The binomial coefficient $\left(\begin{array}{l}99 \\ 19\end{array}\right)$ is $ 107,196,674,080,761,936, x y z $ , Find $x y z$ 
The binomial coefficient $\left(\begin{array}{l}99 \\ 19\end{array}\right)$ is a 21 -digit number:
  $
107,196,674,080,761,936, x y z
$
  Find the three-digit number $x y z$

I showed that $\left(\begin{array}{l}99 \\ 19\end{array}\right) \equiv 2(\bmod 4)$
and $\left(\begin{array}{l}99 \\ 19\end{array}\right) \equiv 19(\bmod 25)$
Now how to combine them to find last two digit (y and z)??
because we can only combine when $a \equiv b(\bmod n)$
$a \equiv b(\bmod m)$ then if (n,m)=1 then
$a \equiv b(\bmod mn)$ but here we have different b's ...
and also can someone tell some easier method to find $\left(\begin{array}{l}99 \\ 19\end{array}\right) \equiv 2(\bmod 4)$
and $\left(\begin{array}{l}99 \\ 19\end{array}\right) \equiv 19(\bmod 25)$  my approach takes me too long, so i want to see some easier method...

Since $99 \equiv -1 \pmod {25}$, we have $99 \cdot 98 \cdots 81 \equiv (-1)^{19}19! \pmod {25}$. What we would like to do is to simply divide by $19!$ and be done, but you'll notice that $19! \equiv 0 \pmod{25}$ because of the multiples of $5$. So instead, we treat the multiples of $5$ separately and this gives
$$ \binom{99}{19} \equiv (-1)^{19 - 3} \frac{95 \cdot 90 \cdot 85}{15 \cdot 10 \cdot 5} \pmod{25}.$$
Now we simplify:
$$ (-1)^{19 - 3} \frac{95 \cdot 90 \cdot 85}{15 \cdot 10 \cdot 5} = 3 \cdot 17 \cdot 19 = 51 \cdot 19 \equiv 19 \pmod{25}.$$
