Zaslavsky A.A.

Martians divide the day into 13 hours. After the Martian Rabbit dropped the clock into the tea, the speed of the second hand changed, while the speed of the other hands remained the same. It is known that all three hands coincide at midnight. How many times in total can such moments occur in a day when all three hands coincide?

![](https://cdn.mathpix.com/cropped/2024_05_06_62351a5cb646b08ac6fbg-03.jpg?height=381&width=374&top_left_y=1010&top_left_x=842)

#
For Martian days, the hour hand makes one revolution, the minute hand - 13. This means that the minute hand overtakes the hour hand (i.e., "coincides" with it) 12 times - every 12th part of the day. According to the condition, the second hand coincides with the hour hand a whole number of times (let's say $n$). If $d=$ GCD(12, $n$), then every $d$-th part of the day, the minute and second hands coincide with the hour hand simultaneously, meaning all three hands coincide $d$ times a day. Depending on $n$, $d$ can be any divisor of the number 12.

## Answer

$12,6,4,3,2,1$.