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 Fun math puzzle 
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Post Fun math puzzle
Found in the exercises of my real analysis book, Abbott's Understanding Analysis. Fantastic book, would highly recommend.

Imagine a clock where the hour hand and the minute hand are indistinguishable from each other. Assuming the hands move continuously around the face of the clock, and assuming their positions can be measured with perfect accuracy, is it always possible to determine the time? (And no, watching which one goes faster doesn't count—imagine we only get their positions at a specific moment.)

The solution:
Spoiler:
No, we cannot.

Let m and h be the positions of the minute and second hands according to where the minute markings are, so h,m are in [0,60). At all times, h mod 5 = m/12. If we can't tell the time, that means that additionally, m mod 5 = h/12; and finally, we must have m not equal to h, because if they're in the same place, it doesn't matter which is which. The first two equations will be satisfied if and only if 12((12(h mod 5))mod 5)-h = 0. You can run a computer program to solve this, but you can also use the Intermediate Value Theorem: the expression on the left, viewed as a function, is continuous except at multiples of 5/12, so we can make it continuous by restricting it to [5/12, 9/12]. You can easily calculate that the expression is negative at h = 5/12 and positive at h = 9/12, so by the IVT, it must be equal to zero somewhere in between. We know that the hands are not overlapping there, since that only happens once per hour and has already occurred this hour at 12:00. Thus, there is some time between 12:25 and 12:45 when we can't tell what time it is.

This isn't necessarily the only time when this is the case—indeed, there must be at least one more, obtained by switching the positions of the minute and hour hands (after all, that's why it's indistinguishable). We can find more by finding other solutions to the equation. However, I'm fairly certain that there are only finitely many solutions, so most of the time, we can in fact tell what time it is.


~dlgn

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Post Re: Fun math puzzle
Hint to get started: There are still some restrictions on where the hands can be. If the minute hand is exactly on 12, then the hour hand must be exactly on one of the 12 integer positions. If the minute hand is not exactly on 12, then the hour hand must be at some point between the 12 integer positions. Thus, an arrangement like one hand on 3 and the other on 6 is impossible because one of these is the minute hand and neither is on 12 and the other is the hour hand and neither is not at an integer position.

Furthermore, the hour hand must be at a point between consecutive integer positions proportional to the position the minute hand is around the clock. Using this rule, I know there are at least some arrangements when you know which hand is which.

Still thinking about it. Thanks for giving my brain something to do.

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