Divisibility and Pigeonhole principle

Question

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.

You may find of interest a similar Pigeonhole argument: if $\rm:n:$ is coprime to $10,$ then every integer with at least $\rm:n:$ digits $\ne 0$ has a contiguous digit subsequence that forms an integer $\ne 0$ divisible by $\rm:n.\ \ $ — Bill Dubuque, Sep 30 '12 at 18:29
See also: http://math.stackexchange.com/questions/1933631/prove-that-there-are-i-and-j-with-1-le-i-le-j-le-n-such-that-a-ia — Martin Sleziak, Nov 21 '16 at 04:14

score 5 · Accepted Answer · answered Sep 30 '12 at 17:14

5

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.

answered Sep 30 '12 at 17:14

Ross Millikan

374,822

A ha! So two of these sums will have the same remainder, so their difference is divisible by p, right? – user1526710 Sep 30 '12 at 17:38
@user1526710: Exactly. Then if the two that match are $k_1$ and $k_2$, the sum from $k1+1 through $k_2$ is the one you want. – Ross Millikan Sep 30 '12 at 17:47

Divisibility and Pigeonhole principle

1 Answers1

Linked