TITLE: Divisibility and Pigeonhole principle
QUESTION [5 upvotes]: Given a sequence of $p$ integers $a_1, a_2, \ldots, a_p$, show that there exist consecutive terms in the sequence whose sum is divisible by $p$.  That is, show that there are $i$ and $j$, with $1 \leq i \leq j \leq p$, such that $a_i + a_{i+1} + \cdots + a_j$ is divisible by $p$.
I'm having trouble with labeling which entities are the pigeons, and which are the pigeonholes.  I think somewhere down the line, there has to be more different sums than $p$, but that is just a guess.

REPLY [5 votes]: Hint:  The holes are remainders on division by $p$.  Consider $\sum_{i=1}^k a_i$ for $k=1,2,3 \ldots,p$  If any are divisible by $p$ you are done.  If not, You have $p$ sums with only $p-1$ values of remainder allowed.