Prove all even perfect numbers will end in 6 or 8.
Since it is even we can write, by Euler,
n=2^{k-1}*(2^k-1) ---> Note k must be a prime.
Trivial for k=2. Assume k>2.
Divide the proof into two cases.
Case I If k=4j+1 then just by working modulo 10 you find that n = 6 (mod 10).
Case II If k=4j+3 then just be working modulo 10
you find that n = 8 (mod 10).