19. Let set $A$ be the set of positive integers whose prime factors are only $2$, $3$, and $5$. If the sum of the reciprocals of all elements in set $A$ can be expressed as $\frac{m}{n}$, that is,
$$
1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\cdots=\frac{m}{n},
$$
where $m, n \in \mathbf{Z}_{+}, (m, n)=1$. Then the value of $m+n$ is ( ).
(A) 16
(B) 17
(C) 19
(D) 23
(E) 36
19. C.

From the problem, we know
$$
\begin{array}{l}
\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\cdots \\
=\left(\sum_{\alpha \in \mathbf{Z}} \frac{1}{2^{\alpha}}\right)\left(\sum_{\beta \in \mathbf{Z}_{4}} \frac{1}{3^{\beta}}\right)\left(\sum_{\gamma \in \mathbf{Z}_{+}}^{5^{\gamma}}\right) \\
=\frac{1}{1-\frac{1}{2}} \times \frac{1}{1-\frac{1}{3}} \times \frac{1}{1-\frac{1}{5}}=\frac{15}{4} \\
\Rightarrow m+n=19 .
\end{array}
$$