$\underline{\text { Folklore }}$

Find the smallest natural $n$, for which the number $A=n^{3}+12 n^{2}+15 n+180$ is divisible by 23.
$n^{3}+12 n^{2}+15 n+180=n^{2}(n+12)+15(n+12)=(n+12)\left(n^{2}+15\right)$. The smallest value of $n$ for which the first factor is divisible by 23 is 11, and for the second factor, such $n$ is 10. This can be verified by direct enumeration or by noting that

$n^{2}+15 \equiv n^{2}-100=(n-10)(n+10)(\bmod 23)$.

## Answer

$n=10$