Hand-Over-Hand

ere is my analysis of the clock-hand puzzle, beginning with First Principles:
• The speed of the second hand (SH) is 360 degrees per minute.
• The speed of the minute hand (MH) is 6 degrees per minute (360 degrees divided by 60 minutes).
 Optional Shortcut (Not as much fun, I don't think.)

Observations: At the beginning of a each hour, both MH and SH will point at the top of the dial.

Exactly one minute later, the SH will arrive back at the top of the dial, but the MH will not be there; the MH will have moved to a place 6 degrees past the top of the dial.

Exactly 1 second later, the SH will have moved to that same place (6 degrees past the top of the dial), but the MH will not be there anymore.

Every time the SH arrives at a place where the MH has been, the MH has already moved on. Will the SH ever catch the MH?  Yes, of course -- but where? {Paradox Found}

hen you don't know something, pretend you do by calling it x.

The SH will catch the MH at x degrees past the top of the dial. One thing for sure, both SH and MH will reach x at exactly the same time: The SH will have traveled a total of 360 plus x degrees, and the MH will have traveled only x degrees. That gives us two questions:

(1) How much time does it take the SH to travel 360 + x
Answer: 360 + x degrees divided by 360 degrees per minute
(2) How much time does it take the MH to travel x?
Answer: x degrees divided by 6 degrees per minute
Both times are the same, so we can write:
(360 + x) / 360 = x / 6
Sophisticated solvers know that with algebra, you can 'solve for x,' which really means that what was unknown becomes known.

Give that a hoo-hah! For this problem, x = 360/59 degrees. That's about 6.1016949153 degrees.

People call 360/59 an 'improper fraction,' but it's really a fine looking number, don't you think?  Let's leave it that way. Besides, sophisticated solvers know that using the fractional form of the number will give an exact answer.
Is 360/59 degrees the solution to the original problem?
No. But we're getting close.
e have figured out 'where' the SH will catch the MH but not how long it will take to get there.
It is easy to figure out how long it will take for the MH to get there...
The MH will travel to a place x degrees past the top of the dial in exactly x / 6 minutes.
So we divide 360/59 by 6 and get 60/59 minutes -- that's 1 minute and 1.0169491524 seconds, for people who don't like 'improper fractions.'
It takes the MH 60/59 minutes to get to a place where the SH just catches up. We know something else, too.
Exactly 60/59 minutes later, the SH will again pass the MH, and exactly 60/59 minutes after that, the SH will again pass the MH, and... you get the idea.

 How many times each day does the second-hand pass over the minute hand?

There are 24 hours in a day and 60 minutes in each hour, so there are 24 times 60 minutes in a day; that's exactly 1,440 minutes per day. You will go ahead and divide that number by the 'time-between-passings,' which is 60/59...

 1,416 times per day

here is at least one really quick way to solve this problem:

In 24 hours, the SH goes around 1,440 times and the MH goes around 24 times. The difference is 1,440 - 24 = 1,416 times.
Question: Is that solution more or less sophisticated than the one provided above?

Sophisticated solvers, by the way, will ask themselves another question:  Here is a solution to a problem: What other problems can I now solve?

If you think of another problem, let us know.