Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Think about what happens when the clock is just barely slow... – Micah Mar 30 '13 at 20:17
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
"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
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).

share|cite|improve this answer

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.