$29 \cdot 9 n$ is a positive integer. If $2 n$ has 28 positive divisors, and $3 n$ has 30 positive divisors, then the number of divisors of $6 n$ is 
(A) 32.
(B) 34.
(C) 35.
(D) 36.
(E) 38.
(47th American High School Mathematics Examination, 1996)
[Solution] Let $n=2^{\alpha} \cdot 3^{\beta} \cdot m$, and $2 \times m$, as well as $3 \times m$, and assume $m$ has $k$ positive divisors, $\alpha, \beta \in N$. From the given conditions,
$$
\left\{\begin{array}{l}
(\alpha+2)(\beta+1) k=28 \\
(\alpha+1)(\beta+2) k=30
\end{array}\right.
$$

From (1), we know $5 \times k$, and $3 \times k$, from (2), we know $7 \times k$,
$\therefore k \mid 2, k$ can only be 1 or 2,
i) When $k=1$, the above two equations become
$$
\left\{\begin{array} { l } 
{ ( \alpha + 2 ) ( \beta + 1 ) = 2 8 , } \\
{ ( \alpha + 1 ) ( \beta + 2 ) = 3 0 . }
\end{array} \text { Solving gives } \left\{\begin{array}{l}
\alpha=5, \\
\beta=3 .
\end{array}\right.\right.
$$

In this case, the number of divisors of $6n$ is $(\alpha+2)(\beta+2)=35$.
ii) When $k=2$, the above two equations become
$$
\left\{\begin{array}{l}
(\alpha+2)(\beta+1)=14, \\
(\alpha+1)(\beta+2)=15 .
\end{array}\right. \text { It has no integer solutions. }
$$

In summary, $6n$ has 35 divisors. Therefore, the answer is $(C)$.