At 12 o'clock, the hour hand and minute hand of the clock can be swapped, and the clock still gives the same time, but at 6 o'clock, it can not be swapped. So in what cases when we swap the hour and the minute hand position does a clock still give a valid time?

valid invalid

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    $\begingroup$ Enumeration works: 12:00, 1:05, 2:10, 3:15... the pattern should be clear. On the other hand, depending on the analog clock's mechanism, the latter times might no longer have the hands to be swappable. $\endgroup$ – J. M. is a poor mathematician Aug 24 '11 at 2:57
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    $\begingroup$ At 1:05, the minute hand's on the 1, the hour hand's a little past it; if you swap, the hour hand's exactly on the 1, the minute hand's a little past it, and that's not a valid position. $\endgroup$ – Gerry Myerson Aug 24 '11 at 3:10
  • $\begingroup$ For the clock I have, that happens on "later times" (e.g. 9:45). I guess it does depend on the clock. $\endgroup$ – J. M. is a poor mathematician Aug 24 '11 at 3:20
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    $\begingroup$ @Gerry Myerson - Hopefully J.M. meant the hands-pointing-in-the-same-direction positions and just rounded (i.e. every 12/11 hours). This is correct but incomplete, as your answer shows (every 13th solution of yours is a same-direction solution). $\endgroup$ – Rex Kerr Aug 24 '11 at 3:22
  • $\begingroup$ Yes, @Rex has it. I wasn't thinking of the "non-coincident hands" solutions, and I'm not in the mood for the needed arithmetic... hence I left it as a comment. $\endgroup$ – J. M. is a poor mathematician Aug 24 '11 at 3:38

Let $x$ be the position of the hour hand, as measured in degrees clockwise from 12 o'clock. So, for example, at 1 o'clock, $x=30$. Let $y$ be the position of the minute hand; then $y\equiv12x\pmod{360}$, because the minute hand spins 12 times as fast as the hour hand. In order for $(y,x)$ to be a valid pair of positions for (hour hand, minute hand), we must also have $x\equiv12y\pmod{360}$. Putting these together, we get $x\equiv144x\pmod{360}$, which is $143x\equiv0\pmod{360}$, which has the solutions $x=0,360/143,720/143,1080/143,\dots$.

$x=360/143$ is $12\times360/143=30.20979\dots$ minutes past 12 o'clock; 30 minutes, 12 and four-sevenths seconds after 12 o'clock. And then any integer multiple of that will do.

EDIT: As Henry points out in a comment, the 2nd paragraph contains an error. $x=360/143$ is $12\times360/143=30.20979\dots$ degrees past 12 o'clock, but it is $2\times360/143$ minutes past 12, which is (as Henry says) 5 minutes, $2{14\over143}$ seconds after 12.

  • $\begingroup$ How many solution in this case? $\endgroup$ – KevinBui Aug 24 '11 at 3:50
  • $\begingroup$ @DKahnh - 143. 360*143/143 = 360 ≡ 0 (mod 360). $\endgroup$ – Rex Kerr Aug 24 '11 at 4:14
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    $\begingroup$ The situation can be viewed geometrically as follows. The handles' position can be specified giving the angles from 12 o' clock. Thus the clock space is the torus $T=S^1\times S^1$ and the valid time positions describe the curve $\phi(\theta)=(\theta, 12\theta)$ in $T$. Swapping the handles gives an automorphism of the clock space $T$ which transforms the curve $\phi(\theta)$ into the curve $\psi(\theta)=(12\theta,\theta)$. The solutions of the problem are the intersections of the two curves. $\endgroup$ – Andrea Mori Aug 24 '11 at 8:46
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    $\begingroup$ I don't understand the second paragraph. $\frac{12}{143}$ hours or $\frac{720}{143}$ minutes or $\frac{43200}{143}$ seconds is $5$ minutes and $2\tfrac{14}{143}$ seconds for the first swappable position after 12 o'clock, and then any integer multiple will do. Thirteen times this ($1$ hour, $5$ minutes and $27\tfrac{3}{11}$ seconds) and you get the first matching position. $\endgroup$ – Henry Aug 24 '11 at 10:30
  • $\begingroup$ @Henry, you're right: at $x=360/143$, the minute hand is at 30.20979... degrees past 12 o'clock; divide by 6 to get the number of minutes past 12 o'clock. I will edit. $\endgroup$ – Gerry Myerson Aug 25 '11 at 4:47

A visual proof. Every intersection point of the black grid is a solution.enter image description here

Explanation. On the $x$-axis the position of the hours clock hand, on the $y$ axis the position of the minutes clock hand. When the short hand goes between hour n to n+1, the long hand makes a complete turn (from 0 to 12). If you exchange the hands, you exchange $x$ and $y$ coordinates... so you look for intersection of the graph with its simmetry with respect to the diagonal of the square domain.

  • $\begingroup$ Beautiful! From this picture, you can easily see that there are 12*12 points of intersection because there are twelve lines crossing twelve lines. This means that there are 144-1 = 143 distinct times where swapping the hour and minute hand give a valid time (because we don't want to count midnight twice). $\endgroup$ – user326210 Jan 18 '18 at 5:47

Define a third imaginary hand at the position the minutes hand would be if the actual minutes hand was the hours hand. This imaginary third hand travels 12 times faster than the actual minutes hand, thus completes 144 cycles in a 12 hour period.

Note that the actual minute and hour hands are swap-able iff the third imaginary hand coincides with the actual hour hand. The actual hour hand is traveling at a speed of 1 cycle per 12 hours, and the third imaginary hand is traveling at a speed of 144 cycles per 12 hours, so their relative speed is 143 cycles per 12 hours. Thus the hands of a clock are swap-able 143 times in a 12 hour period.


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