#### Hide answer & extension

Let the two digit number chosen by \(10a+b\), with \(a\) and \(b\) one digit integers. Reversed this will be \(10b+a\). The sum of these will be \(11a+11b\) which is divisible by 11.

This will work for any integer with an even number of digits. Let our number have \(2n\) digits. It can be written as:

$$\sum_{i=1}^{2n}10^{i-1} a_i $$

Adding it to its reverse, we get:

$$\sum_{i=1}^{2n}10^{i-1} a_i + \sum_{i=1}^{2n}10^{2n-i} a_i = \sum_{i=1}^{2n}(10^{i-1}+10^{2n-i}) a_i $$

\(10^{i-1}+10^{2n-i}\) is divisible by 11 (for \(n \in \mathbb{N}\), \(i \in \mathbb{N}\), \(i\leq 2n\)). This can be shown by induction on \(n\):

If \(n=1\): \(10^{i-1}+10^{2n-i} = 10^{i-1}+10^{2-i}=11\), which is clearly divisible by 11.

Suppose result is true for \(n-1\). Now consider \(10^{i-1}+10^{2n-i}\).

If \(i>1\), then \(10^{i-1}+10^{2n-i}= 10(10^{(i-1)-1}+10^{(2n-2)-(i-1)}\) which is divisible by 11 by the inductive hypothesis.

If \(i=1\), then:

$$10^{i-1}+10^{2n-i} = 1+10^{2n-1} = \sum_{j=1}^{j=2n-2}9\times 10^j +11$$
$$=\sum_{j=1}^{j=n-1}99\times 10^{2j} +11$$
$$=11\left(\sum_{j=1}^{j=n-1}9\times 10^{2j} +1\right)$$

#### Extension

Which numbers with an odd number of digits will be divisible by 11 when added to their reverse?