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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.

shadow10
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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.

user21820
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  • Your solution is really beautiful which I came to see just now but how would I find the optimal bound? Can you shed some light on it? Thanks in advance. And sorry if that is obvious too. I am not too strong in combinatorics so I take unusually long time to understand or think up something trivial. Thanks once more. – shadow10 May 03 '14 at 07:46
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    @Joybangla: See http://math.stackexchange.com/questions/683660/pigeon-hole-principle-with-sum-of-5-integers/683670#683670 where it was a lot of work to get a tight bound for the exact same question for the smaller case of $5$ instead of $7$. – user21820 May 03 '14 at 09:07