Largest n

By long division, $n^3+100=(n^2-10n+100)(n+10)-900$ so $n+10$ must divide 900...highest $n$ is $n=\boxed{890}$

$(n+10)^3=n^3+1000+30n(n+10)$

If $n^3+100$ is divisible by $n+10$

Then $n+10$ must divide $n^3+1000+30n(n+10)-(n^3+100)=30n(n+10)+900$

So $n+10$ must divide $900$

Hence the largest value of $n$ is $900-10=\boxed{890}$

By polynomial division,

$\frac { { n }^{ 3 }+100 }{ n+10 } ={ n }^{ 2 }-10n+100+\frac { -900 }{ n+10 }$

where $\frac { -900 }{ n+10 }$ is the remainder. Since we want to find the largest $n$ , value of denominator must be maximized, while keeping the remainder 0.

Largest number which completely divides -900 is $\left| -900 \right| =900$ .

Therefore,

$n+10=900\\ n=890$