Competition problem — Number theory

Library / /2 of 15

Number theory Difficulty 4.5 AIME Prove it United States

Problem:
Prove that the sum of the $2009$ th powers of the first $2009$ positive integers is divisible by $2009$ .

Solution

Solution:
Using the factoring formula
$x^{2009} + y^{2009} = (x + y)\left(x^{2008} - x^{2007}y + x^{2006}y^{2} - \cdots - x y^{2007} + y^{2008}\right),$
we find that each of the numbers
$0^{2009} + 2009^{2009},\ 1^{2009} + 2008^{2009},\ 2^{2009} + 2007^{2009},\ \ldots,\ 1004^{2009} + 1005^{2009}$
is divisible by $2009$ . Consequently so is their sum.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.