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Prove that given any $17$ integers, there exist nine of them whose sum is divisible by $9.$

I'm pretty sure we have to use the pigeonhole principle, with the possible remainders as the pigeonhole, but that gives a lot of possibilities and I don't know where to start.

I tried to compare it to this question which was easier to solve:

Prove that given any $5$ integers, there exist three of them whose sum is divisible by $3.$

Lily Potter
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  • How'd you do the simpler version. I imagine this is probably exactly the same. In both cases you have $2n -1$ integers and need to prove htat there are $n$ of them whose sum is divisible by $n$. – fleablood Dec 04 '19 at 00:12
  • https://en.m.wikipedia.org/wiki/Restricted_sumset#Cauchy–Davenport_theorem the theorem is called Erdős–Ginzburg–Ziv theorem though I can't find the proof – kingW3 Dec 04 '19 at 00:12
  • @kingW3 Note the answer of my proposed duplicate indicates it's always true for $2n - 1$, i..e, for $n = 9$, it's true for $2n - 1 = 17$, and for $n = 3$ it's true for $2n - 1 = 5$. Also, the answer's comment gives a link to original paper with the theorem you're asking about. – John Omielan Dec 04 '19 at 00:14
  • Yes, but the answer on the original is not easy and refers to the Chevary-Warning theorem without proof. Don't know if this is easier with a specific $n=9$. – fleablood Dec 04 '19 at 00:19
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    @fleablood With a proof for the simpler case the OP asks about, i.e., for $5$ integers having at least $3$ with a sum of $3$, you can prove it for your case of $17$ integers having a subset of $9$ being a multiple of $9$ using the second page of the original paper of the proof where it shows how to fairly easily extend the result to the product of any $2$ values, in particular $3$ and $3$ here. This is after it proves it for all primes, so it can then prove the case for all $n$. – John Omielan Dec 04 '19 at 00:22
  • @LilyPotter You can use the results from Prove that for any set of 5 integers, there is at least one subset of 3 integers whose sum is divisible by 3 to prove your simpler case of $n = 3$ values summing to a multiple of $3$ among $2n - 1 = 5$ integers, then use what I suggest in my comment above to then prove it for $n = 9 = 3 \times 3$. – John Omielan Dec 04 '19 at 00:35
  • I don't think this question is a duplicate. The general $2n$ and $n$ theorem is much more difficult than the specific case in this problem. Voting to reopen. – awkward Dec 04 '19 at 14:01
  • @awkward I agree it's not the best, but I just found Pigeon - Hole principle and remainder modulus, that actually asks the same question, & with your answer basically explaining what I suggest in my comment above. Thus, I'm also voting to reopen & then will vote to close as a duplicate of the other question. – John Omielan Dec 04 '19 at 18:41
  • @JohnOmielan Ha Ha! I had forgotten that I answered that question! Thanks for reminding me. – awkward Dec 05 '19 at 00:26

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