Possible Duplicate:
pigeonhole principle and division
I need a little help in an exercise. Given 100 natural numbers $a_{1},..,a_{100}$ , prove that there is a consecutive sub-sequence $a_{i} , a_{i+1} , ... , a_{j}$ such that $a_{i} + a_{i+1} + ... +a_{j} = 0 \mod 100$
Well, because
$(a_{i} + a_{i+1} + ... +a_{j}) \mod 100 = (a_{i} \mod 100) + (a_{i+1} \mod 100) + ... + (a_{j} \mod 100)$
I can assume that all the elements in the sequence are between $1$ to $100$.
Anyhow, I tried to think about it and I didn't manage to prove that :(
I thought at the beginning I should use pigeonhole principle but it didn't work out for me. Any clues anyone?? Thanks :)