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IMC 2013/2/2

Source: Imc 2013 Problem 7

August 9, 2013
floor functionPutnamnumber theoryrelatively primeIMCcollege contests

Problem Statement

Let p,q\displaystyle{p,q} be relatively prime positive integers. Prove that ∑k=0pq−1(−1)⌊kp⌋+⌊kq⌋={0 if pq is even1if pq odd\displaystyle{ \sum_{k=0}^{pq-1} (-1)^{\left\lfloor \frac{k}{p}\right\rfloor + \left\lfloor \frac{k}{q}\right\rfloor} = \begin{cases} 0 & \textnormal{ if } pq \textnormal{ is even}\\ 1 & \textnormal{if } pq \textnormal{ odd}\end{cases}}
Proposed by Alexander Bolbot, State University, Novosibirsk.
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