From the list of even perfect numbers http://en.wikipedia.org/wiki/List_of_perfect_numbers it can be observed that all of them have either 6 or 8 as a last digit. Is this true for all even perfect numbers? In other words, does one of the congruences $$n\equiv 1 \ (\text{mod 5}), \quad n\equiv 3 \ (\text{mod 5})$$ hold for any even perfect number n? I suppose there are results of this kind but couldn't find any.

closed as off topic by Steven Landsburg, Gjergji Zaimi, Andres Caicedo, Felipe Voloch, Noah Snyder Jun 16 '12 at 15:36
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Every even perfect number is of the form $2^{p1}(2^p  1)$ where $2^p  1$ is a (Mersenne) prime. Note that $p$ must be prime – if $p = ab$ with $a, b > 1$ then $$2^p  1 = 2^{ab}  1 = (2^a)^b  1 = (2^a  1)(1 + 2^a + 2^{2a} + \dots\ + 2^{a(b1)}).$$ If $p = 2$, we obtain the first perfect number $6$ which satisfies $ 6 \equiv 1\ (\text{mod 5})$. Every other prime is odd, so let $p = 2k + 1$. Then $$2^{p1}(2^p  1) = 2^{2k}(2^{2k+1}  1) = 2.2^{4k}  2^{2k} = 2.16^k  4^k \equiv 2  (1)^k\ (\text{mod 5}).$$ So, for $p = 2k + 1$, $$2^{p1}(2^p  1) \equiv \begin{cases} 1 \ (\text{mod 5}) & \text{if }k\text{ is even}\newline 3 \ (\text{mod 5}) & \text{if }k\text{ is odd}. \end{cases}$$ 

