Factorial Divisibility (E)

The numbers are categorised into three groups. (1) Composite numbers which are not squares of prime numbers (2) Squares of prime numbers (3) Prime numbers

For any composite number in (1), $n=pq$ where $1<p<q<n$ , and $(n-1)!=1\times 2\times \cdots \times p \times \cdots \times q \times \cdots \times (n-1)$ Hence, $n | (n-1)!$

For squares of prime numbers we have $n=p^2$ and $(n-1)!=1\times 2\times \cdots \times p \times \cdots \times 2p \times \cdots \times (n-1)$ . Hence, $n | (n-1)!$ UNLESS $n-1 = p^2-1 < 2p$ . The only prime number satisfying this is $p=2$ . Thus, $\boxed{4 \not | 3!}$

If $n$ is prime then, then there would not be any common factor between $n$ and $(n-1)!$ . There are 168 prime numbers less than 1000.

So, there are $168+1=\boxed{169}$ solutons.