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If the hour and minute hand start at 12, how long until they both point to 12 again?

With a normal clock, I know this would take 12 hours.

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    $\begingroup$ Surely you must have tried something ? Tell us ! $\endgroup$ – true blue anil Dec 1 '15 at 11:13
  • $\begingroup$ Lowest common multiple would help here. $\endgroup$ – Element118 Dec 1 '15 at 11:14
  • $\begingroup$ Hint: What is the position of the minute hand after 12 hours? $\endgroup$ – Michael Burr Dec 1 '15 at 11:28
  • $\begingroup$ @MichaelBurr That would help if it ticked 61 minutes in an hour, but instead it ticks an hour in 59 minutes. $\endgroup$ – Ian Dec 1 '15 at 11:45
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  • Since the minute hand takes 59 minutes to complete a cycle, the minute hand will point at 12 after multiples of 59 minutes: $59,\ 118,\ 177$, etc.

  • Since the hour hand takes 12 hours, which is 720 minutes, to complete a cycle, the hour hand will point at 12 after multiples of 720 minutes: $720,\ 1440,\ 2160$, etc.

Since $59$ is a prime number and $720$ is not divisble by $59$, the least common multiple of $59$ and $720$ is $59\cdot720=42480$.

In other words, $42480$ is the first number that is present in both series $$59,\ 118,\ 177,\ \ldots$$ $$720,\ 1440,\ 2160,\ \ldots$$ And it thus takes $\frac{42480}{60}=\textbf{708}$ hours before they simultaneously point at 12 again.

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  • $\begingroup$ It is striking how different this is from the minute hand sweeping out 61 minutes per hour... $\endgroup$ – Ian Dec 1 '15 at 11:44
  • $\begingroup$ How is that? Again, $61$ is prime and $720$ is not divisible by $61$, so $\mathrm{LCM}(61,720)=61\cdot720$. $\endgroup$ – Eric S. Dec 1 '15 at 12:18
  • $\begingroup$ At 61 minutes per hour you get done in 60 hours, since you get ahead by 1 minute each hour. What you just wrote is wrong because you are at 12 not every 61 minutes but rather every 3600/61 minutes. The actual numbers being so different is what struck me. $\endgroup$ – Ian Dec 1 '15 at 12:44
  • $\begingroup$ Ah okay, I understand. I thought you meant the minute hand takes $61$ minutes to complete a cycle. But if it sweeps out $61$ minutes per hour, the it takes $\frac{60\cdot60}{61}$ minutes to complete a cycle. $\endgroup$ – Eric S. Dec 1 '15 at 12:51
  • $\begingroup$ I was typing the same thing :) now that I get your point, It also strikes me, interesting! $\endgroup$ – Eric S. Dec 1 '15 at 12:52

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