My first number theory problem!

We know $n \equiv -10 \mod n+10$ , as

$\dfrac{n}{n+10}=\dfrac{n+10-10}{n+10}=1+\dfrac{-10}{n+10}$

Therefore, $n^3+100 \equiv -900 \mod n+10$ by substitution. For $-900 \equiv 0 \mod n+10$ , $n+10$ must divide $-900$ . The largest number which divides $-900$ is 900, which means

$n+10=900$

$\boxed{890}$

Nice problem!

By division we find that $n^3+ 100 = (n + 10)(n^2 - 10n +100) - 900.$ .Thus, if $n + 10$ divides $n^3 + 100$ , then it must also divide $900$ . Moreover, since $n$ is maximized whenever $n + 10$ is, and since the largest divisor of $900$ is $900$ , we must have $n + 10 = 900$ . Therefore, $\boxed{ n = 890}$