To remember Multiplication Table, consider the sum of multiplicand and multiplier.

Remember the values for the sum 10 (all other values of the multiplication Table) using simple technique from Vedic Mathematics.

The method we follow, here, is very simple to understand and very easy to follow.

The method is based on “Nikhilam” sutra of vedic mathematics.

The method will be clear from the following examples.

**Example 1 : **

Suppose, we have to find 9 x 6.

First we write one below the other.

9

6

Then we *subtract the digits from 10 *and write the values (10-9=1; 10-6=4) to the right of the digits with a ‘-‘ sign in between.

9 – 1

6 – 4

The product has two parts. The first part is *the cross difference *(here it is 9 – 4 = 6 – 1 = 5).

The second part is *the vertical product *of the right digits (here it is 1 x 4 = 4).

We write the two parts separated by a slash.

9 – 1

6 – 4

—–

5/4

—–

So, 9 x 6 = 54.

Let us see one more example.

**Example 2 : **

Suppose, we have to find 8 x 7.

First we write one below the other.

8

7

Then we *subtract the digits from 10 *and write the values (10-8=2; 10-7=3) to the right of the digits with a ‘-‘ sign in between.

8 – 2

7 – 3

The product has two parts. The first part is *the cross difference *(here it is 8 – 3 = 7 – 2 = 5).

The second part is *the vertical product *of the right digits (here it is 2 x 3 = 6).

We write the two parts separated by a slash.

8 – 2

7 – 3

—–

5/6

—–

So, 8 x 7 = 56.

Let us see one more example.

**Example 3 : **

Suppose, we have to find 9 x 9.

First we write one below the other.

9

9

Then we *subtract the digits from 10 *and write the values (10-9=1; 10-9=1) to the right of the digits with a ‘-‘ sign in between.

9 – 1

7 – 1

The product has two parts. The first part is* the cross difference *(here it is 9 – 1 = 9 – 1 = 8).

The second part is *the vertical product *of the right digits (here it is 1 x 1 = 1).

We write the two parts separated by a slash.

9 – 1

9 – 1

—–

8/1

—–

So, 9 x 9 = 81.

In the next examples, the second part has two digits.

Let us see how to handle the issue.

**Example 4: **

To find 7 x 6

First we write one below the other.

7

6

Then we *subtract the digits from 10 *and write the values (10-7=3; 10-6=4) to the right of the digits with a ‘-‘ sign in between.

7 – 3

6 – 4

The product has two parts. The first part is *the cross difference *(here it is 7 – 4 = 6 – 3 = 3).

The second part is *the vertical product *of the right digits (here it is 3 x 4 = 12).

We write the two parts separated by a slash.

7 – 3

6 – 4

—–

3/12

—–

The second part, here, has two digits.

we *retain the units’ digit *(2) and *carry over the other digit *(1) to the left.

7 – 3

6 – 4

————–

(3+1)/2 = 4/2

————-

So, the answer becomes (3+1)/2 = 4/2

Thus, 7 x 6 = 42.

**Example 5 : **

To find 8 x 3

By following the above procedure, we may write as follows.

8 – 2

3 – 7

—–

2/14

—–

The first part = 8 – 7 = 3 – 2 = 1.

The second part here is 2×7 = 14.

It has two digits. we *retain the units’ digit *(4) and *carry over the other digit *(1) to the left.

8 – 2

3 – 7

————–

(1+1)/4 = 2/4

————-

So, the answer becomes (1+1)/4 = 2/4

Thus, 8 x 3 = 24.

let us see one last example.

**Example 6 : **

To find 6 x 5

By following the above procedure, we may write as follows.

6 – 4

5 – 5

—–

1/20

—–

The first part = 6 – 5 = 5 – 4 = 1.

The second part here is 4×5 = 20.

It has two digits. we *retain the units’ digit *(0) and *carry over the other digit *(2) to the left.

6 – 4

5 – 5

————–

(1+2)/0 = 3/0

————-

So, the answer becomes (1+2)/0 = 3/0

Thus, 6 x 5 = 30.

Thus, we can arrive at any values up to 10 x 10.