Finding the remainder with Modular Arthmetic I have this math question that I'm kind of stuck on.

What is the remainder when $1^5+2^5+3^5+\cdots +99^5+100^5$ is divided
   by $4$?

I'm supposed to use modular arithmetic and equivalences. I know that when dividing by $4$ the only possible remainders are $0, 1, 2, 3$ and they cycle, However I'm not sure where to go from here. Thanks.

The even numbers are not of interest here. If $a$ and $b$ are odd integers from $1$ to $99$, call $a$ and $b$ friends if $a+b=100$. 
If $a$ and $b$ are friends, then $a+b\equiv 0\pmod{4}$, so $b\equiv -a\pmod{4}$. It follows that $b^5\equiv -a^5\pmod{4}$, and therefore $a^5+b^5\equiv 0\pmod{4}$.
The sum of the fifth powers of any two friends is therefore congruent to $0$ modulo $4$. It follows that $1^5+3^5+\cdots+99^5\equiv 0\pmod{4}$.
