Wilson's theorem related problem 
prove        $$18! \equiv -1  \pmod{437} $$

I do not want full solution to the above problem but if anybody can tell me  how we can approach to it, I will really appreciate that.
 A: hints :


*

*$437=19\cdot 23$ (as proposed by Sean)

*Wilson's theorem :-)

*$19\cdot 20\cdot 21 \cdot 22=(-4)(-3)(-2)(-1)\pmod{23}$

A: As noted by Sean, $\,427=19\cdot 23\,$, thus using Wilson' theorem twice:
$$(1)\,\,\,18!\cdot 19\cdot 20\cdot 21\cdot 22=22!=-1\pmod {23}\Longrightarrow $$
$$\Longrightarrow 18!=\frac{-1}{(-4)(-3)(-2)(-1)}=-\frac{1}{24}=-1\pmod {23} $$
$$(2)\,\,\,\,18!=-1\pmod {19}$$
A: Wilson's Theorem: (p-1)! = -1 (mod p),   p: is a prime number
Note: When I write the "equal" (=) sign, I mean the congruence sign.....
437 is not a prime number: 437 = 19*23
For: 19
By Wilson's Theorem => 18! = -1 (mod 19)
For 23: 
By W.T => 22! = -1 (mod 23) => 22*21*20*19*18! = -1 (mod 23)
For 22 = -1 (mod 23)
For 21 = -2 (mod 23)
For 20 = -3 (mod 23)
For 19 = -4 (mod 23)
We multiply our reminders => (-1)(-2)(-3)*(-4) = 24
Back to our equation: 22*21*20*19*18! = -1 (mod 23) 
transformed to => (-1)(-2)(-3)*(-4) = 24 * 18! = -1 (mod 23) 
For 24 = 1 (mod 23) (replace 24 by 1)
1 * 18! = -1 (mod 23) 
So, 18! = -1 (mod 23)
Hope it helps you understand it more.
