Any number multiplied by 1 is the number itself. In the same way, any number multiplied by 11 is the number itself repeated twice.

For example:

1 x 3 = 3

11 x 3 = 33

What about multiplying any number higher than 10 with 11?

Do we have any special method that makes the calculation easier and faster?

Yes, we do have a special method.

Let us understand this with an example when a 2 digit number is multiplied by 11:

45 x 11 = ?

In Vedic Method, you write down the number being multiplied and put the total of the digits of that number between the two digits.

multiply-by-11-1

Here, you write 4 and 5 as outer digits, and add pair 4, 5 (4+5=9) and write answer between 4 and 5 as middle digit.

Another example in which a carry generated when adding the digits.

57 x 11 = ?

multiply-by-11-2

Now multiply longer number (3 or more digits) with 11:

243 x 11 = ?

In Vedic Method, you write down the number being multiplied, and add the pair of digits of that number and place between the two digits.

multiply-by-11-3

Here, you write 2 and 3 as outer digits, and add first pair 2, 4 (2+4=6) and write answer between 2 and 3 as middle digit, then add next pair 4, 3 (4+3=7) and write answer between 6 and 3.

Another example in which a carry generated when adding the digits.

561 x 11 = ?

multiply-by-11-4

See the comparison between the Traditional Method with the Vedic Method. The calculation in Vedic Method is so easy that it can be done mentally, in one line and without wasting paper/space.

⇐ Multiplication by 9
Multiplication by 11, 111, 1111 and so on… ⇒