# Meeting of hands on a clock

We have a regular analog clock with three hands - second, minute and hour. On a given day, how many times do:

(a) the minute and hour hands

(b) the minute and second hands

(c) the hour and second hands

(d) all three hands

meet?

One approach would be to just list out all the times they meet, but I'm looking for a more logical/mathematical approach.

I think there is a much simpler solution than all provided so far.

Consider the following three facts:

• hour hand will rotate 2 full times in a day
• minute hand will rotate 24 full times in a day
• second hand will rotate 1440 full times in a day

So then:

(a) the minute and hour hands, will meet exactly:

22 times: 24 - 2 (once every 24/22 hours)

(b) the minute and second hands, will meet exactly:

1416 times: 1440 - 24 (once every 24/1416 hours)

(c) the hour and second hands, will meet exactly:

1438 times: 1440 - 2 (once every 24/1438 hours)

(d) all three hands, will meet exactly:

twice: only at exactly 12:00:00 o'clock (noon and midnight)

Simply because:

The faster hand passes the slower hand by the number of laps it makes minus the number of laps the slower hand makes.

With the special case:

One hand lapping another won't necessarily coincide with the third hand being there. Try to find the common multiples of the three fractions 24/22, 24/1416, and 24/1438 and you will see there are only two at 24/1 and 24/2. (ie. 12 hours and 24 hours after the start).

• I think I've seen one before (or maybe to the similar question of whether they will ever be all three exactly 120 degrees apart), but I don't think you've actually provided a proof of (d). – Random832 Sep 21 '15 at 21:15
• @Random832, that is a fair point and I hope my latest edit has made a better case for (d). – Octopus Sep 21 '15 at 22:24
• I have a question too on point d): Say from 14:00 to 14:15, there will be a time where the (minute hand) stacks on the (hour hand), where the (second hand) pass and they all stack? – Alex Sep 22 '15 at 19:43
• @Alex. Not really. The minute hand will be directly over the hour hand a short moment after 2:10:54, and by the time the second hand gets to where the hour hand is (roughly 17 seconds later), the minute hand will have already passed it. They are close, but they do not all meet. – Octopus Sep 22 '15 at 20:07

Question: At midnight, when all the hands are pointing at twelve, does that count as a meeting? Does it count as a second meeting when they all point at midnight $24$ hours later? I will assume only one of these counts as a meeting. i.e. if one of them is in the $24$ hour time interval, then the other cannot be. For example, let our $24$ hour window go from 12:00:01 to 12:00:01 the following day.

For a)

The hour hand will go around the clock twice during a day, every $12$ hours. During a single rotation, the minute hand will go around $12$ times. The minute hand will pass the hour hand $11$ times in the first $12$ hours, and $11$ times in the second $12$ hours. This means that they will meet $22$ times.

For b)

In a single hour, the second hand will go around the clock $60$ times. The minute hand will go around once. Thus, the second hand will pass the minute hand $59$ times in an hour. This means they will meet $59 \times 24=1416$ times in a day.

For c)

The second hand will go around the clock $60 \times 12=720$ times in $12$ hours. The hour hand will go around once. Thus, they will meet $719$ times in $12$ hours, and $1438$ in $24$ hours.

For d)

This is the trickiest one. First, lets note that the first $12$ hours and the second $12$ hours of the day are symmetrical. Also, in the first 12 hours, we already know that the hour and minute hand only meet 11 times. Since there is the same amount of time between each meeting, we know that there are $\frac{12}{11}$ hours between each meeting. $\frac{12}{11}$ hours $=\frac{12 \times 60}{11}$ minutes $=\frac{12 \times 60 \times 60}{11}$ seconds. Thus, each meeting is at every $3927.28$ seconds. We know the first meeting is around 1:06. The second hand will be pointing at $3927.28 \mod 60 = 27.28$ seconds. Thus, at the first meeting of the minute and hour hand (near the one), the second had will be pointing half way between the 5 and 6. If we continue this, we will see that the second hand never meets they other two at the exact same time. Thus, all three only meet twice - once at noon, and again at midnight.

You can consider the position of the hands as sequences mod 1. By which I mean giving a hand a position $x$ with $0\le x < 1$ which represents its position. When the hand rotates back around to 12 o'clock we assign it zero (so it keeps looping).

(Essentially if $x$ exceeds 1 then we minus an integer so its between 0 and 1 again, this is how the 'mod 1' idea works and compensates for the clock looping. In this case $x= \frac{1}{2}$ would represent the hand at 6 o'clock and in general $x$ represents the hand being at 12x o'clock.

Now we can take the slowest hand, the hour hand and call its position $x$. We know that all three hands coincide at 12 exactly (when $x=0$) and after this the minute hand moves 12 times as fast as the hour hand since one rotation of the hour hand corresponds to 12 hours passing which is 12 rotations of the minute hand. Likewise the seconds hand moves 60 times the speed the minute hand or $60*12$ times faster than the hour hand.

From this we deduce that when the hour hand is at position $x$ mod 1 the minute hand is at position $12x$ mod 1 and the second hand is at position $720x$ mod 1.

Now we can calculate the answers you want by equating these mod 1 sequences:

a) for the minute and hour hand to coincide then $x=12x$ mod 1 and thus $11x=0$ mod 1 which implies $11x$ is an integer meaning $x=k/11$ for $0 \le k \le 10$ giving 11 solutions. (Note the exact time they coincide will be when the hour hand and minute hands are at $60*k/11$). The hour hand goes around twice in a day however giving 22 solutions.

b) for the minute and second hand $12x = 720x$ mod 1 so $x=60x$ mod 1 implying $59x=0$ mod 1 and as before $x=k/59$ for $0 \le k \le 58$ giving 59 solutions in one hour. Therefore $59*24$ in one day.

c) repeating yet again will give 719 solutions to th sequence part and $719*2$ to the total in a day.

d) Finally for d) we note that 11 and 719 are prime so fractions other than 0 of the form $k/11$ and $m/719$ will never coincide (for these fractions less than one) and therefore both a) and c) cannot hold simultaneously(unless we are at o) which is what we need for all hands to coincide. Therefore all three hands only meet when $x=0$ at 12 o'clock, so twice in a day.

Hope this answers the question in a mathematical way!

Minute and hour hands

Since a standard analog clock only shows 12 hours, it's sufficient to solve this for 12 hours and then multiply by 2 to cover the whole day.

The minute hand rotates at a rate of 1 revolution per hour, or 6°/min. The hour hand rotates 12 times more slowly, or 0.5°/min. Suppose N minutes have passed. The positions of the hands are 6N° for the minute hand and 0.5N° for the hour. Thus, we want to solve when these two values are equal. However, since the minute hand effectively restarts every hour, we want to use some modular arithmetic here. Namely, the minute hand's position is actually 6N%60, where % denotes the modulo operator. Additionally, we are constraining the problem to the first 12 hours, or the first 720 minutes. The final equation is then:

6N%60 = 0.5N, N ∈ [0, 720)


There are 11 distinct solutions to this equation in the selected range, so the hands overlap 11 times in each 12-hour period. Therefore they overlap 22 times in each day.

First we find angular velocities of the hands:

\begin{align} H_{\text{speed}}&=360^\circ/12\text h=\tfrac12^\circ/\text{min}\\ M_{\text{speed}}&=360^\circ/\text h=6^\circ/\text{min}\\ S_{\text{speed}}&=360^\circ/\text{min} \end{align}

If a slower hand covers an angle $x$ and another faster one covers angle $360^\circ+x$ in same time, we get

\begin{align} \text{Time 1}&=\text{Time 2}\\[5pt] \frac{\text{Angle 1}}{\text{Speed 1}}&=\frac{\text{Angle 2}}{\text{Speed 2}}\\ \frac{x}{\text{Speed 1}}&=\frac{360^\circ+x}{\text{Speed 2}}\\ \end{align}

Solving for $x$,

$$x=\frac{360^\circ\times \text{Speed 1}}{\text{Speed 2}-\text{Speed 1}}$$

By substituting velocities of any two hands, we get to know what angle the slower hand covers before every time they meet. We divide this angle by the total angle covered during the day (by the slower hand) to find out how many times they meet.