22. For a positive integer $n, 110 n^{3}$ has 110 different positive divisors, including 1 and $110 n^{3}$ itself. Then $81 n^{4}$ has ( ) different positive divisors.
(A) 110
(B) 191
(C) 261
(D) 325
(E) 425
22. D.

By the Fundamental Theorem of Arithmetic, let
$$
110 n^{3}=p_{1}^{n_{1}} p_{2}^{n_{2}} \cdots p_{k}^{n_{k}},
$$

where $n_{i} \in \mathbf{Z}_{+}, i=1,2, \cdots, k ; p_{1}, p_{2}, \cdots, p_{k}$ are prime factors, and $p_{1}<p_{2}<\cdots<p_{k}$.
By the Divisor Counting Theorem, we have
$$
\begin{array}{l}
\left(n_{1}+1\right)\left(n_{2}+1\right) \cdots\left(n_{k}+1\right) \\
=110=2 \times 5 \times 11 \\
\Rightarrow k=3 .
\end{array}
$$

Thus, $110 n^{3}$ has only three prime factors $2, 5, 11$, and $n_{1}, n_{2}, n_{3}$ are $1, 4, 10$.
Therefore, $n=p_{i} p_{k}^{3}\left(p_{i}, p_{k} \in\{2,5,11\}, p_{i} \neq p_{k}\right)$.
Hence, $81 n^{4}=3^{4} p_{i}^{4} p_{k}^{12}$, and it has
$$
(4+1)(4+1)(12+1)=5 \times 5 \times 13=325
$$

distinct positive divisors.