jmc

If $n+10\mid n^3+100$, $\gcd (n^3+100,n+10)=n+10$. Using the Euclidean algorithm, we have $\gcd (n^3+100,n+10)=\gcd (-10n^2+100,n+10)$ $=\gcd (100n+100,n+10)$ $=\gcd (-900,n+10)$, so $n+10$ must divide $900$. The greatest integer $n$ for which $n+10$ divides $900$ is $\boxed{890}$; we can double-check manually and we find that indeed $900\mid 890^3+100$.