What the rest of the division $1^6+2^6+...+100^6$ by $7$? 
What the rest of the division $1^6+2^6+...+100^6$ by $7$?
$1^6\equiv1\pmod7\\2^6\equiv64\equiv1\pmod7\\3^6\equiv729\equiv1\pmod7$
Apparently all the leftovers are $one$, I thought of using Fermat's Little Theorem, however the $(7,7 k) = 7$, so you can not generalize, I think. help please.


The number of multiples of $7$ from $1$ to $100$ is $\left\lfloor\frac{100}{7}\right\rfloor = 14$ so...
By Fermat's Little Theorem :
$7\mid 1^6+2^6+…+100^6 - 86 \implies 1^6+2^6+…+100^6 \equiv 86 \equiv 2 \pmod7   $ 
