2021/038) if $n^2+7n+4$ is divisible by 11 what is the remainder when divided by 121

 Working in mod 11 we have

$n^2+7n + 4 \equiv n^2 + (7-11)n + 4$

$\equiv n^2 - 4n + 4 \equiv (n-2)^2$

so if $n^2 + 7n + 4$ is divisible by 11 then n-2 is divisible by 11

so $n = 11k + 2$

So  $n^2+7n+4 = (11k+2)^2 + 7(11k+2) + 4$

= $121k^2 + 121k + 22$

so remainder when divided  by 121 is 22