3. Petr has a special watch with three hands - the first one completes a full circle of the dial in one minute, the second in three minutes, and the third in 15 minutes. At the beginning, all the hands are in the same position. Determine after how long the hands will divide the dial into three equal parts. Find all solutions.
SOLUTION. Let's imagine a classic clock face with numbers 1-12. Without loss of generality, let's assume that at the beginning all three hands are at the number 12.

If the 15-minute hand rotates by an angle $\alpha$, the 3-minute hand will rotate by an angle $5 \alpha$ and the minute hand by an angle $15 \alpha$. Since any two hands in the desired positions form an angle of $120^{\circ}$ and the 3-minute hand is faster than the 15-minute hand, the desired positions can be obtained as solutions to the equation $5 \alpha - \alpha = k \cdot 120^{\circ}$, which are angles $\alpha = k \cdot 30^{\circ}$, where $k$ takes positive integer values that are not multiples of three, otherwise the hands would overlap.

We can therefore proceed by testing the values $\alpha = k \cdot 30^{\circ}$ for individual values of the number $k$. Let's really start with this and see how, after a few steps, we can obtain all other solutions to the problem due to periodicity.

First, consider $k=1$, i.e., $\alpha=30^{\circ}$. At this value, the fastest hand has rotated by an angle of $450^{\circ}$. At this moment, the slowest hand is at the number 1 on the clock face, the second hand at the number 5, and the fastest hand at the number 3. This case is therefore not a solution to the problem.

Let $k=2$, i.e., $\alpha=60^{\circ}$. At this value, the fastest hand has rotated by an angle of $900^{\circ}$. At this moment, the slowest hand is at the number 2 on the clock face, the second hand at the number 10, and the fastest hand at the number 6. This case is therefore one solution to the problem.

We see that we can create a table from which we can easily read all the solutions:

| positions of the respective hands on the clock face |  |  |  |  |  |  |  |  |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | 15-minute | 3-minute | minute | is it a solution? | time |  |  |  |
| $k=1$ | 1 | 5 | 3 | no | $1.25 \mathrm{~min}$ |  |  |  |
| $k=2$ | 2 | 10 | 6 | yes | $2 \cdot 1.25 \mathrm{~min}$ |  |  |  |
| $k=4$ | 4 | 8 | 12 | yes | $4 \cdot 1.25 \mathrm{~min}$ |  |  |  |
| $k=5$ | 5 | 1 | 3 | no |  |  |  |  |
| $k=7$ | 7 | 11 | 9 | no |  |  |  |  |
| $k=8$ | 8 | 4 | 12 | yes | $8 \cdot 1.25 \mathrm{~min}$ |  |  |  |
| $k=10$ | 10 | 2 | 6 | yes | $10 \cdot 1.25 \mathrm{~min}$ |  |  |  |
| $k=11$ | 11 | 7 | 9 | no |  |  |  |  |
| $k=12$ | 12 | 12 | 12 | no |  |  |  |  |

We have also included the "forbidden" value $k=12$ divisible by three, at which all three hands overlap, so in the next cycle their positions will repeat periodically. The times at which this happens will always be 15 minutes longer. We have thus found that all the desired times are

\[
\begin{aligned}
& t = (12 n + 2) \cdot 1.25 \mathrm{~min} = (15 n + 2.5) \mathrm{~min}, \\
& t = (12 n + 4) \cdot 1.25 \mathrm{~min} = (15 n + 5) \mathrm{~min}, \\
& t = (12 n + 8) \cdot 1.25 \mathrm{~min} = (15 n + 10) \mathrm{~min}, \\
& t = (12 n + 10) \cdot 1.25 \mathrm{~min} = (15 n + 12.5) \mathrm{~min},
\end{aligned}
\]

where $n=0,1,2, \ldots$

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