A number theory problem by Swadesh Rath

I used practically used the same method as Maggie, but I included the substitution step. i.e. $\ u = n+10$ . I made this substitution because it is easier to deal with the fraction if the denominator is u rather than n+10.

For $n\neq10$ , $\frac{n^3+100}{n+10}=n^2-10n+100-\frac{900}{n+10}$ .

Therefore, $\frac{n^3+100}{n+10}$ is an integer if and only if $\frac{900}{n+10}$ is an integer, and the largest such $n$ is $\boxed{890}$ .