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A short small puzzle this week building off an interview inquiry I check out on the internet.

You are watching: How many times do clock hands overlap

The initial question was this:

At what times of the day perform the hour and minute hands of an analog clock perfectly align?

At first glance this can seem choose a trivial question, climate you realize that the hour hand moves smoothly and continuously roughly the dial (albeit in ~ a slow pace than the minute hand), and does no snap to each quantized hour position on every hour. This complicates points a little, however not as well much. (The answers space not 1:05, 2:10, 3:15 …)

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It’s pretty clear the the hands both align once it’s precisely midnight (and midday). Once is the next time?

It’s not 1:05, yet a tiny bit past because, by the time the minute hand is likewise at the 1 o’clock position, the hour hand will certainly have progressed on slightly.

The minute hand spins around the dial twelve times as rapid as the hour hand (it completes one transformation in one hour whilst the hour hand moves one hour, i beg your pardon is 1/12th that the clock face).

In T hours, the minute hand completes T revolutions. In the same amount the time, the hour hand completes the fraction T/12 revolutions. Using degrees, we deserve to see the the minute hand moves at 360° every hour, and the hour hand (360°/12) = 30° every hour.

Below is a graph showing the edge (in degrees) because that both hands for values of T indigenous 0 come 12. Whereby the lines intersect (example displayed with the red circle), is where the hands will coincide.

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This happens at 11 various locations in between midnight and (just before) midday, then recurring again one more 11 time in the afternoon.

We have the right to calculate the exact times by looking for the times once the angle between the 2 hands is zero.

Let"s specify the angle HT, MT to it is in the angle (in degrees) the the hands (from 12 o"clock position), after time T (in hours).

HT = 30T

MT = 360T

For the hand to it is in aligned, the difference in between the hour and also minute hand needs to it is in zero (after an arbitary number of rotations), whereby n is an arbitrarily (integer) variety of rotations.

MT - HT = 360 × n

360T - 30T = 330T = 360n

11T=12n

We can uncover the times, through insterting in n=0,1,2 …

T=12n/11

The hand overlap every (12/11) hours. Right here are the 22 results (rounded to nearest second):

12:00:00 AM12:00:00 PM
1:05:27 AM1:05:27 PM
2:10:55 AM2:10:55 PM
3:16:22 AM3:16:22 PM
4:21:49 AM4:21:49 PM
5:27:16 AM5:27:16 PM
6:32:44 AM6:32:44 PM
7:38:11 AM7:38:11 PM
8:43:38 AM8:43:38 PM
9:49:05 AM9:49:05 PM
10:54:33 AM10:54:33 PM

However, the inquiry I desire to questioning is the next logical development of this problem. What if we add a secs hand?

Hour hand, Minute hand, second hand …

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Image: note Turnauckas

At what times of the day carry out the hour, minute, and 2nd hands all line up?

As before, let"s specify the angles HT, MT, ST to it is in the angles (in degrees) of the hands (from 12 o"clock position), ~ time T (in hours).

HT = 30T

MT = 360T

ST = 360T × 60 = 21600T

For the hands to it is in aligned, the difference between pairs angles needs to be zero (after an arbitary number of rotations).

MT - HT = 360 × n

ST - HT = 360 × m

(Where n and m room integer coefficients). Combining the equations we get:

360T - 30T = 330T = 360n

21600T - 30T = 21570T = 360m

These simplify:

11T=12n

719T=12m

Giving the result:

719n=11m

Both 11 and 719 space Prime, and have no usual factors. For this reason (other than the trivial instance of n=m=0), together n need to be a lot of of 11, speak n = 11x, for some integer x. Then m = 719x, and also T = 12x. That shows that the just time when this happens is ~ an essence multiple the 12 hours, the is, in ~ 12 o"clock.

All the areas where the hour and also minute hand align (angle distinction being a lot of of 360°) are different from all the locations the hour and the 2nd hands align (other than 12:00).

What this way is over there that, other than midnight (and midday), there room no various other times when all 3 hands have exactly the exact same angle.

(This answer provides sense as soon as you think around it, because, if all three were come line-up, it has actually to happen when 2 line-up as well, so it would have to at one of the 11 times calculated previously with simply the hour and also minute hands, and none of this correspond to location of wherein the 2nd hand could be).

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How close deserve to we acquire them?

A fast search of web reveals that Dr Rob, from The mathematics Forum also looked in ~ this problem and also found the closestly he can get all 3 hands is at the time 5:27:27.3, once all the hands space within a 1.0014 degree sector. That goes on come remind united state that this is more than likely still visible to the naked eye together clock hands space thin, and also the edge between 2nd marks is 6°

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