Can anyone help me with this problem;
Let n ≤ 2 be an integer. Given a sequence of n integers a1,
a2,...,an, show that there exist consecutive terms in the sequence whose sum
is divisible by n. That is, show that there are i and j, with 1 ≤i ≤j ≤n,
such that:
sij= ai + ai + 1 + · · · + aj ≡ 0 (mod n)
Thanks
(I am too lazy to read to uncode the sigma's in Paul's post but here is how I would do it).
Consider
....
There are such integers for . Now if any one of these leaves remainder of then the proof is complete. Otherwise let we have remainder (excluding 0) among integers so two of them leave the same remainder by pigeonholing with then is divisible by n.