How to understand the analog clock problem?

Question

I know the answer to this riddle:

How often does the minute hand pass the hour hand on an analog clock?

I know they pass 11 times every 12 hours, but I don't understand how to get to it, no matter how many resources I refer to.

Does anyone have a non-confusing explanation of the solution in simple English?

Thank you.

P.S.

I'm a programmer, so I asked how to implement this in code on Stack Overflow, but they closed my question and referred me to this forum. They said I have to understand the problem before trying to implement it. So someone please help me :)

P.S.P.S.

The real problem I'm having is: what formulas can I use to get that result? Say for example the "x" hand that moves at a certain speed, and a "y" hand that moves at a slower speed. How do I get that constant from the information given?

Answer 1

Start at $12$ at midnight. It depends whether you want to count this as a 'meeting' or not but I'll rather not count this one. Now start the clock.

Since the minute hand is faster, the hour hand remains behind and after one hour when the minute hand is back at $12$ (and the hour hand is at $1$) they never would have met. It is easy to see that the minute hand will always meet the hour hand after this, every hour until $11$ a.m. which brings the count to $10$. After an hour, the minute and the hour hand both simultaneously strike $12$, so you should count this as a 'meeting' which gives us a total of $11$.

Note that if you are asking about the number of passings and not meetings, then I would not count the meeting at $12$ p.m. as a passing, since they never passed each other at that moment. And so the total in this case should be $10$.

Also, give a look here.

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As for finding the constant time between every meeting:

The speed of the hour hand is: $\dfrac{360^\circ}{12\text{ hours}}=0.5^\circ\text{min}^{-1}$.

And that of the minute hand: $\dfrac{360^\circ}{60\text{ min}}=6^\circ\text{min}^{-1}$.

Now change your view point to the hour hand. The relative speed of the minute hand with respect to the hour hand is $(6-0.5)^\circ \text{ min}^{-1}=5.5^\circ\text{ min}^{-1}.$ This is the speed of the minute hand as seen from the hour hand. So you can fix the hour hand to be stationary, and just think of the minute hand to be moving at $5.5^\circ\text{ min}^{-1}$. So the hands meet each other at every, $\dfrac{360^\circ}{5.5^\circ\text{ min}^{-1}} = 65.\overline{45} \text{ min}$.

Answer 2

You're exactly right, the answer is $11$ times every $12$ hours, or $1$ time every $\frac{12}{11}$ of an hour, which works out to about once every $1$ hour, $5$ minutes, and $27.27$ seconds. This isn't an average either; the amount of time that passes between each meeting of the two hands is constant.

Just think about the time 11:59, and count through $12$ hours. You'll notice that there is a meeting for each hour $12,1,2,3,\ldots,10$, but no meeting for the $11$th hour because you stop at 11:59. This makes $11$ meetings in $12$ hours.

It's not too hard to see why the time passing between two meetings is constant. Just think about the rotational symmetry.

Another (and more difficult way) to look at the problem is to calculate the angular speed of each hand. Using convenient units, we have the minute hand traveling at a rate of $60$ units per hour, and the hour hand travels at a rate of $5$ units per hour. Thus, we would like to solve the following equation:

$$60t=5t\operatorname{mod}60$$

$0$ clearly solves the equation, so we look for the smallest positive value of $t$ solving it. Thus we want to solve:

$$55t=60$$

which gives us $t=\frac{12}{11}$ hours as before.

Answer 3

Suppose the faster-moving hand moves clockwise at a rate of $f$ revolutions per hour, and the slower-moving hand moves clockwise at a rate of $s$ revolutions per hour.

Mount the clock on a motor attached to the wall, where the motor turns the entire clock counterclockwise at $s$ revolutions per hour, in order to keep the slower-moving hand always pointing towards the ceiling. The faster-moving hand will then move clockwise (relative to the wall) at $f-s$ revolutions per hour. If it begins in the up position, it will complete one revolution relative to the wall (returning to the up position, where it meets the slow hand) every $\frac{1}{f-s}$ hours.

So for any two speeds in revolutions per hour, $f$ and $s$, where $f>s$, the hands meet every $\frac{1}{f-s}$ hours. For a regular clock, $f=1$ and $s=\frac{1}{12}$, so the hands meet every $\frac{1}{1-\frac{1}{12}}$ hours, which is every $\frac{12}{11}$ hours.