IMO 1993 SL 20
Origin: ROM
Problem
Let c1, . . . , cn \inR (n \geq2) such that 0 \leqn i=1 ci \leqn. Show that we can find integers k1, . . . , kn such that n i=1 ki = 0 and 1 −n \leqci + nki \leqn for every i = 1, . . . , n.
Solution
For every real x we shall denote by \lfloorx\rfloorand \lceilx\rceilthe greatest integer less than or equal to x and the smallest integer greater than or equal to x respectively. The condition ci + nki \in[1 −n, n] is equivalent to ki \inIi = / 1−ci n −1, 1 −ci n . For every ci, this interval contains two integers (not necessarily distinct), namely pi = B 1−ci n −1 C \leqqi = D 1 −ci n E . In order to show that there exist integers ki \inIi with n i=1 ki = 0, it is sufficient to show that n i=1 pi \leq0 \leqn i=1 qi. Since pi < 1−ci n , we have n i=1 pi < 1 − n i=1 ci n \leq1, and consequently n i=1 pi \leq0 because the pi’s are integers. On the other hand, qi > −ci n implies n i=1 qi > − n i=1 ci n \geq−1, which leads to n i=1 qi \geq0. The proof is complete.