[Reprinted] Three-pin match today's beacon star test

I won't repost it.

In the 24 hours of a day, how many times does the hour, minute, and second hands of the clock overlap? What is the time? How did you calculate it?
Reference answer: There are two times when the three needles completely overlap in one day, respectively at noon and AM.
Analysis: the hour, minute, and second hands rotate around the same axis, so they all have their own angular velocity, and there is a certain relationship between the angular velocity. Based on this relationship, we can remove this question.
First, set the angle velocity of the hour to W, then the angle velocity of the minute and second needles is 720 and respectively.
First, we will examine the angle between the hour hand and the minute hand. If it is set to X, there is an equation: X/W = (x + N * 360)/12 W, where N is the number of minutes that exceeds the hour, and N is a positive integer ranging from 1 to 22 (only 22 is obtained because 24 times of the day, but the hour hand also walked two circles, so 24-2 = 22 ).
Then, we can take n value to calculate X. After finding X, we also need to check whether the second is in X at this time. We can see that the time used to walk the hour hand to the X space is X/W, and the total angle of the second hand is 720 w * x/W = 720x. Then, reduce the value to less than 360 to see if it is W. The simple process is as follows:
When n = 1, x = 360/11. 720*360/11 --> 5*360/11. Not overlapped.
When n = 2, x = 2*360/11. 720*2*360/11 --> 10*360/11. Not overlapped.
When n = 3, x = 3*360/11. 720*3*360/11 --> 4*360/11. Not overlapped.
............
When n = 11, x = 11*360/11 = 360. 720*360 --> 360. At noon.
There are two times when three needles are completely overlapped in a day, respectively at noon and AM.
This topic and ideas refer to blog posts ~~

[Reprinted] Three-pin match today's beacon star test