Finding the remainder to an awful division 
If $x=(11+11) \times (12+12) \times (13+13) \times\cdots \times
(18+18) \times (19+19)$, what will be the remainder if $x$ is divided by $100$?

I tried simplifying the expression like this:
$$\frac{2(11)\times2(12)\times2(13)\cdots\times2(19)}{100}$$
$$\frac{2^8(19!\div10!)}{2^2\times5^2}$$
$$\frac{2^6(19!\div10!)}{5^2} ≡x \mod 10$$
Now, am I going in the right direction? I'm completely clueless.
I'm only a 7th grader, so I've very limited knowledge on mathematics. A little help solving this will be really nice. 

We need $$\prod_{r=11}^{19}2r\pmod{100}$$
As $20|(15+15)\cdot(12+12)$
let us find $22\cdot26\cdot28\cdot32\cdot34\cdot36\cdot38\pmod5$
$22\cdot26\cdot28\cdot32\cdot34\cdot36\cdot38\equiv2\cdot1\cdot3\cdot2\cdot4\cdot1\cdot3\equiv4\pmod5$
$$\implies\prod_{r=11}^{19}2r\equiv4\cdot(15+15)\cdot(12+12)\pmod{5\cdot20}$$
$4\cdot(15+15)\cdot(12+12)\equiv80\pmod{100}$
