What is the units digit of 1^1 + 2^2 + 3^4 + 4^8?
Powers of 4
4^1 = 4
4^2 = 16
4^3 = 64
4^4 = 256
In general 4^(2n) ends in 6
So...the units digit of 1^1 + 2^2 + 3^4 + 4^8 =
1 + 4 + 1 + 6 = 2
