Multiplication in vedic maths

Multiplication of two numbers

Today we will learn Multiplication of two numbers . We will take two 3 digit number and see how we multiply it .

Before going forward , go down in this article and read some basic concepts Like Closure and Singleton , which we will be using .

You can releate each step while multiplication with the step shown in the right hand side picture.

1 2 3

X

4 5 6

Ans = XXXXX

Here in each Step the numbers which will come into picture will be

Step 1:

We will take last 1 block of numbers

3

6

and find the closure of it , closure (3,6) = 18

Ans : XXXX8 , CARRY : 1

Step 2 (last 2 blocks):

2 3

5 6

Closure (23 , 56) = 15 + 12 = 27

we add the carry = 1 to 27 and get 28

Ans : XXX88 carry = 2

Step 3 (last 3 blocks)

1 2 3

4 5 6

Closure (123,456 ) = 6 + 10 + 12 = 28

Final = 28 + 2 = 30

Ans = XX088 Carry = 3

Step 4 (First 2 blocks)

When you are done with all the blocks , we take the blocks from left , which keeps getting decreased.

1 2

4 5

Closure(12 , 45) = 5 + 8 = 13

Final = 13 + 3 = 16

Ans = X6088 Carry : 1

Step 5 (Left 1 block) (final step)

1

4

Closure(1,4) = 4

Final = 4 + 1 = 5

Final Answer = 56088

Download the Vedic maths Handout material , it contains some pdf files to learn some basic Vedic maths

Some concepts

Singleton

Singleton of a number is a single digit number which you get after adding all the digits of that number , if the resultant is not a single digit number than you again add it back.

Ex :

Singleton(23453) = 2+3+4+5+3 = 17 => 1+7 => 8

Singleton(343) = 3+4+3 = 10 => 1 + 0 => 1

Singleton(534533256) = 9 or 0

Note : While calculating the Singleton , you can banish those numbers which add up to 9 , Remember : Singleton(9+x) = Singleton(x)

So when you find Singleton(918754) , you can ignore

- 9

- 8 and 1

- 4 and 5

so , you are left with 7 , which is the Singleton

Closure

This is a very important thing in Vedic maths and you require it at almost every point in vedic maths.

Lets consider two numbers N1 = x1x2x3..xn and N2 = y1y2y3..yn (lets take equal digit numbers)

Closure (N1 , N2) = x1 * yn + x2 * yn-1 + x3*yn-2 + .... xn * y1

We have to multiply left most digit of first number with the right most digit of another number and keep on moving ...

Example :

CLOSURE (123 , 456) = 1*6 + 2*5 + 3*4 = 28

CLOSURE(23 , 43) = 8+9 = 17

Note : Closure(N) = Closure(N,N)