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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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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. – J. M. Aug 24 '11 at 2:57
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. – Gerry Myerson Aug 24 '11 at 3:10
For the clock I have, that happens on "later times" (e.g. 9:45). I guess it does depend on the clock. – J. M. Aug 24 '11 at 3:20
@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). – Rex Kerr Aug 24 '11 at 3:22
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. – J. M. Aug 24 '11 at 3:38
up vote 13 down vote accepted

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 thses 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.

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How many solution in this case? – KevinBui Aug 24 '11 at 3:50
@DKahnh - 143. 360*143/143 = 360 ≡ 0 (mod 360). – Rex Kerr Aug 24 '11 at 4:14
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. – Andrea Mori Aug 24 '11 at 8:46
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. – Henry Aug 24 '11 at 10:30
@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. – Gerry Myerson Aug 25 '11 at 4:47

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