What is the remainder when $1^6 + 2^6 + 3^6 + ... + 99^6 + 100^6$ is divided by 5? What is the remainder when $1^6 + 2^6 + 3^6 + ... + 99^6 + 100^6$ is divided by 5?
I think that the only way to solve this would be to applying to the proposition that “the sum/product of congruence classes is equal to the congruence class of the sum/product", however I am unsure how to apply this properly, any suggestions?

Intuitive Solution
The remainder is $\color{blue}{zero}$.
The numbers, $5^6, 10^6, 15^6, \ldots$ are all divisible by 5.
The rest you can pair up as $(5m \pm 1)^6$ which leaves a remainder $1$ and $(5m \pm 2)^6$ which leave a remainder $-1$, so everything cancels out and we get zero.
