Calculate the remainder of $18!$ by $437$

Question

I'm trying to solve by Wilson's theorem.

My attempt

$437 = 19 \cdot 23$

$18!\equiv -1$ $\pmod {19}$

$22!\equiv -$1 $\pmod {23}$

I'm stuck here, I don't know how to combine these 2 informations to solve the problem.

edit

$22 \cdot 21 \cdot 20 \cdot 19 \cdot 18!$ $\equiv$ $-1$ $\pmod {23}$

$(-1)\cdot(-2)\cdot(-3)\cdot(-4)\cdot(18!) \equiv -1$ $\pmod {23}$

$18! \equiv -1$ $\pmod {23}$

Answer 1

Hint: $-1 \equiv 22!\equiv (22\cdot 21\cdot 20 \cdot 19) \cdot 18! \equiv (-1)\cdot (-2)\cdot (-3)\cdot (-4) \cdot 18! \bmod 23$

Answer 2

You have $-1 \equiv 22!$ (mod $23$). Note that $22\times 22 \equiv 21 \times 11 \equiv 20\times 15 \equiv 19\times 17 \equiv 1$ (mod $23$).

One then has $-1\times 22 \times 11 \times 15 \times 17 \equiv (22\times 22)(11\times 19)(20\times 15)(19\times 17) 18!$ (mod $23$), or $-1\times (-1)\times 11 \times (-8)\times (-6) \equiv 18!$ (mod $23$).

So $18! \equiv 22$ (mod $23$). Thus $18! \equiv -1$ (mod $437$).

Answer 3

$437=19×23$

From Wilson's theorem, $18!≡-1\mod 19$ (1)

$$22!≡-1\mod 23\implies 22×21×20×19×18!≡-1\mod 23$$

$$\implies(-1)(-2)(-3)(-4)(18!)≡-1\mod 23$$

$$\implies(24)(18!)≡-1\mod 23$$

$$=>(1)(18!)≡-1\mod 23\tag2$$

Then from equations (1) and (2) using $a≡b\mod m_1$ and $a≡b\mod m_2$ we have $a≡b\mod m_1m_2$

$$\implies \boxed{18!≡-1\mod 437}$$