Competition problem — Number theory

Library / /22 of 120

Number theory Difficulty 4.7 AIME Prove it Saudi Arabia

Let $n$ be a positive integer such that $2011^{2011}$ divides $n!$ . Prove that $2011^{2012}$ divides $n!$ .

Solution

Since $2011$ is a prime and $2011^{2011}$ divides $n!$ , it follows that in $n!$ we have at least $2011$ multiples of $2011$ . These are
$2011, 2 \cdot 2011, 3 \cdot 2011, \ldots, 2011 \cdot 2011.$

Therefore $2011! \cdot 2011^{2011}$ divides $n!$ , hence $2011^{2012}$ divides $n!$ .

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 and solution reproduced as published; topic and difficulty added by this site.