Last two digits of $2^{11212}(2^{11213}-1)$ 
What are the last two digits of the perfect numbers $2^{11212}(2^{11213}-1)$?

I know that if $2^n-1$ is a prime, then $2^{n-1}(2^n-1)$ is a perfect number and that every even perfect number can be written in the form $2^n(2^n-1)$ where $2^n-1$ is prime. I'm not sure how to use this information though.

A little playing with a calculator or a spreadsheet shows that the the last two digits of $2^n$ follow a pattern that repeats with a cycle length of $20$.  So, for example, $2^5$ ends in $32$;  $2^{25}$ ends in $32$, and $2^{45}$ ends in $32$ and so on.
We need to find the last two digits of $$2^{11212}(2^{11213}-1)=2^{22425}-2^{11212}$$ $22425 \mod 20$ is $5$;$2^5=32$. 
$11212\mod 20$ is $12$;$2^{12}=4096$.
So, we need the last two digits of the number you get by subtracting a big number ending in $96$ from a much bigger number ending in $32$.  The answer is $36$
