# Divisible by 11

STATEMENT:

Given number n in decimal form, put alternatively – and + signs between the digits and compute them sum. The number is divisible by 11 if and only if this sum is divisible by 11. Thus a number is divisible by 11 and only if the difference between the sum of the digits in odd places and the sum of digits in even places is divisible by 11.

Example 1: Is the number 23456 divisible by 11?

Solution:

Observe that 2-3+4-5+6 = 4 and hence not divisible by 11. The test indicates that 23456 is not divisible by 11.

A palindrome is a number which leads the same from left to right or right to left. Thus a palindrome is a number n such that by reversing the digits of n, you get back n. For example, 232 is a 3 digit palindrome; 5445 is a 4 digit palindrome.

Example 2: Find all 3 digit palindromes which are divisible by 11.

Solution:

A 3 digit palindrome must be of the form aba, where a≠0 and b are digits. This divisible by 11 if and only if a-b+a = 2a-b is divisible by 11.

This is possible only if 2a-b = 0, 11 or -11. Since a≤1 and b≤9, we see that 2a-b ≥ 2(1)-(9) = 2-9 = -7 > -11. Hence, 2a-b = -11 is not possible.  Suppose 2a-b = 0. Then 2a = b. Thus, a = 1,b = 2 ; a = 2, b = 4; a=3, b = 6; and a = 4, b = 8 are possible.

We get the numbers, 121, 242, 363, 484.

For a = 6, b = 1, we see that 2a-b = 12-1 = 11 and hence divisible by for which 2a-b is divisible by 11. We get four more numbers 616, 737, 858 and 979.

Thus required numbers are 121, 242, 363, 484, 616, 737, 858 979.

Example 3: Prove that 12456 is divisible by 36 without actually dividing it.

Solution:

First notice that 36 = 4 x 9. So it is enough if we prove 12456 is divisible by 4 and 9 both. The last 2 digits of the given number 12456 is 56 which is divisible by 4 and it is left to show that, the given number 12456 is divisible by 9. Let us find the sum of digits of the given number 12456, i.e., 1+2+4+5+6 = 18, which is divisible by 9. Thus, the number 12456 is divisible by 36.