# After swapping the positions of the hour and the minute hand, when will a clock still give a valid time?

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?

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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

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.
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