what is the remainder when the sum 1^5+2^5+3^5+...99^5+100^5 is divided by 4? again a mod problem that I don't know how to work.

Answer 1

Consider the first four terms:

1^5 + 2^5 + 3^5 + 4^5

The next four terms are Mod 4 the same because you can reduce Mod 4 before taking the power. The sum till 100^5 thus contains 25 times the same four terms. Modulo 4, 25 is the same as 1, so the summation is just the sum of the first four terms.

We can caompute the sum of the fist four tems as follows. We have (Modulo 4):

4 = 0, so the last term is zero.

3 = -1 and (-1)^5 = -1, so this cancels against the first term. The second term is, of course zero as it is divisible by 4.

The summation is thus zero.

Answer 2

thank you, count iblis. you have been a tremendous help with homework.

Answer 3

given the answer