Prove that for every composite number $ n>4$, numbers $ kn$ divides $ (n-1)!$ for every integer $ k$ such that $ 1\le k\le \lfloor \sqrt{n-1} \rfloor$.
Well, we know that n can be expressed as $p*q$ where $p>=\sqrt{n}>=q>=0$. If $k$ is not equal to $q$ then the statement is obviously true. If $k=q$, then we must have replace $q$ wit some multiple of $q$. Since we have that $q<\sqrt{n}$ there are atleast $q-1$ other multiples that work.
Case 1: $q$ is not equal to 2. Then the result is obvious since we are guaranteed that we won't select $q$ for the next choice
Case 2: $q=2$. Then since n>4 we always have a multiple of 2 less than $\sqrt{n}$.