I wish to find the remainder of $49!$ when divided by $53$. We have that $52! \equiv -1 \pmod {53}$ by Wilson's Theorem. So we have $52\cdot 51 \cdot 50 \cdot 49! \equiv -1 \pmod {53} \implies 6\cdot 49! \equiv 1 \pmod{53}$. I am not sure what to do here. I do have a feeling that some sort of manual check/trick is needed, but I am unable to see it.
$\endgroup$ 51 Answer
$\begingroup$So, $\displaystyle49!\equiv 6^{-1}\pmod{53}$
Now, as $\displaystyle6\cdot9=54\equiv1\pmod{53}, 6^{-1}\equiv9\pmod{53}$
$\endgroup$ 7
