At what exact times of day do the hour and minute hands of a clock meet?

Question

And by "exact", I mean exact, on an analog clock. No nitpicking... assume the hands are one-dimensional lines, perfect circular motion, yada yada. The answers are NOT 1:05, 2:10, etc. because the hour hand does move slightly as the minute hand moves toward the :05, :10, etc. marks.

Express your answer in any notation you would like, using any numerical form, e.g. fractions, irrational numbers, series, summations, etc. as long as the answer is exact. The best answer will be that which is (a) correct, duh, (b) the most simple expression, and (c) provides a good explanation of why it is correct, in that order.

Hint: The answer is a lot more simple than most people realize at first.

Express as a single function or expression that may be evaluated for all times of day, if possible.

Think more carefully, people. The hands meet many more times during the day than just at noon and midnight. They meet a bit after 1:05, after 2:10, etc.

(Who's the "smart arrse" now, kevsav025?)

Hmm, perhaps some of you misunderstand the question. By "exact time", I don't mean that the time falls exactly at a particular hour or minute mark. I mean, express what the time is exactly when the two hands to meet (or, if you'd rather, the exact moment when they pass one another). They will meet many times during the day.

For example, "H:05 + sin(30/H) minutes, where H is the hour" is a possible answer. Or, "For all hours X, the minute hand will be at X*(5+sqrt(2)/X) minutes when it meets the hour hand" is another possible answer. Both of these answers are absolutely wrong, hehe, but they do show an *exact* time.

Good guesswork, peter_r, very close to the actual values.

Kudos to Eulercrosser, the first correct answer! (Figures you'd be the one, you spend an inordinate amount of time supplying correct answers in the Mathematics topic, pthflpthfth!)

Another way to solve, using a convergent series:

At the hour H, the minute hand is at 12 and the hour hand is at the 5H mark. When the minute hand reaches the 5H minutes mark, the hour hand reaches 5H + H/12. When the minute hand reaches that new mark, the hour hand is now at H/12 + H/12^2. So they meet at the limit of the series:

5H + H/12 + H/12^2 + H/12^3 + H/12^4 + H/12^5 + ...

Which, interestingly enough, converges to H/11. So in terms of minutes, at the hour H the hands will cross at 5H + H/11 minutes.

An even simpler solution when I choose the Best Answer...

Apologies, that last answer's 2nd term is within the 5-minute "zone", so it should read:

5H + 5(H/11) minutes

Answer 1

First of all, lets just look at 12 hrs, as it will be the same for the next twelve:

The minute hand revolves around the clock at 1 rev per 60 minutes, the hour hand revolves around at 1 rev per 60•12 minutes. Knowing that after 60 minutes the minute hand is back to where it started we can say that 2π≡0. We want to rewrite the above relations to give us the positions as a value of time. This can be done by assuming that t is time (in minutes) and using m=2π/60t (minute hand) and h=2π/(12•60)t (hour hand).
Therefore we want to solve m=h+2πk for some integer 0≤k≤10. The reason that we don't have k go to 11 is because if you think about it, the hour and minute hand never cross during the 11th hour, they keep getting closer, but then at 12:00 (during the 12th hour) they meet.

Thus 2π/60t=2π/(12•60)t+2πk ==>> 12t=t+720k ==>>11t=720k ==>> t=720/11k.

Now, this is the time after 12:00 (midnight or noon, I don't care they; are the same).
These times are:

12:00 (k=0)
1:60/11=1:05:300/11≈1:05:27.27 (k=1)
2:10:600/11≈2:10:54.55
3:16:240/11≈3:16:21.81
4:21:540/11≈4:21:49.09
5:27:180/11≈5:27:16.36
6:32:480/11≈6:32:43.64
7:38:120/11≈7:38:10.91
8:43:420/11≈8:43:38.18
9:49:60/11≈9:49:05.45
10:54:360/11≈10:54:32.73

I decided to give you the second values as fractions also, since they can't be given accurately by a terminating decimal.

Answer 2

The next time the minute and hour hand will meet after 12 will be a little after 1:05, call it 1:05+x, then the next time will be 2:10+2x, 3:15+3x, etc. But we must have 11:55+11x = 12.00. So x = 1/11* 5min = 5/11 min = 5*60/11 sec = 27.2727..sec.
Let N = 0 to 10. Then the hands will meet at:

Time = 12:00 + N*(65 5/11)min

or if you want 12:00+ N*3,927.2727..sec

So N=0 gives 12:00
N=1 gives1:05 27.27..sec
2:10 54.54
3:16 21.8181..
4:21 49:0909..
5:27 16.3636..
6:32 43.6363..
7:38 10.9090..
8:43 38.1818
9:49 5.4545..
N =10 gives 10:54 32.7272
repeat

Or alternately, we fast forward to the 1 o'clock position
The minute hand is on the 12 and the hour hand is on the 1. If we assign the hour hand a unit velocity = 1, then the velocity of the minute hand = 12. For simplicity, we'll assign the 5min interval as the unit of distance. Now distance/velocity equals time. In the time it takes them to meet from these positions, the minute hand travels distance 1+x while the hour hand travels distance x. Therefore:
(x+1)/12 = x and 12x = x+1 and x = 1/11. X is 1/11 th of a 5min interval, So x = 5/11 min.

Answer 3

The two hands cross over exactly 12 times in the 12 hour period from 00:00 to 12:00. The times can be described as follows:

let h = the hour (from 0 to 11, no consideration for PM or AM)

the minute hand will be in the following position for each h that they cross:

5*h*(1+1/12)
eg, at just after the 3rd hour it will cross at
5*3*(1+1/12)
= 16.24 min
= 16 min 15 seconds
another example for the 6th hour:
they cross at 6 hr 32 min and 30 seconds

In engineering terms the above will be accurate enough......

Answer 4

Let measure time t in seconds and fix t=0 at 12:00 midnight.

Although we can divide the full angle into 360 degrees
but here for the sake of simplicity we divide the full angles into 60 seconds! abbreviated SEC.
Thus 60 SEC= 360 degrees
or 1 SEC=60 degrees.
(forget all about degrees from now on)

At time t (measured in seconds):
The minute hand shows t/60 SEC
The hour hand shows 5*t/3600 SEC that is t/720

When each turns a complete round that is 60 SEC we reset its value.

At t=0 both hands coincide. (12:00:00)

Next coincidence is at (t-3600)/60=t/720.
t-3600=t/12
12t-43200=t
11t=43200
t=43200/11=3927.27
which is 01:05:27.27

We have computed the kth coincidence for k=0 and k=1.
The general equation for kth coincidence is :
(t-3600k)/60=t/720
which simplifies to:
t=43200k/11 (for k=0,1,2,3,...,10)
There are 11 coincidenced in each 12 hours.

k=0 t=0.
k=1 t=3927.272727
k=2 t=7854.545454
k=3 t=11781.81818
k=4 t=15709.09091
k=5 t=19636.36364
k=6 t=23563.63636
k=7 t=27490.90909
k=8 t=31418.18182
k=9 t=35345.45454
k=10 t=39272.72727

or in more readable form:
0 12:00:00.00
1 01:05:27.27
2 02:10:54.54
3 03:16:21.81
4 04:21:49.09
5 05:27:16.36
6 06:32:43.63
7 07:38:10.90
8 08:43:38.18
9 09:49:05.45
10 10:54:32.72

and most interesting:
k=11 12:00:00 again!

Answer 5

Exact Clock

Answer 6

The watch company's logo is usually right under the twelve. Having the hands at 10:10 on the analog watches/clocks shows the logo, rather than obstructing it.

Answer 7

This is just guess work but it happens at around: -
5 minutes and about 25 seconds after 1
10 minutes and about 51.5 seconds after 2
16 minutes and about 17 seconds after 3
21 minutes and about 43 seconds after 4
27 minutes and about 9 seconds after 5
32 minutes and about 32 seconds after 6
37 minutes and about 58 seconds after 7
43 minutes and about 25 seconds after 8
48 minutes and about 52 seconds after 9
54 minuted and about 17 seconds after 10
12 noon and 12 midnight unless it is a 24 hour clock.

Answer 8

The hands meet at noon and midnight.

Answer 9

according to your conditions they meet only two times a day noon and midnight.

Answer 10

12:00 am or pm