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

$6^{47} \equiv (-1)^{47} = -1^{47}\mod 7$
$5^{47} \equiv (-2)^{47} = -2^{47}\mod 7$
$4^{47} \equiv (-3)^{47} = -3^{47}\mod 7$
Hence $ 1^{47} +2^{47}+3^{47}+4^{47}+5^{47}+6^{47} \equiv 0 \mod 7$.
