Given $n!+(n+1)!+(n+2)!+(n+3)!$ is divisible by $181$ but $n!$ is not divisible by $181$, find three possible values for $n$.

Question

Given $n!+(n+1)!+(n+2)!+(n+3)!$ is divisible by $181$ but $n!$ is not divisible by $181$, find three possible values for $n$.

Answer 1

Hint: $181$ is a prime and $n!+(n+1)!+(n+2)!+(n+3)! = n!\,(n+2)\,(n^2+5n+5)$.

Answer 2

Your first question is the wrong one to ask if you are interested in the second. You care about factorials $\bmod 181$, the remainder on division, not on how many factors there are. For the second, you know $n \lt 181$, so just compute the factorials $\bmod 181$ in a spreadsheet and add sets of four successive terms. Copy down will make it easy.