<s>[INST] What is the remainder of $16!$ is divided by 19? Can anyone share me the trick to solve this problem using congruence?  
Thanks,
Chan
 [/INST] A more explicit answer would go like this:
By Wilson's Theorem, $18!$ is congruent to $-1 \pmod {19}$
$$18! = (18\cdot 17)(16!)$$
Then $(18\cdot 17)(16!)$ is congruent to $-1 \pmod{19}$
But note that $18$ is congruent to $-1 \pmod{19}$ and $17$ is congruent to $-2 \pmod{19}$
Then it follows that $(18\cdot 17)(16!)$ is congruent to $((-1)\cdot (-2))(16!)$ is congruent to $-1 \pmod{19}$
Simplifying, we get, $2\cdot(16!)$ is congruent to $-1 \pmod{19}$. But we need the remainder of $16!$, not $2\cdot 16!$. and $-\frac12$ isn't a valid answer.
Then also notice that $18$ is congruent to $-1 \pmod{19}$. Then $2\cdot 16!$ is congruent to $18 \pmod{19}$ by the fact that if $a$ is congruent to $b \pmod{p}$, then $b$ is congruent to $a \pmod{p}$
So we have that $2\cdot 16!$ is congruent to $18 \pmod{19}$ and from there, its just a matter of dividing both sides by $2$ to get $16!$ is congruent to $9 \pmod{19}$
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