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I was thinking about this expression, and I wondered if it holds true when the clock is slow. I can imagine a slow clock which is not right at all in the span of 12 hours—imagine a clock which ticks 5 minutes every 12 hours, which points to 11:59 at exactly 12:00. 12 hours later it is 12:00 and the clock is pointing to 12:04. The clock won't be right for another 5 minutes or so, and will have gone more than 12 hours without being right.

Is there an upper bound on the amount of time that a clock which moves at a slow but constant rate will spend being wrong before it is right? What would be a good way to model this? How should I tag this question?

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Think about what happens when the clock is just barely slow... –  Micah Mar 30 '13 at 20:17
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If I remember correctly, this question goes back to Lewis Carroll at the very least (or even earlier: the problem might not have been originated by him). He suggested that a broken clock was right twice a day whereas one that ran a minute slow each day would be right once every two years or so. See, for example, here for what Lewis Carroll wrote about the problem. –  Dilip Sarwate Mar 30 '13 at 20:40
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"Even a stopped clock gives the right time twice a day": Ride (once my favourite band), Cool your boots. Not their best song, though. That's much better. –  1015 Mar 30 '13 at 20:50

3 Answers 3

up vote 5 down vote accepted

If the bad clock runs wrong by a factor $k\in\mathbb{R}_+$, for example $k=2$ means that it runs at double speed, then the first time the bad clock is correct is $(12\ \mbox{hours})/|1 - k|$ after the start (where the two clocks agree). By choosing $k$ very close to $1$, you can make that time span as long as you desire (so no, there's not an upper bound).

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Let's let the 'correct' time be modeled by $f=x\pmod {12}$. If we have another clock moving at a constant rate, then that's going to look like $f=\alpha\cdot x+\beta\pmod{12}$, for some $0<\alpha<1$ and $0\leq\beta<12$. $\alpha$ represents the 'slowed' tick rate and $\beta$ the time offset. Is this a good enough model? You can more easily solve $x\equiv\alpha\cdot x+\beta\pmod{12}$.

Edit: or see that this equation has no solution, as André points out.

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André retracted his answer. If we're modeling this with real number $x$, $\alpha$, and so on, there will always be a solution eventually (when $\alpha\ne 1$). Sooner or later, the slow clock will be "overtaken" (again) by the faster clock. –  Jeppe Stig Nielsen Mar 30 '13 at 21:20

Use $12$ hours as time unit, and denote real time by $t$. The time shown by the clock is $$f(t):=\lambda t+ c\ ,$$ where $\lambda>0$ is a constant. When the clock is gradually getting behind this $\lambda$ is a trifle smaller than $1$. The clock shows the correct time whenever $f(t)-t\in{\mathbb Z}$, i.e., $$(\lambda-1) t+c\in{\mathbb Z}\ .$$ The time interval $\Delta t$ between two such incidences is $$\Delta t={1\over |\lambda -1|}\ .$$ An example: When the clock is $1$ minute per day behind then $|\lambda-1|={1\over 24\>\cdot\> 60}$. It follows that $\Delta t=1440$, which corresponds to $720$ days.

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