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.
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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 $p = 2k + 1$, $$2^{p1}(2^p  1) \equiv \begin{cases} 1 \ (\text{mod 5}) & \text{if }k\text{ is even}\newline 3 \ (\text{mod 5}) & \text{if }k\text{ is odd}. \end{cases}$$ 

