In the previous section Multiplication by 11, we learned multiplication of any number by 11.

Now we are going to extend this method for multiplication of any number by another number having all the digits as 1.

Step 1: When any number is multiplied by another number containing 1s, the most significant digit (or leftmost) of the first number becomes the most significant digit of the answer and the least significant digit (or rightmost) of the first number becomes the least significant digit of the answer.

Step 2: The middle number is formed by adding the digits of the first number starting from right in a group of two, then in a group of 3, then in a group of 4 and so on till all the digits of the first number are added in this way and then going opposite, that is adding in a group of 3, group of 2.

Note that we have to keep adding maximum to the depth of the number of 1s in the 2nd number.

Let’s understand this with an example:

**Example 1: 23 x 111 = ?**

Step 1: The most significant digit (or the leftmost) of the answer is the most significant digit of 23, i.e. 2 and the least significant digit (or the rightmost) of the answer is the least significant digit of 23, i.e. 3.

Step 2: Three ones are there in 111. So we have to keep adding maximum to the depth of three. So the middle digits in the answer will be 2+3, 2+3.

=> 2, 5, 5, 3

**Example 2: 87 x 111 = ?**

Note: In this case, carry is generated when adding 8 and 7.

Step 1: The most significant digit (or the leftmost) of the answer is the most significant digit of 87, i.e. 8 and the least significant digit (or the rightmost) of the answer is the least significant digit of 87, i.e. 7.

Step 2: Three ones are there in 111. So we have to keep adding maximum to the depth of three. So the middle digits in the answer will be 8+7, 8+7.

=> 8, (carry 1)5, (carry 1)5, 7

=> 9, 6, 5, 7

**Example 3: 2312 x 1111 = ?**

Step 1: The most significant digit (or the leftmost) of the answer is the most significant digit of 2312, i.e. 2 and the least significant digit (or the rightmost) of the answer is the least significant digit of 2312, i.e. 2.

Step 2: Four ones are there in 1111. So we have to keep adding maximum to the depth of four. So the middle digits in the answer will be 2+3, 2+3+1, 2+3+1+2, 3+1+2, 1+2.

=> 2, 5, 6, 8, 6, 3, 2

Let us practice with a few more examples:

a. 938 x 111 b. 6752 x 111 c. 19385679 x 1111

d. 2382 x 1111 e. 6172 x 11111 f. 123439 x 1111

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