2. (47th American High School Mathematics Examination) $n$ is a positive integer. If $2n$ has 28 positive divisors, and $3n$ has 30 positive divisors, then the number of positive divisors of $6n$ is ( ).
A. 32
B. 34
C. 35
D. 36
E. 38
2. C. Let $n=2^{\alpha} \cdot 3^{\beta} \cdot m$, where $2 \times m, 3 \times m$ and $m$ has $k$ positive divisors. According to the problem, we have $(\alpha+2)(\beta+1) k=28 \quad$ (1), $(\alpha+1)(\beta+2)=30$ (2). From (1), we know $5 \nmid k$, and from (2), we know $7 \times k$, so $k \mid 2$, hence $k$ can only be 1 or 2. When $k=1$, we can solve $\alpha=5, \beta=3$, at this time the number of positive divisors of $6n$ is $(\alpha+2)(\beta+2)=35$. When $k=2$, $\alpha, \beta$ have no integer solutions. Therefore, the answer is C.