How to find the remainder of big number divisions using congruences? I´m asked to find the remainder of dividing $\sum_{i=0}^{1080}i^5$ by $14$. How can I do this using only basic results from modular arithmetic? Only one thing comes to my mind, here's my idea:
we know that each number has a representative $r$ in the class of $\pmod{14}$, that satisfies $0\leq r<14$, so in order to simplify things, it is only needed to find this representative for $n^5$ for the integers $n$ between $0$ and $13$, because, for example, if I wanted to find the residue of $\sum_{i=0}^{27}i^5$ divided by $14$, then assuming I know that $k_n$ is the representative of $n$ for each $n\in\{ 0,\dots,13 \}$, then $14\equiv0\pmod{14}$ implies $14^5\equiv0^5\equiv k_0\pmod{14}$, $15\equiv1\pmod{14}$ implies $15^5\equiv1^5\equiv k_1\pmod{14}$, and so on until $27^5\equiv13^5\equiv k_{13}\pmod{14}$. In this way, it is now possible to know that, given that
\begin{equation*}\sum_{i=0}^{27}i^5=\sum_{i=0}^{13}i^5+(i+14)^5
\end{equation*}, then for $i\in\{ 0,\dots,13 \}$, $i^5\equiv (i+14)^5\pmod{14}\Rightarrow i^5+(i+14)^5\equiv2i^5\equiv2k_i\pmod{14}$, which implies that:
\begin{equation*}\sum_{i=0}^{13}i^5+(i+14)^5\equiv\sum_{i=0}^{13}2k_i\equiv R_k\pmod{14}
\end{equation*}
where $R_k$ is the representative of $\sum_{i=0}^{13}2k_i$ in $\mathbb{Z}_{14}$. That's the residue we're looking for.
Following the pattern, I'd need to find out how many times a number which is congruent to a number between $0$ and $13$ in $\mathbb{Z}_{14}$ appears between $0$ and $1080$, and then rewrite the original sum in terms of how many times the "repeated" (by this I mean in the sense that they are equivalent in $\mathbb{Z}_{14}$) numbers appear. 
(I already did this, but as I said, it's a very long and tedious process). 
Another idea is to use the formula for the sum of the first $n$ fifth powers.
Any other idea of an easier process, or a check to mine would be really appreciated. Thanks in advance.

We have
$\tag 1 \displaystyle \sum_{i=0}^{13}i^5 \equiv 0 + + 7^5  +\sum_{i=1}^{6}\bigr(i^5+(-i)^5\bigr)  \equiv 7^5 \equiv 7 \pmod{14} $
Also, since
$\quad 1081=77\times14+3$ we can write
$ \tag 2 \displaystyle \sum_{i=0}^{1080}i^5 \equiv 77\cdot7 + 0^5 + 1^5 + 2^5 \equiv 7 + 0 + 1 + 4 \equiv 12\pmod{14} $
