TITLE: In every set of $14$ integers there are two that their difference is divisible by $13$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: I'm glad I now know what the Pigeonhole Principal is after having heard it so many times. :)
There being $14$ distinct numbers$\in N\gt 13$, simply apply the Division Algorithm to get the remainders $0-12$ for $13$ of those numbers.  So produce set $\{13q_0 + r_0,13q_1 + r_1,...,13q_{12}+r_{12}\}$.  Remainders $r_1-r_{12}$ are the numbers $0$ through $12$ and are distinct.  if any were the same the difference would be divisible by $13$ and done.  now the $14th$ number must have the same remainder as one of the other $13$ and so the difference is divisible by $13$.  I actually stumbled on this in an induction proof.  Induction on the range of numbers/greatest number $n$.
Base case $n=14$:must have all numbers $1$ through $14$.  $14-1=13$.
Assume true for $n$.  For $n+1$ can apply what is said above.  If all $14$ chosen from set through range $n$ done/true, by inductive hypothesis.
Otherwise if $n+1$ chosen have set/subset $\{\{i_1,i_2,...,i_{13}\},n+1\}$ and can apply what is said above.  That is, if none of $\{i_1,i_2,...,i_{13}\}$ have same remainder mod $13$ they cover the entire range of remainders mod $13$ and therefore $n+1$ mod $13$ must be same as one of $\{i_1,i_2,...,i_{13}\}$mod $13$.  This completes the induction. (if same mod $13$ difference between the two is divisible by $13$) 
Induction is not needed for this problem, though.