Remainder of $(1\cdot2\cdots102)^3$ modulo $105?$

Question

I am having trouble in finding the remainder of $(1\cdot2\cdots102)^3\mod 105$

It is not possible to apply Wilson's Theorem here because 105 is composite. Can anybody help me?

Answer 1

$105 = 3 \cdot 5 \cdot 7, 102!$ is divisible all of $3,5$ and $7,$ therefore $102!^3$ is divisible $105$

Answer 2

Hint: Yes, $105$ is composite. So while you can't use Wilson's theorem, it actually makes the problem a lot easier, rather than more difficult. Think about exactly why Wilson's theorem fails for composite numbers: What is $(n-1)!\pmod n$ in those cases?