# 1. Prove: Any set of n integers has a subset whose sum is divisible by n.

Observe the useful divisibility definition: ${n}$ divides ${a}$ when ${a \equiv 0 \pmod n}$.

From the set of ${n}$ integers ${\{x_i\}}$, construct a set of ${n}$ sums ${y_i}$:

${y_1 =x_1}$
${y_m=y_{m-1}+x_m, 1

If ${y_i \equiv 0 \pmod n}$ for some ${i}$, ${y_i}$ is a sum divisible by ${n}$, which was to be shown.

Else, we must have ${y_k \equiv y_j \pmod n}$ for some ${j, either because the sums ${y_i}$ are not unique ${\pmod n}$ , or if they are so unique, because then there are ${n}$ sums in ${n-1}$ congruence classes (no ${y_i \equiv 0 \pmod n}$) and so the Pigeonhole Principle will require it.

In either case, ${y_k-y_j \equiv 0 \pmod n}$ implies ${y_k-y_j = x_{j+1} + ... + x_k}$ is a sum divisible by ${n}$.
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Aside: I saw this in a problem-solving exercise book and its immediate appeal caused me to choose it as an example problem. Later, I found that Erdős also selected it for inclusion in ‘Proofs from THE BOOK’, which made me smile. The Pigeonhole principle seems too obvious to be of much use, but it insinuates itself into a variety of proofs.

Note to students surfing the web to find answers to assigned problem sets: best not to copy my stuff; I am terrible at math.