Seems straightforward pigeonhole If we are given $37$ integers then show that it is possible to choose $7$ of them with sum divisible by $7$.
 I have tried this problem but with no avail. If we assume there are no integers with remainder $0\mod 7$ then we get there are at least $7$ with same remainder. But when we have more than $1$ integers with remainder $0\mod 7$ I don't see any other way to solve it without considering cases. But that is a tedious job to do and maybe some slick application of PHP will do this instantly. Any and all help is welcome. Thanks in advance. 

By your argument, if there are no integers with remainder $k$ mod $7$ for any $k \in [0..6]$, then we are done. So there must be at least one integer for each remainder mod $7$. What is their sum mod $7$?
But as you might have suspected, if you want a tight bound on the minimum number of integers needed to guarantee such a set, it will be much less than $37$ and proving it will need a proliferation of cases.
