1. (1986 4th American Invitational Mathematics Examination) Find the largest positive integer $n$ such that $n^{3}+100$ is divisible by $n+10$.
1. We try to transform $n^{3}+100$ to make $n+10$ appear. For this purpose,
$$
n^{3}+100=n^{3}+1000-900=(n+10)\left(n^{2}-10 n+100\right)-900 \text {. }
$$

If $n+10$ divides $n_{3}+100$, then $n+10$ can divide 900. To make $n$ as large as possible, we can let
$$
n+10=900,
$$

so $n=890$.
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