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

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.

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 $6a\equiv 1\pmod{53}$?.
• $a=9$ since $6\cdot9=54=1(mod53)$? – Diya Oct 10 '14 at 21:59
• so finally $49! \equiv 9(mod43)$? – Diya Oct 10 '14 at 22:05
• @Diya Yes ($\mod{53}$ of course) – Hagen von Eitzen Oct 10 '14 at 22:31
Hint: Find the multiplicative inverse, modulo $53$, of $51\cdot 50 \equiv (-2)(-3)=6$.
• The multiplicative inverse is 9. So $52\times 9 \times 49= -9(mod 53)$. Then $(-1) \times 9 \times 49! = -9(mod53)$. So $9\times 49! = 9(mod 53)$. Then? – Diya Oct 10 '14 at 22:03