MathDB
m does not divide f(n) - Austria 2010

Source:

May 13, 2010
number theory proposednumber theory

Problem Statement

Let f(n)=∑k=02010nkf(n)=\sum_{k=0}^{2010}n^k. Show that for any integer mm satisfying 2⩽m⩽20102\leqslant m\leqslant 2010, there exists no natural number nn such that f(n)f(n) is divisible by mm.
(41st Austrian Mathematical Olympiad, National Competition, part 1, Problem 1)
Back to ProblemsView on AoPS