the last decimal digit of every even perfect number is always 6 or 28.

Question

how to Prove that the last decimal digit of every even perfect number is always $6$ or $28$.

Answer 1

The even perfect numbers are of the form $x_n=2^{n-1}(2^n-1)$, where both $n$ and $2^n-1$ must be primes.

Your claim obviously holds, if $n=2$, so we can assume $n\ge 3$. Let us split our consideration according to the residue class of $n$ modulo $4$.

If $n\equiv 1\pmod{4}$, then $2^{n-1}\equiv1\pmod5$, $2^n\equiv2\pmod5$, and hence $x_n\equiv1\pmod5$. Obviously $2\mid x_n$, so the Chinese Remainder Theorem (applied to moduli $2$ and $5$) tells us that $x_n\equiv 6\pmod{10}$.

Let us then assume that $n\equiv3\pmod4$. In this case $2^n-1\equiv2\pmod5$, $2^{n-1}\equiv4\pmod5$, and $x_n\equiv3\pmod5$. This will not be enough, because we need to say something about the two last digits of $x_n$, and need to know its residue class modulo $25$. Consider the polynomial $$p(x)=x(2x-1)-3=2x^2-x-3=(x+1)(2x-3).$$ We easily see that when $x\equiv 4\pmod5$ both factors, $x+1$ and $2x-3$, are divisible by $5$. Consequently $$ 25\mid p(2^{n-1})=x_n-3 $$ whenever $n\equiv 3\pmod4$. We have shown that $x_n\equiv3\pmod{25}$. Obviously $4\mid x_n$, so another application of the Chinese Remainder Theorem (applied to moduli $4$ and $25$) tells us that $x_n\equiv28\pmod{100}$.

Note that the argument didn't use the full power of the assumptions that $n$ and $2^n-1$ must be prime. The argument works for all odd $n$.