Ireland

Note that $(n-1)(n^5+n^4+n^3+n^2+n+1)=n^6-1=(n^3-1)(n^3+1)=(n-1)(n+1)(n^2-n+1)(n^2+n+1)$, and so $$n^5+n^4+n^3+n^2+n+1=(n+1)(n^2-n+1)(n^2+n+1).$$ Because 199 is a prime number, this expression is divisible by 199 iff one of the factors $n+1$, $n^2-n+1$ or $n^2+n+1$ is divisible by 199. Divisibility of $n+1$ by 199 is equivalent to $n\equiv 198(\mathrm{mod}199)$. To solve the quadratic congruence $n^2-n+1\equiv 0(\mathrm{mod}199)$, we observe $n^2-n+1\equiv n^2-200n+1\equiv (n-100{)}^2-100^2+1(\mathrm{mod}199)$. Because $100^2-1=200\cdot 50-1=199\cdot 50+49\equiv 49(\mathrm{mod}199)$, the congruence $n^2-n+1\equiv 0(\mathrm{mod}199)$ is equivalent to $(n-100{)}^2\equiv 49(\mathrm{mod}199)$. Again, because 199 is a prime, this has exactly two solutions (mod 199) which are determined by $n-100\equiv \pm 7(\mathrm{mod}199)$. These two solutions are $n\equiv 107(\mathrm{mod}199)$ and $n\equiv 93(\mathrm{mod}199)$. Note now that that $(n-1{)}^2+(n-1)+1=n^2-n+1$, hence $n\equiv 106(\mathrm{mod}199)$ and $n\equiv 92(\mathrm{mod}199)$ are the solutions of $n^2+n+1\equiv 0(\mathrm{mod}199)$. This shows that $n^5+n^4+n^3+n^2+n+1$ is divisible by 199 iff $n$ is congruent to 92, 93, 106, 107 or 198 (mod 199).