57^100 + 56^100 + 55^100 = N

All positive integer powers of positive integers ending in 6 end in 6.
Therefore 56^100 ends in 6.

All positive integer powers of positive integers ending in 5 end in 5.
Therefore 55^100 ends in 5.

So the only difficulty here is with positive integer powers of positive
integers ending in 7.  So we write down a few:

7^1 = 7
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 must end in 7, and the units digits must start over.

So the units digits of powers of positive integers ending 
in 7 go 7,9,3,1,7,9,3,1,...

which means every 4th one will be a 1.

Since the 100th one is a 4th one, 57^100 will end in a 1.

So now we know that: 

57^100 ends in a 1, 
56^100 ends in a 6, and
55^100 ends in a 5,

Therefore we know that when we add them we will begin by adding their
last digits 1+6+5 = 12, and we will put down the 2 and carry the 1.
So the last digit of 57^100 + 56^100 + 55^100 = N must be 2.

Edwin