**Dr. V.K.Maheshwari****, M.A(Socio, Phil) B.Sc. M. Ed, Ph.D**** **

**Former Principal, K.L.D.A.V.(P.G) College, Roorkee, India**

**E-mail ****ID-mahesh42n@rediffmail.com**

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.

**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.

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

** **

**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 multiplication.

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 —–

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 _____________

/ (3×2) = 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 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 + – 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.

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

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.

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 / 4×2 = 10608 ( rule -f )

‾‾‾‾‾‾‾‾‾‾‾‾

Ex. 10: 1275X1004. Base is 1000.

1275 275

1004 004

‾‾‾‾‾‾‾‾‾‾‾‾

1279/ 275×4 = 1279 / 1100 ( rule -f )

____________ = 1280100

**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) :

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)

Ex.3

41

X 41

‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾

16 : 4 + 4 : 1 = 1681.

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) : 2 X 4 = 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.

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.

Ex.6

47

‾‾‾‾‾‾

1606

65

‾‾‾‾‾‾‾

2256

Step (i): 8 X 7 = 56; 5, the carried over digit is placed below the second digit.

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.

**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.

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 Quotient.

Thus Q = 102 andR = 1

Example 2 : Divide 1697 by 14.

14 1 6 9 7

- 4 -4–8–4

‾‾‾‾ ‾‾‾‾‾‾‾

1 2 1 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 digits of the dividend for the remainder.

1 2 3 25 98 Step ( 1 ) & Step ( 2 )

-2-3

‾‾‾‾‾ ‾‾‾‾‾‾‾‾

Now proceed the sequence of steps write –2 and –3 as follows :

1 2 3 25 98

-2-3 -4 -6

‾‾‾‾‾ -2–3

‾‾‾‾‾‾‾‾‾‾

21 1 5

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.

**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 follows.

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’

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 denominators

i.e., 3x – 2 + 2x – 1 = 5x – 3 = 0

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)

a2x

2

+ 2abx + b2

= a2x

2

+ 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.

**6. Anurupye – Sunyamanyat**

The Sutra Anurupye Sunyamanyat says : ‘If one is in ratio, the other one is zero’.

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 variable.

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

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

x2

+ 5xy + 4y2

+ 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 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

**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

**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 Equations’.

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 Calculus’.

**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

Example 2: 15 x 99 Step ( a ) : 15 – 1 = 14

Step ( b ) : 99 – 14 = 85 ( or 100 – 15 )

Step ( c ) : 15 x 99 = 1485

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

**11. Anurupyena**

The upa-Sutra ‘anurupyena’ means ‘proportionality’. This Sutra is highly useful to find products of two numbers when both of them are near the Common bases i.e powers of base 10 . It is very clear that in such cases the expected ‘Simplicity ‘ in doing problems is absent.

Example 1: 46 X 43

As per the previous methods, if we select 100 as base we get 46 -54 This is much more difficult and of no use.

43 -57

‾‾‾‾‾‾‾‾

Now by ‘anurupyena’ we consider a working base In three ways. We can solve the problem.

Method 1: Take the nearest higher multiple of 10. In this case it is 50.

Treat it as 100 / 2 = 50. Now the steps are as follows:

i) Choose the working base near to the numbers under consideration. i.e., working base is 100 / 2 = 50

ii) Write the numbers one below the other

i.e. 4 6

4 3

‾‾‾‾‾‾‾

iii) Write the differences of the two numbers respectively from 50 against each number on right side

i.e. 46 -04

43 -07

‾‾‾‾‾‾‾‾‾

iv) Write cross-subtraction or cross- addition as the case may be under the line drawn.

v) Multiply the differences and write the product in the left side of the answer.

46 -04

43 -07

____________

39 / -4 x –7

= 28

vi) Since base is 100 / 2 = 50 , 39 in the answer represents 39X50.

Hence divide 39 by 2 because 50 = 100 / 2

Thus 39 ÷ 2 gives 19½ where 19 is quotient and 1 is remainder . This 1 as

Reminder gives one 50 making the L.H.S of the answer 28 + 50 = 78(or

Remainder ½ x 100 + 28 ) i.e. R.H.S 19 and L.H.S 78 together give the answer1978 We represent it as

46 -04

43 -07

‾‾‾‾‾‾‾‾‾

2) 39 / 28

‾‾‾‾‾‾‾‾‾

19½ / 28

= 19 / 78 = 1978

Example 2: 42 X 48.

With 100 / 2 = 50 as working base, the problem is as follows:

42 -08

48 -02

‾‾‾‾‾‾‾‾‾

2) 40 / 16

‾‾‾‾‾‾‾‾‾

20 / 16

42 x 48 = 2016

Method 2: For the example 1: 46X43. We take the same working base 50. We treat it as 50=5X10. i.e. we operate with 10 but not with 100 as in method now

(195 + 2) / 8 = 1978

[Since we operate with 10, the R.H.S portion shall have only unit place .Hence out of the product 28, 2 is carried over to left side. The L.H.S portion of the answer shall be multiplied by 5, since we have taken 50 = 5 X 10.]

Now in the example 2: 42 x 48 we can carry as follows by treating 50 = 5 x 10

**12. Adyamadyenantya – mantyena**

The Sutra ‘ adyamadyenantya-mantyena’ means ‘the first by the first and the last by the last’. Suppose we are asked to find out the area of a rectangular card board whose length and breadth are respectively 6ft . 4 inches and 5 ft. 8 inches. Generally we continue the problem like this.

Area = Length X Breath

= 6’ 4″ X 5’ 8″ Since 1’ = 12″, conversion

= ( 6 X 12 + 4) ( 5 X 12 + in to single unit

= 76″ 68″ = 5168 Sq. inches.

Since 1 sq. ft. =12 X 12 = 144sq.inches we have area

5168 = 144) 5168 (35

‾‾‾‾

144 432

‾‾‾‾

848

720 i.e., 35 Sq. ft 128 Sq. inches

‾‾‾‾‾

128

By Vedic principles we proceed in the way “the first by first and the last by last”

i.e. 6’ 4″ can be treated as 6x + 4 and 5’ 8″ as 5x + 8,

Where x= 1ft. = 12 in;x2

is sq. ft.

Now ( 6x + 4 )(5x + 8 )

= 30×2

+ 6.8.x + 4.5.x + 32

= 30×2

+ 48x + 20x + 32

= 30×2

+ 68. x + 32

= 30×2

+ ( 5x + 8 ). x + 32 Writing 68 = 5 x 12 + 8

= 35×2

+ 8. x + 32

= 35 Sq. ft. + 8 x 12 Sq. in + 32 Sq. in

= 35 Sq. ft. + 96 Sq. in + 32 Sq. in

= 35 Sq. ft. + 128 Sq. in

** **

**13. Yavadunam Tavadunikrtya Varganca Yojayet**

The meaning of the Sutra is ‘what ever the deficiency subtract that deficit from the number and write along side the square of that deficit’.

This Sutra can be applicable to obtain squares of numbers close to bases of powers of 10.

Method-1 : Numbers near and less than the bases of powers of 10.

Eg 1: 92

Here base is 10.

The answer is separated in to two parts by a’/’

Note that deficit is 10 – 9 = 1

Multiply the deficit by itself or square it

12

= 1. As the deficiency is 1, subtract it from the number i.e., 9–1 = 8.

Now put 8 on the left and 1 on the right side of the vertical line or slash i.e., 8/1.

Hence 81 is answer.

Eg. 2: 962

Here base is 100.

Since deficit is 100-96=4 and square of it is 16 and the deficiency subtracted from the number 96 gives 96-4 = 92, we get the answer 92 / 16

Thus 962

= 9216.

**14. Antyayor Dasakepi**

The Sutra signifies numbers of which the last digits added up give 10. i.e. the Sutra works in multiplication of numbers for example: 25 and 25, 47 and 43, 62 and 68, 116 and 114. Note that in each case the sum of the last digit of first number to the last digit of second number is 10. Further the portion of digits or numbers left wards to the last digits remain the same. At that instant use

Ekadhikena on left hand side digits. Multiplication of the last digits gives the right hand part of the answer.

Example 1 : 47 X 43

See the end digits sum 7 + 3 = 10 ; then by the sutras antyayor dasakepi and ekadhikena we have the answer.

47 x 43 = ( 4 + 1 ) x 4/ 7 x 3

= 20 / 21

= 2021.

Example 2: 62 x 68

2 + 8 = 10, L.H.S. portion remains the same i.e.,, 6.

Ekadhikena of 6 gives 7

62 x 68 = ( 6 x 7 )/ ( 2 x 8 )

= 42 / 16

= 4216.

It is further interesting to note that the same rule works when the sum of the last 2, last 3, last 4 – - – digits added respectively equal to 100, 1000, 10000 — – - . The simple point to remember is to multiply each product by 10, 100, 1000, – - as the case may be . Your can observe that this is more convenient while working with the product of 3 digit numbers.

Eg. 1: 292 x 208

Here 92 + 08 = 100, L.H.S portion is same i.e. 2

292 x 208 = ( 2 x 3 )/ 92 x 8

60 / =736 ( for 100 raise the L.H.S. product by 0 )

= 60736.

**15. Antyayoreva**

‘Atyayoreva’ means ‘only the last terms’. This is useful in solving simple equations of the following type.

The type of equations are those whose numerator and denominator on the L.H.S. bearing the independent terms stand in the same ratio to each other as the entire numerator and the entire denominator of the R.H.S. stand to each other.

Let us have a look at the following example.

Example 1:

x2 + 2x + 7 x + 2

__________ = _____

x2 + 3x + 5 x + 3

In the conventional method we proceed as

x2 + 2x + 7 x + 2

__________ = _____

x2 + 3x + 5 x + 3

(x + 3) (x2 + 2x + 7) = (x + 2) (x2 + 3x + 5)

x3 + 2×2 + 7x +3×2 + 6x + 21 = x3 + 3×2 + 5x + 2×2 +6x + 10

x3 + 5×2 + 13x + 21 = x3 + 5×2 + 11x+ 10

Canceling like terms on both sides

13x + 21 = 11x + 10

13x – 11x = 10 – 21

2x = -11

x = -11 / 2

Now we solve the problem using anatyayoreva.

x2 + 2x + 7 x + 2

__________ = _____

x2 + 3x + 5 x + 3

Consider

x2 + 2x + 7 x + 2

__________ = _____

x2 + 3x +5 x + 3

Observe that

x2 + 2x x (x + 2) x + 2

______ = ________ = _____

x2 + 3x x (x + 3) x + 3

This is according to the condition in the sutra. Hence from the sutra

x + 2 7

_____ = __

x + 3 5

5x + 10 = 7x + 21

7x – 5x = -21 + 10

2x = -11

x = -11 / 2

**16. Lopana Sthapanabhyam**

Lopana sthapanabhyam means ‘by alternate elimination and retention’. Consider the case of factorization of quadratic equation of type ax2 +by2 + cz2 +dxy + eyz + fzx This is a homogeneous equation of second degree in threevariables x, y, z. The sub-sutra removes the difficulty and makes thefactorization simple. The steps are as follows:

i) Eliminate z by putting z = 0 and retain x and y and factorize thus obtained aquadratic in x and y by means of ‘adyamadyena’ sutra.;

ii) Similarly eliminate y and retain x and z and factorize the quadratic in x and z.

iii) With these two sets of factors, fill in the gaps caused by the elimination

process of z and y respectively. This gives actual factors of the expression.

Example 1: 3×2

+ 7xy + 2y2

+ 11xz + 7yz + 6z2

.

Step (i): Eliminate z and retain x, y; factorize

3×2

+ 7xy + 2y2

= (3x + y) (x + 2y)

Step (ii): Eliminate y and retain x, z; factorize

3×2

+ 11xz + 6z2

= (3x + 2z) (x + 3z)

Step (iii): Fill the gaps, the given expression

= (3x + y + 2z) (x + 2y + 3z)

Example 2: 12×2

+ 11xy + 2y2

- 13xz – 7yz +3z2

.

Step (i): Eliminate z i.e., z = 0; factorize

12×2

+ 11xy + 2y2

= (3x + 2y) (4x + y)

Step (ii): Eliminate y i.e., y = 0; factorize

12×2

- 13xz + 3z2

= (4x -3z) (3x – z)

Step (iii): Fill in the gaps; the given expression

= (4x + y – 3z) (3x + 2y – z)

Example 3: 3×2+6y2+2z2+11xy+7yz+6xz+19x+22y+13z+20

Step (i): Eliminate y and z, retain x and independent term

i.e., y = 0, z = 0 in the expression (E).

Then E = 3×2

+ 19x + 20 = (x + 5) (3x + 4)

Step (ii): Eliminate z and x, retain y and independent term

i.e., z = 0, x = 0 in the expression.

Then E = 6y2

+ 22y + 20 = (2y + 4) (3y + 5)

Step (iii): Eliminate x and y, retain z and independent term

i.e., x = 0, y = 0 in the expression.

Then E = 2z2

+ 13z + 20 = (z + 4) (2z + 5)

Step (iv): The expression has the factors (think of independent terms)

= (3x + 2y + z + 4) (x + 3y + 2z + 5).

In this way either homogeneous equations of second degree or general equations of second degree in three variables can be very easily solved by applying ‘adyamadyena’ and ‘lopanasthapanabhyam’ sutras.

**17. Vilokanam**

The Sutra ‘Vilokanam’ means ‘Observation’. Generally we come across problems which can be solved by mere observation. But we follow the same conventional procedure and obtain the solution. But the hint behind the Sutra enables us to observe the problem completely and find the pattern and finally solve the problem by just observation.

Let us take the equation x + ( 1/x ) = 5/2 Without noticing the logic in the problem, the conventional process tends us to solve the problem in the following way.

1 5 x + __ = __ x 2 x2 + 1 5

_____ = __

x 2 2×2 + 2 = 5x 2×2 – 5x + 2 = 0

2×2 – 4x – x + 2 = 0

2x (x – 2) – (x – 2) = 0

(x – 2) (2x – 1) = 0

x – 2 = 0 gives x = 2

2x – 1 = 0 gives x = ½

But by Vilokanam i.e.,, observation

1 5 x + __ = __ can be viewed as x 2 1 1

x + __ = 2 + __ giving x = 2 or ½. x 2

Consider some examples.

Example 1 :

x x + 2 34

____ + _____ = ___

x + 2 x 15

In the conventional process, we have to take L.C.M, cross-multiplication.

simplification and factorization. But Vilokanam gives

34 9 + 25 3 5

__ = _____ = __ + __

15 5 x 3 5 3

x x + 2 3 5

____ + _____ = __ +__

x + 2 x 5 3

gives

x 3 5

_____ = __ or __

x + 2 5 3

5x = 3x + 6 or 3x = 5x + 10

2x = 6 or -2x = 10

x = 3 or x = -5

Example 2 :

x + 5 x + 6 113

____ + _____ = ___

x + 6 x + 5 56

Now,

113 49 + 64 7 8

___ = _______ =___ + ___

56 7 x 8 8 7

x + 5 7 x+5 8

____ = __ or ____ = __

x + 6 8 x+6 7

8x + 40 = 7x+ 42 7x + 35 = 8x + 48

or

x = 42 – 40 =2 -x = 48 –35 = 13

x = 2 or x = -13.

**8. Gunita Samuccayah : Samuccaya Gunitah**

In connection with factorization of quadratic expressions a sub-Sutra, viz. ‘Gunita samuccayah-Samuccaya Gunitah’ is useful. It is intended for the purpose of verifying the correctness of obtained answers in multiplications, divisions and factorizations. It means in this context:

‘The product of the sum of the coefficients sc in the factors is equal to the sum of the coefficients sc in the product’ Symbolically we represent as sc of the product = product of the sc (in the factors)

Example 1: (x + 3) (x + 2) = x2 + 5x + 6

Now ( x + 3 ) ( x + 2 ) = 4 x 3 = 12 : Thus verified.

Example 2: (x – 4) (2x + 5) = 2×2 – 3x – 20

Sc of the product 2 – 3 – 20 = – 21

Product of the Sc = (1 – 4) (2 + 5) = (-3) (7) = – 21. Hence verified.

In case of cubics, biquadratics also the same rule applies.

We have (x + 2) (x + 3) (x + 4) = x3 +9×2 + 26x + 24

Sc of the product = 1 + 9 + 26 + 24 = 60

Product of the Sc = (1 + 2) (1 + 3) (1 + 4) = 3 x 4 x 5 = 60. Verified.

So far we have considered a majority of the upa-sutras as mentioned in the

Vedic mathematics book. Only a few Upa-Sutras are not dealt under a separate

heading . They are

2) S’ISYATE S’ESASAMJ ÑAH

4) KEVALAIH SAPTAKAMGUNYAT

5) VESTANAM

6) YAVADŨNAM TAVADŨNAM and

10) SAMUCCAYAGUNITAH already find place in respective places.

**REFERANCE:**

http://www.vedamu.org/Veda/1795$Vedic_Mathematics_Methods.pdf