\section*{Problem 8}

Find all natural numbers \(n\) such that \(\mathrm{n}^{2}\) does not divide \(\mathrm{n}!\).


\section*{Solution}

Answer: \(\mathrm{n}=4\) or prime.

If \(\mathrm{n}=\mathrm{rs}\), with \(12\), then \(r<2 r<n\), and hence \(n^{2}\) divides \(n!\). This covers all possibilities except \(n=4\) or \(n=\) prime, and it is easy to see that in these cases \(n^{2}\) does not divide \(n!\).

