 23 and subtract to get the value of x and

then substitute the value of x in one of the equations to get the value of y or

we have to multiply equation ( 1 ) by 23 and equation ( 2 ) by 45 and then

subtract to get value of y and then substitute the value of y in one of the

65

3. 4x – 6y = 24 4. ax + by = bm

7x – 9y = 36 cx + dy = dm

In solving simultaneous quadratic equations, also we can take the help of the

‘sutra’ in the following way:

Example 3 :

Solve for x and y

x + 4y = 10

2 2

x + 5xy + 4y + 4x - 2y = 20

x2 + 5xy + 4y2 + 4x - 2y = 20 can be written as

( x + y ) ( x + 4y ) + 4x – 2y = 20

10 ( x + y ) + 4x – 2y = 20 ( Since x + 4y = 10 )

10x + 10y + 4x – 2y = 20

14x + 8y = 20

Now x + 4y = 10

14x + 8y = 20 and 4 : 8 :: 10 : 20

from the Sutra, x = 0 and 4y = 10, i.e.,, 8y= 20 y = 10/4 = 2½

Thus x = 0 and y = 2½ is the solution.

7. Sankalana - Vyavakalanabhyam

This Sutra means 'by addition and by subtraction'. It can be applied in solving

a special type of simultaneous equations where the x - coefficients and the y

- coefficients are found interchanged.

Example 1:

45x – 23y = 113

23x – 45y = 91

In the conventional method we have to make equal either the coefficient of x

or coefficient of y in both the equations. For that we have to multiply equation

( 1 ) by 45 and equation ( 2 ) by 23 and subtract to get the value of x and

then substitute the value of x in one of the equations to get the value of y or

we have to multiply equation ( 1 ) by 23 and equation ( 2 ) by 45 and then

subtract to get value of y and then substitute the value of y in one of the

65

68.
equations, to get the value of x. It is difficult process to think of.

From Sankalana – vyavakalanabhyam

add them,

i.e., ( 45x – 23y ) + ( 23x – 45y ) = 113 + 91

i.e., 68x – 68y = 204 x–y=3

subtract one from other,

i.e., ( 45x – 23y ) – ( 23x – 45y ) = 113 – 91

i.e., 22x + 22y = 22 x+y=1

and repeat the same sutra, we get x = 2 and y = - 1

Very simple addition and subtraction are enough, however big the

coefficients may be.

Example 2:

1955x – 476y = 2482

476x – 1955y = -4913

Oh ! what a problem ! And still

just add, 2431( x – y ) = - 2431 x – y = -1

subtract, 1479 ( x + y ) = 7395 x+y=5

once again add, 2x = 4 x=2

subtract - 2y = - 6 y=3

Solve the following problems usingSankalana – Vyavakalanabhyam.

1. 3x + 2y = 18

2x + 3y = 17

2. 5x – 21y = 26

21x – 5y = 26

3. 659x + 956y = 4186

956x + 659y = 3889

66

From Sankalana – vyavakalanabhyam

add them,

i.e., ( 45x – 23y ) + ( 23x – 45y ) = 113 + 91

i.e., 68x – 68y = 204 x–y=3

subtract one from other,

i.e., ( 45x – 23y ) – ( 23x – 45y ) = 113 – 91

i.e., 22x + 22y = 22 x+y=1

and repeat the same sutra, we get x = 2 and y = - 1

Very simple addition and subtraction are enough, however big the

coefficients may be.

Example 2:

1955x – 476y = 2482

476x – 1955y = -4913

Oh ! what a problem ! And still

just add, 2431( x – y ) = - 2431 x – y = -1

subtract, 1479 ( x + y ) = 7395 x+y=5

once again add, 2x = 4 x=2

subtract - 2y = - 6 y=3

Solve the following problems usingSankalana – Vyavakalanabhyam.

1. 3x + 2y = 18

2x + 3y = 17

2. 5x – 21y = 26

21x – 5y = 26

3. 659x + 956y = 4186

956x + 659y = 3889

66

69.
8. Puranapuranabhyam

The Sutra can be taken as Purana - Apuranabhyam which means by the

completion or non - completion. Purana is well known in the present

system. We can see its application in solving the roots for general form of

quadratic equation.

We have : ax2 + bx + c = 0

x2 + (b/a)x + c/a = 0 ( dividing by a )

x2 + (b/a)x = - c/a

completing the square ( i.e.,, purana ) on the L.H.S.

x2 + (b/a)x + (b2/4a2) = -c/a + (b2/4a2)

[x + (b/2a)]2 = (b2 - 4ac) / 4a2

________

- b ± √ b2 – 4ac

Proceeding in this way we finally get x = _______________

2a

Now we apply purana to solve problems.

Example 1. x3 + 6x2 + 11 x + 6 = 0.

Since (x + 2 )3 = x3 + 6x2 + 12x + 8

Add ( x + 2 ) to both sides

We get x3 + 6x2 + 11x + 6 + x + 2 = x + 2

i.e.,, x3 + 6x2 + 12x + 8 = x + 2

i.e.,, ( x + 2 )3 = ( x + 2 )

this is of the form y3 = y for y = x + 2

solution y = 0, y = 1, y = - 1

i.e.,, x + 2 = 0,1,-1

which gives x = -2,-1,-3

Example 2: x3 + 8x2 + 17x + 10 = 0

We know ( x + 3)3 = x3 + 9x2 + 27x + 27

So adding on the both sides, the term (x2 + 10x + 17 ), we get

x3 + 8x2 + 17x + x2 + 10x + 17 = x2 + 10x + 17

i.e.,, x3 + 9x2 + 27x + 27 = x2 + 6x + 9 + 4x + 8

i.e.,, ( x + 3 )3 = ( x + 3 )2 + 4 ( x + 3 ) – 4

67

The Sutra can be taken as Purana - Apuranabhyam which means by the

completion or non - completion. Purana is well known in the present

system. We can see its application in solving the roots for general form of

quadratic equation.

We have : ax2 + bx + c = 0

x2 + (b/a)x + c/a = 0 ( dividing by a )

x2 + (b/a)x = - c/a

completing the square ( i.e.,, purana ) on the L.H.S.

x2 + (b/a)x + (b2/4a2) = -c/a + (b2/4a2)

[x + (b/2a)]2 = (b2 - 4ac) / 4a2

________

- b ± √ b2 – 4ac

Proceeding in this way we finally get x = _______________

2a

Now we apply purana to solve problems.

Example 1. x3 + 6x2 + 11 x + 6 = 0.

Since (x + 2 )3 = x3 + 6x2 + 12x + 8

Add ( x + 2 ) to both sides

We get x3 + 6x2 + 11x + 6 + x + 2 = x + 2

i.e.,, x3 + 6x2 + 12x + 8 = x + 2

i.e.,, ( x + 2 )3 = ( x + 2 )

this is of the form y3 = y for y = x + 2

solution y = 0, y = 1, y = - 1

i.e.,, x + 2 = 0,1,-1

which gives x = -2,-1,-3

Example 2: x3 + 8x2 + 17x + 10 = 0

We know ( x + 3)3 = x3 + 9x2 + 27x + 27

So adding on the both sides, the term (x2 + 10x + 17 ), we get

x3 + 8x2 + 17x + x2 + 10x + 17 = x2 + 10x + 17

i.e.,, x3 + 9x2 + 27x + 27 = x2 + 6x + 9 + 4x + 8

i.e.,, ( x + 3 )3 = ( x + 3 )2 + 4 ( x + 3 ) – 4

67

70.
y3 = y2 + 4y – 4 for y = x + 3

y = 1, 2, -2.

Hence x = -2, -1, -5

Thus purana is helpful in factorization.

Further purana can be applied in solving Biquadratic equations also.

Solve the following using purana – apuranabhyam.

1. x3 – 6x2 + 11x – 6 = 0

2. x3 + 9x2 + 23x + 15 = 0

3. x2 + 2x – 3 = 0

4. x4 + 4x3 + 6x2 + 4x – 15 = 0

9. Calana - Kalanabhyam

In the book on Vedic Mathematics Sri Bharati Krishna Tirthaji mentioned

the Sutra 'Calana - Kalanabhyam' at only two places. The Sutra means

'Sequential motion'.

i) In the first instance it is used to find the roots of a quadratic equation7x2 –

11x – 7 = 0. Swamiji called the sutra as calculus formula. Its application at

that point is as follows.Now by calculus formula we say: 14x–11 = ±√317

A Note follows saying every Quadratic can thus be broken down into two

binomial factors. An explanation in terms of first differential, discriminant with

sufficient number of examples are given under the chapter ‘Quadratic

ii) At the Second instance under the chapter ‘Factorization and Differential

Calculus’ for factorizing expressions of 3rd, 4th and 5th degree, the procedure

is mentioned as'Vedic Sutras relating to Calana – Kalana – Differential

Further other Sutras 10 to 16 mentioned below are also used to get the

required results. Hence the sutra and its various applications will be taken up

at a later stage for discussion.

But sutra – 14 is discussed immediately after this item.

68

y = 1, 2, -2.

Hence x = -2, -1, -5

Thus purana is helpful in factorization.

Further purana can be applied in solving Biquadratic equations also.

Solve the following using purana – apuranabhyam.

1. x3 – 6x2 + 11x – 6 = 0

2. x3 + 9x2 + 23x + 15 = 0

3. x2 + 2x – 3 = 0

4. x4 + 4x3 + 6x2 + 4x – 15 = 0

9. Calana - Kalanabhyam

In the book on Vedic Mathematics Sri Bharati Krishna Tirthaji mentioned

the Sutra 'Calana - Kalanabhyam' at only two places. The Sutra means

'Sequential motion'.

i) In the first instance it is used to find the roots of a quadratic equation7x2 –

11x – 7 = 0. Swamiji called the sutra as calculus formula. Its application at

that point is as follows.Now by calculus formula we say: 14x–11 = ±√317

A Note follows saying every Quadratic can thus be broken down into two

binomial factors. An explanation in terms of first differential, discriminant with

sufficient number of examples are given under the chapter ‘Quadratic

ii) At the Second instance under the chapter ‘Factorization and Differential

Calculus’ for factorizing expressions of 3rd, 4th and 5th degree, the procedure

is mentioned as'Vedic Sutras relating to Calana – Kalana – Differential

Further other Sutras 10 to 16 mentioned below are also used to get the

required results. Hence the sutra and its various applications will be taken up

at a later stage for discussion.

But sutra – 14 is discussed immediately after this item.

68

71.
Now the remaining sutras :

10. YĀVADŨNAM ( The deficiency )

11. VYAŞłISAMAŞłIH ( Whole as one and one as whole )

12. ŚEŞĀNYAŃ KENA CARAMEĥA ( Remainder by the last digit )

13. SOPĀNTYADVAYAMANTYAM ( Ultimate and twice the penultimate )

15. GUĥITASAMUCCAYAH ( The whole product is the same )

16. GUĥAKA SAMUCCAYAH ( Collectivity of multipliers )

The Sutras have their applications in solving different problems in different

contexts. Further they are used along with other Sutras. So it is a bit of

inconvenience to deal each Sutra under a separate heading exclusively and

also independently. Of course they will be mentioned and also be applied in

solving the problems in the forth coming chapter wherever necessary. This

decision has been taken because up to now, we have treated each Sutra

independently and have not continued with any other Sutra even if it is

necessary. When the need for combining Sutras for filling the gaps in the

process arises, we may opt for it. Now we shall deal the fourteenth Sutra, the

Sutra left so far untouched.

10. Ekanyunena Purvena

The Sutra Ekanyunena purvena comes as a Sub-sutra to Nikhilam which gives

the meaning 'One less than the previous' or 'One less than the one before'.

1) The use of this sutra in case of multiplication by 9,99,999.. is as follows .

Method :

a) The left hand side digit (digits) is ( are) obtained by applying the ekanyunena

purvena i.e. by deduction 1 from the left side digit (digits) .

e.g. ( i ) 7 x 9; 7 – 1 = 6 ( L.H.S. digit )

b) The right hand side digit is the complement or difference between the

multiplier and the left hand side digit (digits) . i.e. 7 X 9 R.H.S is 9 - 6 = 3.

c) The two numbers give the answer; i.e. 7 X 9 = 63.

Example 1: 8 x 9 Step ( a ) gives 8 – 1 = 7 ( L.H.S. Digit )

Step ( b ) gives 9 – 7 = 2 ( R.H.S. Digit )

Step ( c ) gives the answer 72

69

10. YĀVADŨNAM ( The deficiency )

11. VYAŞłISAMAŞłIH ( Whole as one and one as whole )

12. ŚEŞĀNYAŃ KENA CARAMEĥA ( Remainder by the last digit )

13. SOPĀNTYADVAYAMANTYAM ( Ultimate and twice the penultimate )

15. GUĥITASAMUCCAYAH ( The whole product is the same )

16. GUĥAKA SAMUCCAYAH ( Collectivity of multipliers )

The Sutras have their applications in solving different problems in different

contexts. Further they are used along with other Sutras. So it is a bit of

inconvenience to deal each Sutra under a separate heading exclusively and

also independently. Of course they will be mentioned and also be applied in

solving the problems in the forth coming chapter wherever necessary. This

decision has been taken because up to now, we have treated each Sutra

independently and have not continued with any other Sutra even if it is

necessary. When the need for combining Sutras for filling the gaps in the

process arises, we may opt for it. Now we shall deal the fourteenth Sutra, the

Sutra left so far untouched.

10. Ekanyunena Purvena

The Sutra Ekanyunena purvena comes as a Sub-sutra to Nikhilam which gives

the meaning 'One less than the previous' or 'One less than the one before'.

1) The use of this sutra in case of multiplication by 9,99,999.. is as follows .

Method :

a) The left hand side digit (digits) is ( are) obtained by applying the ekanyunena

purvena i.e. by deduction 1 from the left side digit (digits) .

e.g. ( i ) 7 x 9; 7 – 1 = 6 ( L.H.S. digit )

b) The right hand side digit is the complement or difference between the

multiplier and the left hand side digit (digits) . i.e. 7 X 9 R.H.S is 9 - 6 = 3.

c) The two numbers give the answer; i.e. 7 X 9 = 63.

Example 1: 8 x 9 Step ( a ) gives 8 – 1 = 7 ( L.H.S. Digit )

Step ( b ) gives 9 – 7 = 2 ( R.H.S. Digit )

Step ( c ) gives the answer 72

69

72.
Example 2: 15 x 99 Step ( a ) : 15 – 1 = 14

Step ( b ) : 99 – 14 = 85 ( or 100 – 15 )

Step ( c ) : 15 x 99 = 1485

Example 3: 24 x 99

Answer :

Example 4: 356 x 999

Answer :

Example 5: 878 x 9999

Answer :

Note the process : The multiplicand has to be reduced by 1 to obtain the LHS

and the rightside is mechanically obtained by the subtraction of the L.H.S from

the multiplier which is practically a direct application of Nikhilam Sutra.

Now by Nikhilam

24 – 1 = 23 L.H.S.

x 99 – 23 = 76 R.H.S. (100–24)

_____________________________

23 / 76 = 2376

Reconsider the Example 4:

356 – 1 = 355 L.H.S.

x 999 – 355 = 644 R.H.S.

________________________

355 / 644 = 355644

70

Step ( b ) : 99 – 14 = 85 ( or 100 – 15 )

Step ( c ) : 15 x 99 = 1485

Example 3: 24 x 99

Answer :

Example 4: 356 x 999

Answer :

Example 5: 878 x 9999

Answer :

Note the process : The multiplicand has to be reduced by 1 to obtain the LHS

and the rightside is mechanically obtained by the subtraction of the L.H.S from

the multiplier which is practically a direct application of Nikhilam Sutra.

Now by Nikhilam

24 – 1 = 23 L.H.S.

x 99 – 23 = 76 R.H.S. (100–24)

_____________________________

23 / 76 = 2376

Reconsider the Example 4:

356 – 1 = 355 L.H.S.

x 999 – 355 = 644 R.H.S.

________________________

355 / 644 = 355644

70

73.
and in Example 5: 878 x 9999 we write

0878 – 1 = 877 L.H.S.

x 9999 – 877 = 9122 R.H.S.

__________________________

877 / 9122 = 8779122

Algebraic proof :

As any two digit number is of the form ( 10x + y ), we proceed

( 10x + y ) x 99

= ( 10x + y ) x ( 100 – 1 )

= 10x . 102 – 10x + 102 .y – y

= x . 103 + y . 102 – ( 10x + y )

= x . 103 + ( y – 1 ) . 102 + [ 102 – ( 10x + y )]

Thus the answer is a four digit number whose 1000th place is x,100th place is

( y - 1 ) and the two digit number which makes up the 10th and unit place is the

number obtained by subtracting the multiplicand from 100.(or apply nikhilam).

Thus in 37 X 99. The 1000th place is x i.e. 3

100th place is ( y - 1 ) i.e. (7 - 1 ) = 6

Number in the last two places 100-37=63.

Hence answer is 3663.

Apply Ekanyunena purvena to find out the products

1. 64 x 99 2. 723 x 999 3. 3251 x 9999

4. 43 x 999 5. 256 x 9999 6. 1857 x 99999

We have dealt the cases

i) When the multiplicand and multiplier both have the same number of digits

ii) When the multiplier has more number of digits than the multiplicand.

In both the cases the same rule applies. But what happens when the multiplier

has lesser digits?

i.e. for problems like 42 X 9, 124 X 9, 26325 X 99 etc.,

71

0878 – 1 = 877 L.H.S.

x 9999 – 877 = 9122 R.H.S.

__________________________

877 / 9122 = 8779122

Algebraic proof :

As any two digit number is of the form ( 10x + y ), we proceed

( 10x + y ) x 99

= ( 10x + y ) x ( 100 – 1 )

= 10x . 102 – 10x + 102 .y – y

= x . 103 + y . 102 – ( 10x + y )

= x . 103 + ( y – 1 ) . 102 + [ 102 – ( 10x + y )]

Thus the answer is a four digit number whose 1000th place is x,100th place is

( y - 1 ) and the two digit number which makes up the 10th and unit place is the

number obtained by subtracting the multiplicand from 100.(or apply nikhilam).

Thus in 37 X 99. The 1000th place is x i.e. 3

100th place is ( y - 1 ) i.e. (7 - 1 ) = 6

Number in the last two places 100-37=63.

Hence answer is 3663.

Apply Ekanyunena purvena to find out the products

1. 64 x 99 2. 723 x 999 3. 3251 x 9999

4. 43 x 999 5. 256 x 9999 6. 1857 x 99999

We have dealt the cases

i) When the multiplicand and multiplier both have the same number of digits

ii) When the multiplier has more number of digits than the multiplicand.

In both the cases the same rule applies. But what happens when the multiplier

has lesser digits?

i.e. for problems like 42 X 9, 124 X 9, 26325 X 99 etc.,

71