29-63 Let $n^{3}+p$ be divisible by $n+q$ $(n, p, q$ are all positive integers $)$. When $n$ is the largest, the corresponding values of $p$ and $q$ are
(A) $p=100, q=10$.
(B) $p=5000, q=20$.
(C) $p=50, q=12$.
(D) $p=300, q=15$.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1986)
[Solution] From $\frac{n^{3}+p}{n+q}=n^{2}-q n+q^{2}+\frac{p-q^{3}}{n+q}$, if $(n+q) \mid\left(n^{3}+p\right)$, then $(n+q) \mid\left(p-q^{3}\right)$. For $(p, q)$ in $(A),(B),(C),(D)$, $p-q^{3}<0$, and
\begin{tabular}{c|c|c|c|c}
\hline$(p, q)$ & $(100,10)$ & $(5000,20)$ & $(50,12)$ & $(300,15)$ \\
\hline$p-q^{3}$ & -900 & -3000 & -1678 & -3075 \\
\hline
\end{tabular}

Clearly, when $n=3060$, $\frac{-3075}{3060+15}=-1$, and $n$ is the largest. Therefore, the answer is $(D)$.