<s>[INST] 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?
 [/INST] We have  $p=5$ is prime, then by Fermat's little theorem: for every integer $a$ such that $p \nmid a$ we have  $$ a^{p-1} \equiv 1 \mod{p}$$
Now for any  $x$ integer in  $[1,100]$,  if $5$ divides  $x$ then the remainder is zero, if  $5\nmid  x$ , then  $x^{4} \equiv 1 \mod{5}$, and so   as $$y=1^6+ 2^6+3^6+...+99^6+100^6 =  1^4\cdot1^2+2^4\cdot2^2+3^4\cdot3^2+...+99^4\cdot99^2+100^4\cdot100^2 $$
then  $$ y \equiv  1^2+ 2^2+3^2+4^2+6^2+...+99^2 $$
Now for any integer $x$, such that $5\nmid x$, we have  $x^2 \equiv \pm 1 \mod{5}$. More precisely, if $ x \equiv 2,3 \mod{5}$ then $x^2 \equiv  -1 \mod{5}$, and  if   $ x \equiv 1,4 \mod{5}$ then $x^2 \equiv  1 \mod{5}$. Thus summing the squares of any four consecutive numbers will be congruent to zero  $\mod{5}$.
Hence summing the squares from   $1$ to $99$, the remainder would be zero.
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