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: 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 $100$. 
A: 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 $36$
A: 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.
