1^2014 + 2^2014 + 3^2014 + 4^2014 + 5^2014 divided by 5, remains ______
powers of the numbers end in
1: 1,...
2: 2,4,8,6,...
3: 3,9,7,1,...
4: 4,6,...
5: 5,...
Note that all can be grouped in periods of 4.
2014 = 4*53 + 2
So, the 2014th powers end in
1,4,9,6,5
Their sum ends in 5
So, the remainder is zero.