Question

1^2014 + 2^2014 + 3^2014 + 4^2014 + 5^2014 divided by 5, remains ______

Answer 1

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.