Given a sequence of $10$ integers, show that there is a subset of consecutive integers whose sum is divisible by $10$

Suppose I have subsets




$$\{a_1,a_2,a_3,a_4, \dots, a_{10}\}$$

I am stuck on what to do to prove this. Am I suppose to somehow use $a_i \equiv a_j \pmod{10}$ for $i \neq j $?

  • $\begingroup$ Why have you assumed that the first of the consecutive integers is always $a_1$? $\endgroup$ – Shaun Aug 21 at 22:26

Consider the sums $a_1$, $a_1+a_2$, ..., $a_1+a_2+...+a_{10}$. If any of those are divisible by 10, you are done. Otherwise, the ten have the remainders 1 through nine, and therefore two of them have the same remainder modulo 10 by the Pigeonhole Principle. Let $i$ and $j$ be those two numbers. Then $$a_{i+1}+...+a_j=(a_1+...+a_j)-(a_1+...+a_i)$$ is surely divisible by 10.

  • $\begingroup$ They will all have remainders, and since there's 10 subsets two will have the same remainder. Where do i go from here. Is the difference of these subsets divisible by 10 or is this the wrong way to go $\endgroup$ – Kevin G Aug 21 at 22:16
  • $\begingroup$ Proof expanded. $\endgroup$ – Matthew Daly Aug 21 at 22:23
  • $\begingroup$ There's actually 1023 nonempty subsets @KevinG $\endgroup$ – Roddy MacPhee Sep 1 at 23:32

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