From: cd@newsun-graphe.lri.fr (Charles Delorme) Subject: Re: Subset sum divisible by n Date: 16 Jan 2000 09:53:21 GMT Newsgroups: sci.math Summary: Erdos-Ginzburg-Ziv theorem In article <85ql7b$jb2$1@uwm.edu>, David G Radcliffe writes: |> Does every set of 2n-1 integers contain a subset |> with n elements whose sum is divisible by n? |> |> Yes, this is known as Erdos-Ginzburg-Ziv theorem, Here is a reference Eros, Ginzburg, Ziv a Theorem in additive number theory Bull. Res. Council Israel 10 (1961) Sincerely yours -- Charles Delorme tous les mégalomanes LRI ont une signature cd@lri.fr à étages ============================================================================== From: Robin Chapman Subject: Re: Subset sum divisible by n Date: Sun, 16 Jan 2000 15:08:11 GMT Newsgroups: sci.math In article <85ql7b$jb2$1@uwm.edu>, David G Radcliffe wrote: > Does every set of 2n-1 integers contain a subset > with n elements whose sum is divisible by n? Yes. It's quite easy to see that if this is valid for n = r and n = s then it's also true for n = rs. I'll omit this. So we can reduce to the case where n = p is prime, which is not so trivial. Let a_1, ..., a_{2p-1} be the numbers and let I = {1,..., 2p-1} be the index set. For a subset J of I let S(J) be the sum of the a_j for j in J. We count the sum T of the S(J)^{p-1} (mod p) over all J with |J| = p in two ways. First of all each term in S(J)^{p-1} is a monomial in the a_i s. Each such monomial contains k different a_i where 1 <= k < p. This monomial occurs in (2p-1-k choose p - k) = (2p-1-k)!/[(p-k)!(p-1)!] of the S(J) and this number is divisible by p. Hence T = 0 (mod p). Now if all of the S(J) are non-zero then S(J)^{p-1} = 1 (mod p). Thus T = (2p-1 choose p) =/= 0 (mod p), a contradiction. -- Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html "`The twenty-first century didn't begin until a minute past midnight January first 2001.'" John Brunner, _Stand on Zanzibar_ (1968) Sent via Deja.com http://www.deja.com/ Before you buy.