The hour, minute, and second hands of a clock are on a common axis. At 12 o'clock, they overlap. When will they next overlap?
Let the time until the next meeting be $x$ hours $y$ minutes $z$ seconds, where $x$ and $y$ are positive integers, and $z$ is non-negative but not necessarily an integer. (We assume that the clock hands move continuously.)

The second hand moves $360^{\circ}$ in 1 minute, and $6^{\circ}$ in 1 second, so it will move $6z$ degrees until the meeting.

The minute hand moves $6^{\circ}$ in 1 minute, and $\frac{6^{\circ}}{60}=\frac{1}{10}$ degrees in 1 second. Thus, in $y$ minutes and $z$ seconds, it will move

$$
6y + \frac{1}{10}z \text{ degrees.}
$$

The hour hand moves $\frac{360^{\circ}}{12}=30^{\circ}$ in 1 hour, $\frac{30^{\circ}}{60}=\frac{1}{2}$ degrees in 1 minute, and $\frac{1}{120}$ degrees in 1 second. Therefore, in $x$ hours, $y$ minutes, and $z$ seconds, it will move

$$
30x + \frac{1}{2}y + \frac{1}{120}z
$$

degrees.

According to the problem, the angles of rotation are equal:

$$
6y + \frac{1}{10}z = 6z
$$

from which

$$
60y = 59z
$$

and

$$
30x + \frac{1}{2}y + \frac{1}{120}z = 6z
$$

Substituting $60y = 59z$ into the above equation, we get

$$
11y = 59x
$$

Since $x$ and $y$ are integers, and 11 and 59 are relatively prime, the smallest non-zero solution is $y = 59$, $x = 11$, and in this case $z = 60$.

Therefore, the next meeting will occur in 11 hours, 59 minutes, and 60 seconds, which is 12 hours later. So if it is noon, it will be midnight, and if it is midnight, it will be noon.

Hablicsek Márton (Fazekas M. Főv. Gyak. Gimn., 8th grade)