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?

$$1^6+2^6+3^6+4^6+5^6+\dots+100^6\equiv20(1^6+2^6+3^6+4^6+5^6)\\\equiv0\text{ (mod }5)$$
The first conguruence can be applied since $1\equiv6\equiv11\equiv\dots\equiv96\text{ (mod }5)$, $1^6\equiv6^6\equiv11^6\equiv\dots\equiv96^6\text{ (mod }5)$
Likewise, $2^6\equiv7^6\equiv12^6\equiv\dots\equiv97^6\text{ (mod }5)$, and the same can be done with the numbers which are congruent to 3, 4 and 0 modulo 5.
Hence, we can replace all these terms with $1^6+2^6+3^6+4^6+5^6$ repeated 20 times and therefore the sum is a multiple of 20 and is divisible by 5 exactly. (Comment made by David K.)
