In every set of $14$ integers there are two that their difference is divisible by $13$ 
Prove that in every set of $14$ integers there are two that their difference is divisible by $13$

The proof goes like this, there are $13$ remainders by dividing by $13$, there are $14$ numbers so from the pigeon hole principle, there are two that have the same remainder so their difference is divisible by $13$.
But what if we'll take a set of numbers that has nothing in common with $13$? 
Like $\{10,10^2,10^3...,10^{14}\}$ or a prime that's further away from 13: $\{89,89^2,...,89^{14}\}$ how is it possible that those numbers and their differences have something in common with a totally different prime?

You can break up the set of integers in to $13$ sets which do not intersect.
There are the multiples of of $13$, numbers whose remainder when divided by $13$ is $1$, numbers whose remainder is $2$ and so on.  If you pick $14$ distinct numbers, at least two of them have to belong to set same set by the pigeonhole principle. The difference of these two will be divisible by $13$.
