$3 \cdot 23$ On an escalator, the visible part consists of $n$ steps, which descend at a fixed speed. Two boys, $A$ and $Z$, walk down the moving escalator at a steady pace. $A$ walks down twice as many steps per minute as $Z$. $A$ reaches the bottom after 27 steps, while $Z$ reaches the bottom after 18 steps. Then $n$ is
(A) 63.
(B) 54.
(C) 45.
(D) 36.
(E) 30.
(16th American High School Mathematics Examination, 1965)
[Solution 1] Let the escalator rolling one step be one unit of time, and let $Z$ walk $k$ steps per unit time, i.e., it takes $\frac{1}{k}$ unit time to walk one step. Therefore, the speed of $Z$ going downstairs is $k+1$.
Thus,
$$
\frac{18}{k}(k+1)=n \text {. }
$$

Similarly for $A$, we have $\quad \frac{27}{2 k}(2 k+1)=n$.
$\therefore \quad \frac{18}{k}(k+1)=\frac{27}{2 k}(2 k+1)$, which gives $\quad k=\frac{1}{2}$.
Thus,
$$
n=54 \text {. }
$$

Therefore, the answer is $(B)$.
[Solution 2] Let $Z$ walk $x$ steps per minute, then $A$ walks $2 x$ steps per minute. Also, let the escalator walk $y$ steps per minute.
$Z$ takes $\frac{18}{x}$ minutes to walk down, and $A$ takes $\frac{27}{2 x}$ minutes to walk down. According to the problem, we have
$$
\left\{\begin{array}{l}
(x+y) \cdot \frac{18}{x}=n, \\
(2 x+y) \cdot \frac{27}{2 x}=n,
\end{array}\right.
$$

Solving these equations gives $n=54$.
Therefore, the answer is $(B)$.