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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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    @JM: I don't see the phrase "later time" that you said?!?!? I think your clock has something wrong.2011-08-24

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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.

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    @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.2011-08-25
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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.

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    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).2018-01-18