Find the remainder of $49!$ modulo $53$

Question

Since $53$ is prime, from Wilson's theorem, $52! \equiv -1\pmod{53}$,

i.e. $52 \times 51 \times 50 \times 49! \equiv -1\pmod {53}$

I don't understand how to take it from here.

The other form I have is $51! \equiv 1\pmod{53}$ but again I don't know how to use that.

Answer 1

Well, $52\cdot51\cdot 50\equiv (-1)\cdot(-2)\cdot(-3)\equiv -6$. Hence you know that $49!\cdot(-6)\equiv -1$ (or equivalently and from the second form $49!\cdot 6\equiv 1$. Know any $a$ with $6a\equiv 1\pmod{53}$?.

Answer 2

Hint: Find the multiplicative inverse, modulo $53$, of $51\cdot 50 \equiv (-2)(-3)=6$.