# 16 Sutra Formulas of Vedic Mathematics

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Vedic mathematics these days is gaining popularity because of its speedy and accurate calculations. Calculations are an integral part of any profession today and the ability to do it quickly and accurately definitely is an important skill that anyone would desire to have.

Vedic Maths is based on sixteen sutras, which were unravelled by Swami Bharati Krishna tirtha Ji Maharaj, the Shankaracharya of Govardhan peeth, puri. The Atharvaveda is said to have lots of information on science and mathematics and it is from there, Shankaracharya of Goverdhan peeth puri, decoded the whole information and presented it in the form of 16 sutras.

Through this article, an attempt has been made to bring this knowledge to you in a simple and lucid language, with examples.

Very useful in finding the product of numbers, if the sum of unit digits of the two numbers totals to 10.

e.g 24 x 26=?

=(first digit x one more than first digit) (product of unit digits of both the number)

=( 2
x 3 ) ( 4x6)

=6 24

So, 24 x 26=624. The answer has come without doing any elaborate calculation.

This sutra is very commonly used in the subtraction of a number from the powers of 10.

Eg. 10000

- 7688

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

2312

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

- The last number is 8 and this is subtracted from 10 and
the next 8 is subtracted from 9, all other numbers are subtracted by 9 and the result comes out almost orally.

This is used for multiplications and the formula used is explained below.

ab x cd=(ac) (ad + bc ) (bd)

e.g

24 x 12=(2x1) (2x2 + 4 x1) (4 x 2 )

2 8 8

And the answer is 288.

This is used to solve division problems when the divisor is a little greater than the nearest power of 10.

Let us see this division 434 / 12.

The first thing to be noted is that divisor 12, is near 10, so the above sutra can be applied. Look at the divisor and check how many digits it has. In this case, it is having 2 digits. So according to this sutra, the dividend should be split into two parts 43 and 4 and the working is as below.

As shown above, the divisor is written and leaving 1 apart, 2 is taken down as 2 bar, i.e. vinculum 2.

The dividend is divided into two parts 43 and 4. 4 of 43 is taken down and to this four, the vinculum 2 is multiplied to get vinculum 8 which is written under 3 of 43. 3 vinculum8 would be vinculum 5 which is taken down. Vinculum 2 of the divisor is multiplied with this vinculum 5 and the result 10 is written under 4 and totaled to 14.

14 is taken down as it is. Now 45 is a vinculum number because 5 is vinculum. According toVedic maths rules vinculum, 5 is complemented with 10 to get normal 5 which is taken down. The number next to the vinculum number should be reduced by 1. So, 4 becomes 3 and comes down

And the answer is quotient=35 and remainder=14 when 434/12.

This is used to solve equations in the form

a. ax + b=cx +d

can

So x=d-b/a-c

b. (x+a)(x+b)=(x+c)(x+d)

So x=cd-ab/a+b-c-d

1. A term which occurs as a common factor in all the terms is equated to zeroe.g. : 14x + 9x=4x + 12x

Here x occurs as a common factor with all terms and hence the value of x according to this sutra is zero.

2. If the product of the independent term on either side of the equation is equal the value of the variable will be zero, which is the second interpretation of this sutra.

Eg.(x +8) (x+3)=(x + 12 ) (x + 2 )

8 x 3=24=12 x 2 and hence value of x in this equation would be 0

Suppose:

2x + 4y=8 and

4x + 6y=16, the ratio of terms with x=2x/4x=½

The ratio of the R.H.S term is also 8/16=½

Therefore, the other variable, in this case y=0

Substituting this value of y in any other of the
two equations, we can get value of x

2x + 4 (0)=8

2x=8

Therefore x=8/4=2.

This sutra is used to solve equations. (if the coefficient of 1 variable in same in both the equation irrespective of the sign)

What it means is that the coefficient of the 1 variable in equation 1 should be equal to the
coefficient of the 2nd variable in the second equation and the coefficient of the 2nd variable in equation 1 should be equal to the coefficient of 1st variable in equation 2. Then the two equations can be added and subtracted
and solved for variables

For e.g

4x + 2y=6………… equation1 and

2x + 4y=7 ………….equation 2. Now add equation 1 and 2 we get

6x + 6y=13 or

6 (x + y)=13 or

X + y=13/7…………eq 3. Subtract equation 2 for eq 1

2x
-2y=-1

2 (x -y )=-1 or

X – y=- 1 / 2 …… eq 4…….therefore

Y=x + 1 / 2…………eq 5 substitute this in equation 3. So we get

X + (x + ½)=13/7… solving for x, we get

X=19/7=2.71…….

And y=x + 0.5….. from eq 5

So,
y=2.71 +0.5=3.21

This can be used to solve addition problems when the unit digits of the numbers add up to 10 for e.g. number 22 and 18 the unit digits add up to 10. Let try to add

295 + 46 + 28 + 15 + 44 + 22=?

Now we need to check
and number and pair them in such a way that their unit places add up to 10. So….

295 + 46 + 28 + 15 + 44 + 22=?

Rearrange to put the paired number together.

(295 + 15) + (46 + 44) + (28 + 22)

300 + 90 + 50=440.

This
happened in easy steps instead of long calculations

The application of this sutra can be found in calculus to find roots of a quadratic equation and the second application is in differential calculus for factorizing 3rd, 4th, and 5 degrees expression. This sutra finds very specialized applications in the area of higher mathematics .

This is used to find squares of numbers that are close to the powers of base 10. Compare the number with the closed base to it and find the deficiency or excess. Square the difference and this is one part of the answer, reduce
the given number or increase it by the difference it has to the power of base 10

Let us understand this with an example.

Let us try to find the square of 12

12 is near to 10 and it is 2 excess than 10.

1. Square
the difference (excess in this case). so 2 x2=4….this is the unit place

2. Now add the excess to the number. the number is 12 so 12 + 2=14…this is the left part of the answer

3. Combining both of them we get=144

4.
Solving it in equation form

5. 122=(12 + 2) ( 2)2=144

This helps in the factorization of quadratic equations.

This sutra gives you the process of converting fractions to decimals.

For eg 1/29

1. The last digit of the divisor should be 9. It is in this case, now increase the value by 1 of the number next to 9. So, the number is
2 and increasing it by 1 makes it 3

2. The dividend is 1 now it has to be divided by 3 so,

3. 1 / 3

4. Doing it mentally it will be 0.0 and remainder 1 and it is written as

5. 0.10 and 10 is divided by 3 and
it will be written as 3 and remainder 1 written to left

6. 0.1 01 3 now 13 is to be divided by 3 and it will be written as 4 and remainder 1 written to left

7. 0.101314 and keep on dividing it by 3 to as many decimal
places as needed. For three decimal places the answer is 0.034

This sutra is used to find solution of equations in the following form

1/ ab + 1/ac=1/ad + 1/bc

Where a, b, c and d are in arithmetic progression

b=a + z

c=a + 2z

d=a + 3z

solution for such equations
is 2c + d=0

e. g.

1/ (x+1)(x+2) + 1/ (x +1)(x+3)=1/ (x+1)(x + 4) + 1/ (x+2)(x +3)

Now according to the above sutra the solution would be

2(x +3) + (x+4)=0

2x +6+x+4=0

3x + 10=0

X=-10/3

Multiplication can be done using this sutra.

The product of two number can be calculated using this sutra when the multiplier consists of only 9

For example 12 x 99=?

The process to do it is

1. Reduce 1 from multiplicand
ie. 12-1=11

2. The other part of the answer would be 99-11=88 (complement of 99)

Hence the answer is 1188

It is used to find the correctness of the answers in factorization problems and it states that the sum of the coefficients in the product is equal to the sum of coefficients of the factors and if this condition is satisfied
then the equation can be considered to be balanced.

For eg let us consider a quadratic equation

8x2 + 11x + 3=(x+1)(8x+3)

In this case, the sum of coefficients is 8+11+3=22

Product of the sum of coefficients
of the factors=2 (8+3)=2 x 11=22

Since both, the totals tally the equation is balanced and correct.

This sutra holds good for a perfect number.

Let us find the factors of number 28,

1 x28=28

2 x 14=28

4 x 7=28

So, in this case, the sum of factors is 1+2+4+7+14=28

The sum of factors equals the factor
of the sums, so 28 is said to be a perfect number.

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