A problem dealing with even perfect numbers. Question: Show that all even perfect numbers end in 6 or 8.
This is what I have. All even perfect numbers are of the form $n=2^{p-1}(2^p -1)$ where $p$ is prime and so is $(2^p -1)$.
What I did was set $2^{p-1}(2^p -1)\equiv x\pmod {10}$ and proceeded to show that $x=6$ or $8$ were the only solutions.
Now, $2^{p-1}(2^p -1)\equiv x\pmod {10} \implies 2^{p-2}(2^p -1)\equiv \frac{x}{2}\pmod {5}$, furthermore there are only two solutions such that $0 \le \frac{x}{2} < 5$. So I plugged in the first two primes and only primes that satisfy. That is if $p=2$ then $\frac{x}{2}=3$ when $p=3$ then $\frac{x}{2}=4$. These yield $x=6$ and $x=8$ respectively. Furthermore All solutions are $x=6+10r$ or $x=8+10s$.
I would appreciate any comments and or alternate approaches to arrive at a good proof.

$p$ is prime so it is 1 or 3$\mod 4$.
So, the ending digit of $2^p$ is (respectively) 2 or 8
(The ending digits of powers of 2 are $2,4,8,6,2,4,8,6,2,4,8,6...$
So, the ending digit of $2^{p-1}$ is (respectively) 6 or 4; and
the ending digit of $2^p-1$ is (respectively) 1 or 7.
Hence the ending digit of $2^{p-1}(2^p-1)$ is (respectively) $6\times1$ or $4\times7$ modulo 10, i.e., $6$ or $8$.
