Given a collection of (not necessarily distinct) integers, how large must the collection be to guarantee that for a particular $n$, there is a subcollection of $n$ numbers summing to a multiple of $n$?

A glance shows that for any $n$, at least $2n-1$ are required, since if we have a collection of $n-1$ 0's and $n-1$ 1's, then there is no way to sum to a multiple of $n$. Is there an elegant way to prove $2n-1$ is always sufficient?

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    $\begingroup$ If we have a subcollection of $0$s... then isn't $0$ a multiple of $n$? $\endgroup$ – Ben Grossmann Jul 23 '17 at 20:38
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    $\begingroup$ Just to clarify: you require that we use $n$ numbers, yes? Not $≤n$. Thus , having $n-1$ $0's$ doesn't solve the problem. $\endgroup$ – lulu Jul 23 '17 at 20:40
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    $\begingroup$ If I have understood you correctly, this is the content of the Erdős–Ginzburg–Ziv theorem. $\endgroup$ – lulu Jul 23 '17 at 20:42
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    $\begingroup$ Here is the original Erdos paper bolyai.hu/~p_erdos/1961-25.pdf $\endgroup$ – Raffaele Jul 24 '17 at 9:31
  • $\begingroup$ Yes, I did mean exactly $n$ numbers, no less. Thanks everyone for linking me to resources! $\endgroup$ – Bolton Bailey Jul 24 '17 at 19:24

There is an elegant proof of this using the Chevalley Warning theorem which easily proves it for prime $n$ (then a multiplicative property is proved for the rest).

See this, section 4. (page 11)


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