A cute problem from Peter Winkler's collection of math puzzles.
$$ S = \{n! | 1 \leq n \le 100, n \in N\}$$
Can we remove a single element from $S$ such that the product of the elements of the resulting set is a perfect square?
Scroll down for a solution.
Product of elements of $S$ is
$$ P = 1! 2! \dots 99! 100! $$
Pair up terms $(2n-1)! (2n)! = ((2n-1)!)^2 2n$
Thus
$$P = (1! 3! 5! \dots 99!)^2 (2 . 4 . 6 \dots 100) $$
$$ = (1! 3! 5! \dots 99!)^2 . 2^{50} . 50!$$
Thus removing $50!$ from $S$ will give us the desired result.
