A position of the hands of a (12-hour, analog) clock is called valid if it occurs in the course of a day. For example, the position with both hands on the 12 is valid; the position with both hands on the 6 is not. A position of the hands is called bivalid if it is valid and, in addition, the position formed by interchanging the hour and minute hands is valid. Find the number of bivalid positions.
Let $h$ and $m$ denote the respective angles of the hour and minute hands, measured clockwise in degrees from 12 o'clock $(0 \leq h, m<360)$. Since the minute hand moves twelve times as fast as the hour hand, we have

$$
m=12 h-360 a
$$

with $a$ an integer, for any valid time. Conversely, it is clear that any ordered pair $(h, m)$ satisfying (1) represents the valid time of $h / 30$ hours after 12 o'clock.

The condition that a time be valid after switching the hour and minute hands is, of course,

$$
h=12 m-360 b \text {, }
$$

with $b$ an integer. Thus the problem of finding bivalid times is reduced to finding pairs $(h, m)$ satisfying (1) and (2). Note that if $h$ is known, $m$ is uniquely determined from (1) and the condition $0 \leq m<360$. Note also that the truth of (2) is not
affected by increasing or decreasing $m$ by 360 , as this simply increases or decreases $b$ by 12 , giving a new integer value $b^{\prime}$. Consequently we substitute $12 h$ for $m$ in (2), getting

$$
\begin{gathered}
h=144 h-360 b^{\prime} \\
143 h=360 b^{\prime} \\
h=\frac{360 b^{\prime}}{143} .
\end{gathered}
$$

This equation has 143 solutions in the range $0 \leq h<360$, each of which gives rise to one solution of (1) and (2). We conclude that there are 143 bivalid positions.