A clock runs irregularly but after 24 hours it has neither gained nor lost overall. Find a way the clock can run irregularly such that there is no continuous 576-minute period across which the clock shows that 576 minutes have passed. The clock time should not discontinuously jump.

I know that 24 hours = 1440 minutes and $576=\frac{2}{5}1440$

The likely idea is to divide the 24 hours/1440 minutes in a number of intervals and have the clock run faster in odd intervals (say running one hour in 30 minutes), then slower in even intervals (say running one hour in 90 minutes) so that overall it still runs 1440 minutes. I cant figure out which function would work for 576 minutes though.

  • $\begingroup$ You may want to phrase this other than "non-continuous", because if the clock is just dead, and after 24 hours it's considered to have jumped 24 hours forward, then this solves the question as stated. It seems you want the function from actual time to clock time $f(t)$ to be continuous but nonlinear. $\endgroup$
    – obscurans
    Dec 24, 2018 at 22:19
  • $\begingroup$ Yes the function $f(t)$ should be continuous. $\endgroup$
    – user35202
    Dec 24, 2018 at 22:24
  • 4
    $\begingroup$ A related paper. $\endgroup$
    – Micah
    Dec 24, 2018 at 23:11
  • $\begingroup$ Can this clock run backwards? $\endgroup$ Dec 25, 2018 at 4:46

1 Answer 1


Presuming you want to find a function from real time to clock time $f:\left[0,1440\right]\rightarrow\mathbb{R}$ such that

  • $f(0)=0$
  • $f(1440)=1440$
  • For all $t\in\left[0,1440-576\right]$ we have $f(t+576)-f(t)\neq576$
  • $f$ is continuous

If $f$ is allowed to be discontinuous as in the problem statement, then $$f(t)=\begin{cases}0&t<1440\\1440&t=1440\end{cases}$$ works. Hilariously, this is a dead clock, which is right once a day.

The first part of your solution idea is sound: make the clock run fast at times and slow at other times. The difficult part is to force the average to be nowhere correct.

Suppose we construct $f$ such that $f(576)-f(0)=f(576)\neq576$. Then one way to ensure sliding the window forwards cannot result in a time difference of exactly $576$ is to not allow the difference to change at all.

So, suppose we next construct $f(t+576)=f(t)+f(576)$ for some range of $t$. Nothing stops us from literally requiring $f(t+576)=f(t)+f(576)$ for all $0\leq t\leq1440-576$.

The last part is to fit to the requirement that $f(1440)=1440$. We do this by splitting the clock gain within the 576 minutes into two periods of movement: one from $0$ to $\epsilon$, and the other from $1440\mod576-\epsilon$ to $1440\mod576=288$.

This particular solution is as follows:

  • The clock runs extremely fast from $0$ to $\epsilon$, clocking 4 hours
  • The clock stays dead from $\epsilon$ until $288-\epsilon$
  • The clock runs extremely fast from $288-\epsilon$ to $288$, clocking 4 more hours (now 8)
  • The clock stays dead from $288$ to $576$
  • This cycle repeats per $576$ minutes
  • Note the third (partial) cycle ends with clock running extremely fast from $576\times2+288-\epsilon=1440-\epsilon$ to $1440$, clocking 4 more hours (now $6\times4=24$ hours as desired).

By construction, $f(t+576)-f(t)=f(576)=480\neq576$ for all $t$.

  • 4
    $\begingroup$ Just for illustrative purposes, here's a graph of $f(t)$ for $\epsilon=96$. $\endgroup$ Dec 25, 2018 at 2:05

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