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Consider a generalized clock, where the minute hand goes n times as fast as the hour hand, where n is a positive integer. The standard clock has n=12 (sometimes n=24).

As which times can swapping the hour and minute hands result in a legal time?

In particular, for each hour h from 1 to n, for which minutes does this happen?

This obviously happens when the hands point in the same or opposite directions.

Are there any other times?

marty cohen
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    https://math.stackexchange.com/questions/59379/after-swapping-the-positions-of-the-hour-and-the-minute-hand-when-will-a-clock?rq=1 – Alex R. Jun 12 '17 at 23:51
  • What are your thoughts? Have you considered simply doing this for n = 12? It's pretty straight forward. – fleablood Jun 13 '17 at 15:07

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Let the hour hand be pointing at an exact value $H \in [0,1]$ ($0$ represents "12 o'clock" or the angle zero. $.5$ represents "6 o'clock" or the angle 180 or $\pi$ and $1$ represents 360, "12 o'clock" or $2\pi$ or "full circle"). Then the minute hand must be pointing at a precise value $M = \{nH\}$. i.e. the fractional part of $nH$ that is $nH = \lfloor nH\rfloor + \{ nH\}$ where $\lfloor nH\rfloor\in \mathbb Z$ ad $0 \le \{nH\} < 1$.

That's what a legitimate time is: if $M = \{nH\}$ the time is legitimate.

So we need times where $M = \{nH\}$ and $H = \{nM\}$ or $H = \{n\{nH\}\}$

$\{nH\} = nH - \lfloor nH\rfloor$; $n\{nH\} = n^2H - n\lfloor nH\rfloor=\{n^2H\} + \lfloor n^2H\rfloor- n\lfloor nH\rfloor$. $0 \le \{n^2H\} < 1$ and $\lfloor n^2H\rfloor- n\lfloor nH\rfloor\in \mathbb Z$.

So $\{n\{nH\}\} = \{n^2H\}$.

If $H = \{n^2H\}$ then $H = n^2H - k; H = \frac {k}{n^2-1} \in \mathbb Q$.

And, that's that. If $H = \frac {k}{n^2 - 1}$ then $H$ is a "reversible time".

Example: for the 12 hour clock, there are $143$ of these times. (Which actually surprises me as I assumed there were only 11.)

fleablood
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