I know that for n to be an even perfect number greater than 6 it has the form $2^{m-1}(2^m-1)$ where $m$ is prime. I also know that since n is an even perfect number, it is congruent to 1 (mod 9).
What I have worked out is: $n=9s+1$. Since $n$ is even, $s$ must be odd giving $n=9(2r+1)=18r+10$ So $n$ is congruent to 10 (mod 18).
I was hoping I might be able to use the Chinese remainder theorem on the congruences, but I see now that won't work. I have been working on this for a couple days and I am stuck! I don't know how to change (mod 9) and (mod 18) into (mod 12) or if I'm going about this in completely the wrong manner.