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.