# $\exists V {\subseteq} \lbrace v_1, \dots, v_n \rbrace \colon \sum V \equiv 0 \pmod{n}$

Last week I stumbled upon a nice little math problem: given a set of $n$ integers, show that there exists a nonempty subset $V \subseteq \lbrace v_1,v_2,\dots,v_n \rbrace$ such that $\sum V \equiv 0 \pmod{n}$.

Can you solve it? On the surface, this one feels intimidating — there’re $2^n$ subsets to consider. However, a nice simple technique made it as easy as a pie. So, maybe give it a shot?

**Solution**: Let $s_k$ be the sum of the first $k$ elements, that is

We are also allow an empty sum, $s_0=0$. Thus, there are $n{+}1$ sums. However, since we’re working in $\mathbb{Z}/n\mathbb{Z}$, there are only $n$ distinct possible result. By pigeonhole principle, there must exists $s_i=s_j$ whose $0 \le i <j \le n $. Hence,

\[s_j - s_i \equiv v_{i+1} + v_{i+2} + \cdots + v_j \equiv 0 \pmod{n}\]is an answer to the problem.

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