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. 
 A: $$
x=(20+2) \times (20+4) \times (20+6) \times\cdots \times
(20+16) \times (20+18)
$$
After the multiplication is expanded, we get many terms. But the remainder is driven by the last two terms, namely
$$
S = S_1 + S_2 = \left(2\cdot4\cdot\dots\cdot18\right) + \left(20\cdot2\cdot4\cdot\dots\cdot18\left(\frac12+\dots+\frac1{18}\right)\right),
$$
as the others are divided b $20^2$. In fact, the second term is 
$$
S_2 = \left(20\cdot2\cdot4\cdot\dots\cdot18\left(\frac12+\dots+\frac18+\frac1{12}\dots+\frac1{18}\right)\right) + \left(20\cdot2\cdot4\cdot\dots\cdot18\left(\frac1{10}\right)\right),
$$
where the first term is divided by $20\times10$ and does not influence the remainder. And the other is just $2S_1$. So we should find 
$$S_1 = 2^99!\mod100 = 2^99!\mod100 = 512\cdot9!\mod100 = 12\cdot 362880\mod100 = $$$$= 12 \cdot 80\mod100 = 60$$
Then 
$$S\mod100 = 3S_1 \mod 100 = 80$$
The result is verified by Python :)
A: We need $$\prod_{r=11}^{19}2r\pmod{100}$$
As $20|(15+15)\cdot(12+12)$
let us find $22\cdot26\cdot28\cdot32\cdot34\cdot36\cdot38\pmod5$
$22\cdot26\cdot28\cdot32\cdot34\cdot36\cdot38\equiv2\cdot1\cdot3\cdot2\cdot4\cdot1\cdot3\equiv4\pmod5$
$$\implies\prod_{r=11}^{19}2r\equiv4\cdot(15+15)\cdot(12+12)\pmod{5\cdot20}$$
$4\cdot(15+15)\cdot(12+12)\equiv80\pmod{100}$
