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?
Answer
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.
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