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.

HINT:
Note that $2^n-1$ is prime implies $n$ is prime.  Consider what that means for the form of $n$ as it applies to $2^n-1$.
Find the value of $2^n-1$ for several small values of $n$, enough to determine whether there is a pattern and if so, what it might be.  Apply the same logic to $2^{n-1}$, and then to the multiplication of the two.
Note that for $n\gt 2$ this implies that $n$ is odd.  So consider $2^{2k+1}-1$ and $2^{2k}$ for several small values of $k$ to see if a pattern develops.
