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.

We want the remainder when the product is divided by $100$. Th remainder on division by $4$ is $0$, so all we need is the remainder on division by $25$. 
Note that $\varphi(25)=20$. So by Euler's Theorem, $2^{20}\equiv 1\pmod{25}$. It is easier to note that $2^{10}\equiv -1\pmod{25}$. 
It follows that $2^{11212}\equiv -4\pmod{25}$. The same idea shows that $2^{11213}-1\equiv -9\pmod{25}$. Thus our product is congruent to $36$, or equivalently $11$, modulo $25$.
Now solve $x\equiv 0\pmod{4}$, $x\equiv 11\pmod{25}$. The solution is $36$ modulo $100$. 
