If $m \equiv 5\mod 10 $ prove that $1991 \mid 12^m + 9^m + 8^m + 6^m$ I tried to find the remainder of each one of $12$,$9$,$8$ and $6 \mod 5$ and then combine them but I didn’t get the answer 

If $m/5$ is odd, $11|2^5+3^5|2^m+3^m$ while $181|3^5+4^5|3^m+4^m$, so$$\operatorname{lcm}(11,\,181)=1991|(2^m+3^m)(3^m+4^m)=6^m+8^m+9^m+12^m.$$
