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.

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.

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.
42
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
43
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.

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.

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, 1000 i.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).


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.

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

sutra i.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.

Example :- 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 fromthe 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 number 8 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 + 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

Multiplication Method 2

METHOD 2

Sutra :"EKANYUNENA POORVENA"

This sutra means :"ONE LESS THAN ONE BEFORE"

STEPS

1.We get answer in two parts

2. Reduce 1 from multiplication and (EKANYUNENA POORVENA) viz 8-1=7

write 7 as the LHS part of the answer

3. subtract 7 from 9 viz 9-7 =2 and write it as the RHS

The answer is 72 

Multiplication Method 1

METHOD 1.

Sutra 1) "EK ADHIKENA POORVENA"

(sutra=Vedic word formula)

This sutra literally menas : BY ONE MORE THAN ONE BEFORE


2) "ANTYAYORDASHAKEPT"

This sutra literally means :"END TO SUM AS TEN"

Eg:- 65 *  65

HOW TO SOLVE

STEP 1) We write answer in two parts. We multiply R.H.S digit of multiplier viz 5. It Means 5 * 5 = 25 . This is the R.H.S answer

NOTE :-  IN ANY MULTIPLICATION OF DIGITS ENDING WITH 5 WE USE THIS METHOD ONLY TO MULTIPLY.

STEP 2) We add one to L.H.S digit viz 6 + 1 = 7 and multiply it with the L.H.S. this forms the L.H.S of the answer

so ,  6 + 1 = 7 * 6 = 42

So the total of answer by joining LHS and RHS is

4225 

What is Vedic Mathematics

Vedic mathematics is a magical method of fast calculation. It is a new and unique system based on simple rule  and principals which enable mathematics problem of all kinds to be solved easily and efficiently .

Vedic mathematics is a list of sixteen basic sūtras, or aphorisms, presented by a Hindu scholar and mathematician, Bharati Krishna Tirthaji Maharaja, during the early part of the 20th century. While its author claimed it to be a system of mathematics, this is not generally accepted, and it is more generally regarded as a set of strategies for calculation. These are said to be creative and useful, and can be applied in a number of ways to calculation methods in arithmetic and algebra, most notably within the education system. Some of its methods share similarities with the Trachtenberg system.