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.
 A: The remainders modulo $4$ of the fifth powers are
$$1,0,3,0\cdots$$
This sequence goes repeating, as
$$(m+4k)^5\bmod4=(m^5+5\cdot4m^4k+10\cdot4^2m^3k^2+10\cdot4^3m^2k^3+5\cdot4^4mk^4+4^5k^5)\bmod 4=m^5\bmod4.$$
Then the accumulated sequence yields the remainders
$$1,1,0,0\cdots$$
A: Usual arithmetic operations such as addition, multiplication and raising to a natural power commute with taking remainders modulo 4. You can simplify the calculation by replacing each of the numbers 1...100 with its remainder modulo 4 before taking 5th powers.
Then construct a little table of 5th powers (modulo 4) for the numbers 0, 1, 2 and 3.
Then find simplifications so the result becomes obvious without actually doing many additions.
A: 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 $1^5+3^5+\cdots+99^5\equiv 0\pmod{4}$.
