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?
Best Answer
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.