<s>[INST] 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.
 [/INST] In this proof, n is any integer(not just primes)  
The number 2^(n-1) last digit repeats in the four number cycle(2,4,8,6)
The number 2^(n) -1 last digit repeats in the four number cycle (3,7,5,1)
The product's last digit repeats in the four number cycle (6,8,0,6)  
If n is odd, the pattern repeats (8,6)  
You will find the number end in 6 or 8 for all odd numbers(not just primes)
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