Show that $7 \mid( 1^{47} +2^{47}+3^{47}+4^{47}+5^{47}+6^{47})$ I am solving this one using the fermat's little theorem but I got stuck up with some manipulations and there is no way I could tell that the residue of the sum of each term is still divisible by $7$. what could be a better approach or am I on the right track? Thanks
 A: One more solution.
By Fermat's Little Theorem,
$a^{p - 1} \equiv 1 \pmod{p}$ for $a \not\equiv 0$.  Thus,
$$
a^{48} = \left(a^6\right)^8 \equiv 1^8 = 1 \pmod{7}
$$
for each $a \in \{1, 2, \ldots, 6\}$.
As a consequence, $a^{47} \equiv a^{-1}$, so
$$
\begin{align}
1^{47} + 2^{47} + 3^{47} + 4^{47} + 5^{47} + 6^{47} &\equiv 1^{-1} + 2^{-1} + 3^{-1} + 4^{-1} + 5^{-1} + 6^{-1} \\
&\equiv 1 + 4 + 5 + 2 + 3 + 6 \\
&\equiv 0 \pmod{7}.
\end{align}
$$
A: HINT Make use of the fact that
$$(a+b) \vert (a^{2n+1} + b^{2n+1})$$
A: The map $x \mapsto x^{47}$ is injective on the nonzero classes mod $7$ because they form a group of order $6$. Hence this map is a permutation and so
$1^{47} +2^{47}+3^{47}+4^{47}+5^{47}+6^{47} \equiv 1+2+3+4+5+6 \equiv 0 \mod 7$.
A: Using this, $$\sum_{1\le r\le 6}r^{2k+1}$$ is divisible by  $$\frac{6(6+1)}2=21$$
A: $6^{47} \equiv (-1)^{47} = -1^{47}\mod 7$
$5^{47} \equiv (-2)^{47} = -2^{47}\mod 7$
$4^{47} \equiv (-3)^{47} = -3^{47}\mod 7$
Hence $ 1^{47} +2^{47}+3^{47}+4^{47}+5^{47}+6^{47} \equiv 0 \mod 7$.
A: HINT Find the mod cycle of 7 with the powers of each of the numbers. Then simply add them together. 
