Question: What is the remainder when $40!$ is divided by $1763$? 

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**My try :**

Factorizarion of $1763 = 41 \* 43$

By using Wilson's theorem, I can say

$40! = -1 (mod 41) $

and $42.41.40! = -1(mod 43)$

=> $40! = -22(mod 43)$

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How can I combine both the results ?

Answer: 
Letting $x=40!$, you have
$$\begin{align}
x&\equiv40\bmod41\\
x&\equiv21\bmod43
\end{align}$$
You can use the Chinese Remainder Theorem to solve this system of modulo equations.
We have $21\cdot41-20\cdot43=1$.

$40!=41q\_1-1$ $\implies$ $(-20)\cdot43\cdot40!=-20\cdot1763q\_1+20\cdot43=1763Q\_1+860$

$40!=43q\_2-22$ $\implies$ $21\cdot41\cdot40!=21\cdot1763q\_2-21\cdot41\cdot22 = 21\cdot1763q\_2-18942=1763Q\_2+451$

Adding:

$(21\cdot41-20\cdot43)\cdot40!=1763(Q\_1+Q\_2)+860+451$

i.e. $40!\ =\ 1763Q+1311$

So the remainder is $1311$.