Proving that among any $2n - 1$ integers, there's always a subset of $n$ which sum to a multiple of $n$

Question

How can one prove that among any $2n - 1$ integers, there's always a subset of $n$ which sum to a multiple of $n$?

It is not hard to see this is equivalent to show that among $2n-1$ residue classes modulo $n$ there are $n$ whose sum is the zero-class. Thus, this problem is an example of a Zero sum problem.

Also, the general case was first proven in the $1961$ paper of Erdős, Ginzburg and Ziv.

This is a resource intended to be part of the ongoing effort to deal with abstract duplicates. There are quite a few posts here related to proving that among any $2n - 1$ integers, there's always a subset of $n$ which sum to a multiple of $n$, with varying degrees of generality from using only specific values of $n$ to proving it for all cases. Each of my following answers deal with a degree of generality by explaining it and then linking to the related existing posts.

However, there are many ways to deal with this problem, including some which may not yet be handled by any posts on this site. Some examples, as suggested by quid's question comment, include:

• What are some different ways to prove the results? Basically all solutions use the pigeon-hole principle in some way, so can this be solved without using that principle? Also, as MathOverflow's EGZ theorem (Erdős-Ginzburg-Ziv) asks, can the general solution, as in the EGZ theorem, be proven without using the Chevalley–Warning theorem (or a variant of its proof)?
• The answer links to Use Pigeonhole to show of any set of $2^{n+1}-1$ positive integers is possible choose $2^{n}$ elements such that their sum is divisible by $2^{n}$. where one answer shows how you can prove it using induction. Are there any other special cases of subsets of $n$ which can be solved on their own apart from the linked ones related to powers of $2$?
• The answer gives a proof for $n = 2$, and gives examples of specific $n$ which have been asked on this site of $3$, $4$, $5$, $6$ and $9$. However, are there any other small values of $n$ which can also reasonably be handled explicitly?

Answer 1

Most of the questions on this site involve asking to prove the result for a specific, relatively small, value of $n$ (although sometimes the question specifies a larger value than $2n-1$ for the number of integers to choose from). The answers for $n$ being prime usually involve some sort of sets of cases and using the pigeon-hole principle, while the non-prime values involve handling each of the prime factor(s) separately and then showing how they can be combined to get the final result.

• For $n = 2$, there are no posts I could find, possibly because it's very simple to do. Since there are only $2$ parities, i.e., even & odd, then among any $3$ integers, at least $2$ must have the same parity, so the sum of those $2$ integers will be even.
• For $n = 3$, there are:
  • How do I prove that among any $5$ integers, you are able to find $3$ such that their sum is divisible by $3$?
  • Prove that for any set of 5 integers, there is at least one subset of 3 integers whose sum is divisible by 3
• For $n = 4$, there is:
  • Given 7 arbitrary integers,sum of 4 of them is divisible by 4
• For $n = 5$, there are:
  • Given any nine numbers, prove there exists a subset of five numbers such that its sum is divisible by $5$.
  • Pigeon hole principle with sum of 5 integers
• For $n = 6$, there is:
  • Among any $11$ integers, sum of $6$ of them is divisible by $6$
• For $n = 9$, there is:
  • Pigeon - Hole principle and remainder modulus

Answer 2

There are sometimes posts involving asking for proving the result for some subset of possible values of $n$. This would normally involve using some specific property of the subset to prove the result. The only posts I could find which involves this are for powers of $2$:

• Use Pigeonhole to show of any set of $2^{n+1}-1$ positive integers is possible choose $2^{n}$ elements such that their sum is divisible by $2^{n}$.
• West German Mathematics Olympiad 1981
• Prove for any number n, it is possible to select $X = 2^n$ numbers from $2^{n+1}$ numbers s.t. the sum of X is divisible by $2^n$

Answer 3

Posts can also show how to prove you can multiply results with $2$ known cases which work to get a larger case which also works, e.g., if the result works for $n = i$ and $n = j$, then it also works for $n = ij$. From this, you can extend known results for a few specific cases only to show it works for an infinite set of values.

This answer proves it for the specific case of when $n = 3$. Also, A question relevant to EGZ theorem? shows how to prove it in the general case.

Answer 4

There are several posts which deal with proving the general result. These are:

• Given any 2n-1 integers, prove that there are always n of them which add up to a multiple of n
• Given $n$ integers, is it always possible to choose $m$ from them so that their sum is a multiple of $m$?
• Show that in any set of $2n$ integers, there is a subset of $n$ integers whose sum is divisible by $n$.

There's also a post about this on MathOverflow:

• EGZ theorem (Erdős-Ginzburg-Ziv)

Answer 5

Posts may potentially alter the general conditions, such as restrict the set of available congruences and use a set of available integers which is considerably larger than necessary, with the idea being that a specific method can be used to solve the problem. The only such post I know of is the following one which deals with choosing $19$ integers from a set of $181$ integers which only include the $10$ square congruences modulo $19$, with this being solved directly using the pigeon-hole principle on those available congruences:

• Can't understand the solution of this INMO problem