We cannot clearly read the time from this clock if the hour and minute hands are in interchangeable positions. Let the angles between the two hands and "12 o'clock" be represented by $\alpha$ and $\beta$ respectively, where $0^{\circ}<\alpha, \beta<360^{\circ}$. Since we cannot distinguish between the positions, the following equations hold:

$$
12 \alpha=k \cdot 360^{\circ}+\beta, \quad \text { and } \quad 12 \beta=\ell \cdot 360^{\circ}+\alpha
$$

where $0 \leq k, \ell<12$ are natural numbers.

![](https://cdn.mathpix.com/cropped/2024_05_02_cf8cd9a45e98e12cd453g-1.jpg?height=374&width=377&top_left_y=491&top_left_x=863)

Express $\beta$ from the first equation and substitute it into the second equation:

$$
\begin{aligned}
\beta & =12 \alpha-k \cdot 360^{\circ}, \\
12\left(12 \alpha-k \cdot 360^{\circ}\right) & =\ell \cdot 360^{\circ}+\alpha
\end{aligned}
$$

From this, $\alpha$ and then $\beta$ can be expressed in terms of $k$ and $\ell$:

$$
\alpha=\frac{(12 k+\ell) \cdot 360^{\circ}}{143}, \quad \beta=\frac{(12 \ell+k) \cdot 360^{\circ}}{143} \text {. }
$$

The hands can be interchanged if $k \neq \ell$. (If $k=\ell$, then the two hands overlap and the time can be precisely determined.)

The natural numbers $k$ and $\ell$ can each take 12 different values $(0 \leq k, \ell<12)$, so there are 144 possibilities in total. However, we must subtract the 12 cases where $k=\ell$. Therefore, the total number of ambiguous time points is $144-12=132$.

II. Solution. Consider another - new - clock that runs exactly 12 times faster than the clock in the problem, and start both clocks simultaneously at 12:00. On the new clock, the hour hand moves at the same speed as the minute hand on the original clock. We are interested in the situations where the hour hand on the original clock is in the same position as the minute hand on the new clock (since these are the situations where, with the same minute and hour hands, it cannot be determined which of the two variants shows the actual time).

By the time the hour hand on the original clock completes one full circle, the minute hand on the new clock will have completed 144 circles, and the two hands will be in the same position 144 times. However, there will be 12 occasions when all four hands on both clocks are in the same position. On these 12 occasions, the time can be determined unambiguously, so $144-12=132$ positions of the hands exist where the time shown by our clock cannot be determined unambiguously.

Note. Many people forgot that when the two hands overlap, the time is unambiguous despite the interchangeability.
A precisely running clock has its hour and minute hands exactly the same, and it has no second hand. How many times are there in a day between 12 hours after and 24 hours before, when we cannot clearly read the time?