A Pigeonhole Principle problem:
101 positive integers are placed on a circle whose sum is 300. Prove
that it is possible to choose from these numbers some consecutive
numbers whose sum is equal to 200.
(I don't know if the word 'consecutive' is appropriate in this case ,I mean that these numbers follow each other on that circle)
Best Answer
Start at a certain position and form sums of subsequences of length $1, 2, \dotsc, 101$ starting at that position and going in clockwise direction. This is an increasing sequence of $101$ numbers so there are two different entries that are equal $\bmod$ $100$ (end in the same two digits). The difference between those entries is a positive multiple of $100$ and less than $300$ so either $100$ or $200$. This difference corresponds to a subsequence of numbers on the circle with sum either $100$ or $200$. If it is $200$ we're done, otherwise take the complement of that sequence.