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Is the following statement true? If so, how can it be proved?

Every set of $2^n-1$ positive integers, $n\in\{1,2,\dots\}$, has a subset of size $2^{n-1}$ that sums to a multiple of $n$.

An attempt at a proof

I was able to prove a similar, but weaker statement:

If $n\in\{1,2,\dots\}$ is a power of $2$, then every set of $2^n-1$ positive integers has a subset of size $2^{n-1}$ that sums to a multiple of $n$.

Indeed, noting that $k\leq2^{k-1}$ for every $k\in\{1,2,\dots\}$ (this can be shown by induction), we must have $n\leq 2^{n-1}$, and therefore $n=2^{k-1}$ for some $k\in\{1,2,\dots,n\}$, hence $2^{n-1}=mn$ for some $m\in\{1,2,\dots\}$ (namely, $m=2^{n-k}$), whence $2^n-1\geq m(2n-1)$. By the Erdős–Ginzburg–Ziv theorem, there are $m$ disjoint subsets each of size $n$, each of which sums to a multiple of $n$. Their union is a subset of size $mn=2^{n-1}$ that sums to a multiple of $n$. Q.E.D.

Evan Aad
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    If the $2^n-1$ numbers are all congruent to $1\bmod{n}$ then any $2^{n-1}$ of them have a sum congruent to $2^{n-1}\bmod{n}$ hence we need $n\mid 2^{n-1}$, i.e. the problem is false when $n$ is not a power of two. When $n$ is a power of two, you correctly solve it. – oVlad May 29 '23 at 15:04
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    @oVlad This solves my question. Thanks. If you post it as an answer, I will accept it. – Evan Aad May 29 '23 at 15:21
  • Glad I helped :) – oVlad May 29 '23 at 15:50

1 Answers1

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If the $2^n-1$ numbers are all congruent to $1\bmod{n}$, then any $2^{n−1}$ of them have a sum congruent to $2^{n−1}\bmod{n}$, hence we need $n\mid 2^{n−1}$, i.e. the problem is false when $n$ is not a power of two.

When $n$ is a power of two, you correctly solve it.

oVlad
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