Find $m$ such that $m \equiv 18( mod 19)$ and $m \equiv 22(mod 23)$

Question

I actually need to find a number $m$ which satisifes $m \equiv 18\,( mod\ 19)$ and $m\equiv 22\,(mod\ 23)$

I know Wilson theorem, which says $18! \equiv 18(mod\ 19)$ and $ 22! \equiv 22(mod\ 23)$. So I'm getting two distinct numbers, need to choose a single one which works for both.

Answer 1

We have $19|m+1$ and $23|m+1$

Let $n=m+1$, then $lcm(19,23)=437|n$

Thus n can be $437k$ for all $k \in Z$

Hence $m=437k-1$