What the rest of the division $1^6+2^6+...+100^6$ by $7$?

Question

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.

Answer 1

Split the sum into $1^6$ to $7^6$, then $8^6$ to $14^6$, and so on up to $92^6$ to $98^6$, plus the little tail $99^6+100^6$.

Each of the full chunks of length $7$ has shape $(7k+1)^6+(7k+2)^6+\cdots +(7k+6)^6+(7k+7)^6$. So the sum of the remainders of such a chunk on division by $7$ is the same as the remainder of $1^6+2^6+\cdots+6^6+7^6$, which is $6$. (The $7^6$ term is congruent to $0$ modulo $7$, and each of the $6$ other terms is congruent to $1$ modulo $7$, by Fermat's Theorem, or by direct calculation.)

There are $14$ full chunks, whose remainder is the same as the remainder of $(6)(14)$, which is $0$.

That leaves $99^6$ and $100^6$, which each have remainder $1$. Thus the remainder of the full sum is $2$.

Answer 2

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 $