Your buddy Frankie sold you a shoddy clock: it keeps good time, but the minute and hour hands look exactly the same! Both of these hands move continuously, and there is no second hand.

How many times a day is it impossible to tell what time it is?

  • $\begingroup$ Surely it's only possible to tell what time it is when the two hands are on top of each other? $\endgroup$ Commented Jun 26, 2015 at 21:56
  • $\begingroup$ When I saw this I though "I know this puzzle!" then realized I didn't. This isn't re-wording of "How many times do the hands cross?" is it!? $\endgroup$
    – Bob
    Commented Jun 26, 2015 at 21:57
  • $\begingroup$ How Many Clock Hand Positions Swap to a Valid Position? Maybe you know it from there? $\endgroup$
    – A. P.
    Commented Jan 7, 2018 at 22:17

4 Answers 4


When the hour hand has moved $x$ degrees around the clock from the top, the minute hand has moved $y = 12x$ degrees. If the time is still a valid configuration when the hands are switched around then $x = 12y$ as well.

Therefore, we want the values of $x, y$ that satisfy the following two equations:

\begin{align} 12x &\equiv y \pmod {360} \\ 12y &\equiv x \pmod {360} \end{align}

Conveniently, this reduces to $x \equiv 144x \pmod {360}$ or $143x \equiv 0 \pmod {360}$, so whenever the hour hand moves exactly $x/143$ of the way around the clock where $x$ is an integer, it's impossible to tell which hand is which.

There's just one problem, though. The above doesn't take into account the times when the hour and minute hands are in the exact same position, in which case it doesn't matter which hand is which. This occurs whenever $x \equiv 12x \pmod {360}$, or $11x \equiv 0 \pmod {360}$. Naturally this is a total of 11 times, so there are $143 - 11 = 132$ times when the time is actually ambiguous in a 12-hour period, making it $264$ times a day.

  • $\begingroup$ Can you work out what those times actually are? $\endgroup$
    – Bob
    Commented Jun 26, 2015 at 22:17
  • $\begingroup$ @Bob, Starting from noon, every 5 minutes and 2.0979 seconds or so, except for those times when the hands overlap. $\endgroup$
    – user88
    Commented Jun 26, 2015 at 22:18
  • 1
    $\begingroup$ I think I understand the calculations. But even I know whether it is midday or early evening (most of the time :D). How many time a day do think the ambiguity would actual matter to working out what the correct time must be? $\endgroup$
    – Bob
    Commented Jun 26, 2015 at 22:28
  • 9
    $\begingroup$ Or trim the hand that moves slowest. LOL $\endgroup$
    – Bob
    Commented Jun 26, 2015 at 22:30
  • 2
    $\begingroup$ This all presumes our measurement methods are accurate enough, of course. $\endgroup$ Commented Jun 26, 2015 at 23:07

The answer is:

All the time


The clock only has 12 hours, and there are 24 hours in a day. Every orientation has at least 2 possible times.

  • 2
    $\begingroup$ That is true of all clocks and obviously not the spirit of this puzzle. $\endgroup$
    – Trenin
    Commented Feb 29, 2016 at 16:06
  • 3
    $\begingroup$ I know. Just thought it was a fun answer. $\endgroup$
    – Kruga
    Commented Mar 1, 2016 at 8:37
  • $\begingroup$ @Kruga try better for the homour next time $\endgroup$ Commented May 29, 2016 at 4:32
  • 5
    $\begingroup$ Any level of humor is preferable to humorlessness IMO. $\endgroup$
    – BobRodes
    Commented May 29, 2016 at 8:23

If $x$ is the position of the hour hand, $0\le x < 12$, then the position of the minute hand $y$ is $y =f(x) \colon= 12 \{ x\}$ ( $\{\cdot\}$ the fractional part). In base $12$ $x = a_0.a_1 a_2 \ldots _{(12)}$, $f(x) =12\{x\}=a_1.a_2 a_3 \ldots _{(12)}$ (a shift ) One cannot tell the time for a pair of hand positions $x$, $y$ such that $y=f(x)$, and $x=f(y)$, and $x$, $y$ not equal. This means $x = a.bab\ldots_{(12)}$,$y = b.aba\ldots_{(12)}$ with $a$, $b$ nonequal base $12$ digits. So there are $12^2-12$ times in the morning, and double that in a day, that is $2 \cdot 132 = 264$ (we assume the person can tell between morning and afternoon)

Example: the time $12 \cdot \frac{3\cdot 12 + 7}{143}$ can be mistaken for $12\cdot \frac{7\cdot 12 +3}{143}$, that is $3\colon36\colon30$ for $7\colon18\colon02$.


You can actually tell the correct time all the time. Let's say it's 3:30. One hand is on 6 and the other is exactly between 3 and 4. You can tell it's NOT 6:15 because one hand is exactly on 6, and for it to be exactly 6:00 the other hand would have to be on 12 ... which it isn't. This can be similarly surmised around the dial for any time if you look carefully at the hands.


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