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Suppose you have a clock that is set at the twelve o' clock position. Given one rotation how many times will both the minute and hour hand coincide?

The answer is pretty simple, it's every 12/11 hours, which makes sense because the hour hand moves at 30 degrees/hr while the minute hand moves at 360 degrees/hr so 360/330 = 12/11

One can also determine the answer by taking the sum $\sum_{x=0}^{\infty}(1/12)^x$, which also gives 12/11. Why did we take the geometric series of 1/12?

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    $\begingroup$ Title question: 11 times, second question: google geometric series, third question: The hour arm travels at 1/12 the speed of the minute arm $\endgroup$ – imranfat Aug 20 '17 at 0:43
  • $\begingroup$ I get that the hour arm travels at 1/12 the speed but how does this relate to the geometric series? Why would we take the geometric series of 1/12? I understand how we got the 12/11 by taking the geometric series but I don't understand why we took 1/12^x from 0 -> Infinity $\endgroup$ – John Aug 20 '17 at 0:47
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    $\begingroup$ Think about the tortoise and the hare problem (Zeno's paradox). When the arms coincide at noon, after 60 minutes it is 1 o clock. When the minute arm moves to the 5, the hours arm has traveled 1/12 of the distance. When the minute arm moves to the position of the hour arm, the hour arm again has moved 1/12 of the distance, etc.That way you can find that from noon to the next "eclipse" (sorry, no pun intended) the time is 1hr, 5 minutes, 27 and 3/11 seconds $\endgroup$ – imranfat Aug 20 '17 at 0:50
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    $\begingroup$ @Idéophage Meeting once on every 12/11 hour is at a rate of 11/12 times an hour. The answer is that in 12 hours they meet 11 times. $\endgroup$ – Graham Kemp Aug 20 '17 at 0:57
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    $\begingroup$ @imranfat You should turn your comment into an answer. $\endgroup$ – amd Aug 20 '17 at 1:05
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From an infinite series point of view, as indicated in my comment, the minute arm travels at 1/12 the speed of the hour arm, so $\sum=\frac{1}{1-\frac{1}{12}}$=$\frac{12}{11}$hour. So this is 1 hour and 1/11 of an hour. You can figure out that this corresponds 5 minutes, 27 and 3/11 seconds. But there is another way to calculate this (hence my post), without infinite series and that is just using a basic linear equation, as time continues linearly. If it is 1 o clock, the hour arm is 30 degrees ahead of the minute arm. Let $t$ be the time in seconds. The amount of degrees the minute arm covers in 1 second is $t/10$ and for the hour arm $t/120$. From here you can set up the equation $\frac{t}{10}=\frac{t}{120}+30$. Solving this equation results in $\frac{11t}{120}=30$ or $t=3600/11$ from which the 5 minutes, 27 +3/11 seconds follow. Using this principle over and over again, you can see on the arms eclipse 11 (not 12) times on a 12 hour cycle.

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