36 - 55 A watch is slow by $2 \frac{1}{2}$ minutes each day. Suppose it was set right at 1:00 PM on March 15. Let $n$ be the number of minutes that must be added to the time indicated by the watch to get the correct time. If on March 21 the watch indicates 9:00 AM, then $n$ is equal to
(A) $14 \frac{14}{23}$.
(B) $14 \frac{1}{14}$.
(C) $13 \frac{101}{115}$.
(D) $13 \frac{83}{115}$.
(E) $13 \frac{13}{23}$.
(15th American High School Mathematics Examination, 1964)
[Solution] Since there are $60 \cdot 24=1440$ minutes in a day, and the clock is slow by $2 \frac{1}{2}$ minutes each day, it actually only walks
$$
1440-2 \frac{1}{2}=1437 \frac{1}{2} \text { (min). }
$$

Therefore, the ratio of actual time to the time the clock has walked is $\frac{1440}{1437 \frac{1}{2}}=\frac{576}{575}$.
If we calculate one minute as one minute and one day as one day according to the clock, from 1 PM on March 15 to 9 AM on March 21, it is $5 \frac{5}{6}$ days, or $60 \cdot 24 \cdot 5 \frac{5}{6}$ minutes.
Thus, $60 \cdot 24 \cdot 5 \frac{5}{6}+n=60 \cdot 24 \cdot 5 \frac{5}{6} \cdot \frac{576}{575}$,
$$
\begin{aligned}
\therefore \quad & n=60 \cdot 140 \cdot\left(\frac{576}{575}-1\right) \\
= & 14 \frac{14}{23} .
\end{aligned}
$$

Therefore, the answer is $(A)$.