Prove that for any $x \in \mathbb N$ such that $xProve that every positive integer $x$ with $x<n!$ is the sum of at most $n$ distinct divisors of $n!$.

Hint: Note that $x = m (n-1)! + r$ where $0 \le m < n$ and $0 \le r < (n-1)!$.  Use induction.
EDIT: Oops, this is wrong: as Steven Stadnicki noted, $m (n-1)!$ doesn't necessarily divide $n!$.
