Following is from Elementary Number Theory - David Burton, P.203, theorem 10.2

sigma(n) = 2n =>
n = 2^(p-1).(2^p-1)	[p prime]

suppose p=2, result true by inspection

suppose p=4m+1
n = 2^4m.(2^(4m+1)-1)
= 2^(8m+1) - 2^4m
= 2.16^2m -16^m
But 6^t == 6 [mod 10], so
n == 6 [mod 10]

suppose p=4m+3
n = 2^(4m+2).(2^(4m+3)-1)
= 2^(8m+5) - 2^(4m+2)
= 2.16^(2m+1) - 4.16^m
as before, therefore,
n == 8 [mod 10]