Find the remainder when $45!$ is divided by $47$?

Question

Find the remainder when $45!$ is divided by $47$?

My approach

I am using Wilson's theorem to solve the problem.

I reduced the expression into ($47$-$1$-$1$)!/$47!$=$(47$-$1$)/$47$!-$1$/$47!$=-$1$-$1$=-$2$

Am I right in my approach.Please correct me if I am wrong?

Please correct me how to approach towards the problem.

Answer 1

$2*24=3*16=4*12=5*19=6*8=7*17=...=1\mod47$
so the product of all of them is $1\mod 47$

Answer 2

$$-1\equiv46!\equiv46\cdot45!\equiv(-1)\cdot45!\pmod{47}$$

$$\implies1\equiv45!$$

Answer 3

Note that for any prime $(p-1)^2=+1\pmod p$. Now start with Wilson's theorem $(p-1)!=-1\pmod p$ and multiply both sides by $(p-1)$, use the above fact, to arrive at $(p-2)!=+1\pmod p$

Answer 4

Wilson's theorem says $46!\pmod{47}\equiv -1\pmod{47}$. But then

$$ \begin{align*} 45!\pmod{47}&\equiv [46!\cdot(46)^{-1}]\pmod{47}\\ &\equiv [46!\pmod{47}]\cdot[(46)^{-1}\pmod{47}]\\ &\equiv [-1\pmod{47}]\cdot[46\pmod{47}]\\ &\equiv -46\pmod{47}\\ &\equiv 1\pmod{47} \end{align*} $$

Answer 5

By Wilson's Theorem: 46!≡-1(mod 47)≡46(mod 47).

1. 46!≡46 x 45!≡46(mod 47).
2. Thus, 45!≡1(mod 47)