All values of a variable in given conditions

Question

Find all $n$ for which $10\mid(n-1)!+1$.

I think we might use Wilson's theorem by: $(n-1)!\equiv-1\pmod{10}$ and using the fact that $5!$ or above are always divisible by $10$ and nor $3!$ neither $2!$ gives remainder $9$ when divided by $10$, so there should be no solution.

Can someone provide any other way to solve it and is my solution even correct? I would be grateful for any hints or solutions.

Answer 1

Wilson's theorem is $p\mid (p-1)!+1$ for prime $p$. That looks typographically like what you want, but it is mathematically too different to be of great help.

On the other hand, once you know that all factorials from $5!$ and up are divisible by $10$, and therefore can't be solutions, you can just check the remaining five cases one-by-one by hand.

Answer 2

As @Arthur mentioned, you should realize that $(n-1)!$ is always even for $n\geq 3$, hence $(n-1)!+1$ is odd, so $10$ couldn't divide it. For $n=1$ and $n=2$, its more than obvious that it is not the case.