## Problem of the Day #23: Divisibility of Factorials and Factorial SumsApril 11, 2011

Posted by Albert in : potd , trackback

Let $x$ be the highest positive integer such that $x!$ divides $$\sum\limits_{n=k}^{2011} n!$$ for $0 \leq k \leq 2011$. Find the number of integers $k$ such that $x \neq k$.