# A math question that's stumped me since 10th grade

When I was in 10th grade, our school had a monthly math competition of some sort. To this day, some 20 years later, I can recall a question that baffled me then, and, on the occasions since then when I have had time to think about it, continues to baffle me now. I’m sure that some of you will make short work of it, and perhaps allow me to reallocate the mental real estate it occupies to something more useful, like the birthdates of my children.

The question is simple: when a clock is set at 12:00.00, the hour, minute, and second hands are all precisely aligned atop each other. At what time will this condition occur again?

(Obviously, the problem can be solved by trial and error, but the test required a mathematical way of solving it, so I’m looking for an equation or set of equations.)

At what time will this first happen again? That’s the important part.

Realistically, I think it would depend on the clock and how the gears are set up. . .sometimes the second hand “jumps” from second to second, sometimes it moves smoothly–same with the minute hand, etc.

I would assume for the purposes of the problem, though, that one would assume they all move smoothly and regularly, ie you could figure out the speed of the hands and determine where they will be at any particular moment. . .

The hour and minute hand cross each other 11 times every 12 hours. Excluding 12:00, let’s consider when this happens. Let h and m be the speed of the hour and minute hands. m = 12h (and s = 12m). They cross at time t when mt=ht+n for n=1,2,…,10. This happens exactly when n=11ht, when t=n/(11h), i.e. at 1+1/11 hours, 2+2/11 hours, … 10+10/11 hours. (11+11/11 = 12).

Now we need to figure out where the second hand is at these times. Remember, the relationship between the minute hand and the hour hand is exactly the same as the relationship between the second hand and the minute hand (just 12x slower), so it’s easy to see that the second hand will be in on it whenever the other two hands meet…
Between 12:01 and 11:59, the minute had crosses the hour hand

Right around 1:06 and 6 seconds. I leave it to the more mathematical dopers to put some rigor into this estimate

12:00 (again, maybe?) Can all three line up at any other timt?

The hour and second hands will line up 1 and 1/11 o’clock, 2 and 2/11 o’clock, thru 11 and 11/11 o’clock (or 12:00).

Will the second hand be anywhere near at any other times?

Express the positions relative to the whole circle.

Second hand position = sec / 60
Minute hand position = (min + sec/60) / 60
Hour hand position = (hour + min/60 + sec/60) / 12

Set these equal to each other, and solve for an integer solution.

(Plus in this problem, you can assume hour = 1.)

Oops. I left a dangling fragment on the end of my post. If you consider the second hand to be discrete (instead of gliding from second to second) then it only happens at 12:00.

Oops! That should be:

Hour hand position = (hour + min/60 + sec/3600) / 12

The minute hand moves 12 times as fast as the hour hand - there are 11 other places on the clock face where the minute hand crosses the hour hand - these are equally spaced (in fact, one way of looking at the problem is the hour hand moves to 1 oc’lock - then 5 minnutes later the minute hand gets to where it was - but the hour hand has moved 1/12 of the way toward 2 o’clock, so the minute hand has to go that distance, but the hour hand moves - so it’s an infinite series sum - but I digress).

The second hand moves 60 times as fast as the minute hand. There are 59 places on the clock face, other than 12 o’clock, where they cross -they, too, are equally spaced.

11 and 59 are relatively (and absolutely, for that matter) prime - therefore, none of the hour/minute crossings correspond to the minute/second crossings.

BTW, the next time the minute/hour cross is 1:05:27.272727…

Whoops. Let me correct my phrasing. There are 10 other places where the hour/minute cross, and 58 other places where the minute/second cross.

If you think of like a pie, though, one pie has 11 equal pieces, the other has 59 equal pieces.

No piece edges correspond except the one at 12, so the answer stands (whether or not the hands ‘sweep’).

bup’s answer is correct from a pure mathematics perspective, but assumes three infinitely narrow clock hands. The exact centers only line up every twelve hours, but real world clock hands would have an instant of at least partial overlap for all three, eleven times every twelve hours.

The separate issue of discrete versus continuous movement of the second hand is a red herring. The stops are discrete, but its position is continuous. It still has to pass through every point in the arc between one second and the next, it just does so pretty quickly (pause, lurch, pause, lurch). That is, unless it slips past time-space through some sort of wormhole every time it advances. I’m looking at my own watch right now, and it appears to stay in real space pretty consistently.

Hmm. Actually, if the clock is some sort of lcd analog display, the second hand could be truly discrete. In that case, though, you’re probably only looking at 60 possible values for any hand, so synchrony would be guaranteed anyway.

At any rate, would the OP be so kind as to submit arc measures for all three hands, so that we may offer a more responsible calculation?