$1 \cdot 25$ Find the largest positive integer $n$, such that $n^{3}+100$ is divisible by $n+10$.
(4th American Invitational Mathematics Examination, 1986)
［Solution］We attempt to transform $n^{3}+100$ to make $n+10$ appear, for this purpose
$$
\begin{aligned}
& n^{3}+100 \\
= & n^{3}+1000-900 \\
= & (n+10)\left(n^{2}-10 n+100\right)-900 .
\end{aligned}
$$

Thus, if
$$
n+10 \mid n^{3}+100 \text {. }
$$

then
$$
n+10 \text { divides } 900 \text {. }
$$

Since
$$
n+10 \leqslant 900 \text {. }
$$

Therefore, to make $n$ as large as possible, we take $n+10=900$, then $n=890$.