Clock confusion

We all know that homemade presents are supposed to be the best, but the clock your Aunty Mabel made you is a little hard to get used to — the hour and minute hands are exactly the same! You can muddle through most of the time but sometimes, say 26 minutes past 2 or just after 12 minutes past 5, you can't tell which of the two times it is. In the 12-hour period between noon and midnight, how many moments are there when it is not possible to tell the time on this clock?

Solution

To solve this puzzle we need to figure out exactly which of the possible positions of the hands of the clock are indistinguishable. That is, find all those times such that if you swapped the minute and the hour hands over you also get a valid time.

Let's call the angle the hour hand makes with a line through 12, $\alpha$, and the angle the minute hand makes with the line through 12, $\beta$. There is an angle of $30^{\circ}$ between each number on the clock face. We are actually interested how far past the last number on the clock face each hand has gone, so we can write $$\alpha = \alpha_0 + 30k$$ and $$\beta = \beta_0 + 30l$$ for some integers $k$ and $l$ where $0\leq k,l \leq 11$. Then $\alpha_0$ and $\beta_0$ are the amount that each hand has moved past the last number on the clockface, and $k$ and $l$ give the last number each hand has passed.

To make sure that swapping the hour and minute hands will give a valid time, we are interested in those times when the proportion of the hour the minute hand has moved through ($\beta / 360$) is the same as the proportion of the gap between two adjacent numbers the hour hand has moved through ($\alpha_0 / 30$). Then this time is indistinguishable from the time with the hour and minute hands swapped over. Therefore we can't tell the time whenever \begin{equation} \beta / 360 = \alpha_0 / 30 = \alpha / 30 - k \end{equation} or \begin{equation} \alpha / 360 = \beta_0 / 30 = \beta / 30 - l. \end{equation}We can rearrange equation (1) to get: \begin{equation} \beta = 12 \alpha - 360k. \end{equation} And we can substitute this into equation (2) to get: \begin{eqnarray*} \alpha /360 & = (12 \alpha - 360k)/30 - l \\ \alpha & = 12(12 \alpha - 360k) - 360l \\ \alpha & = 360(12k + l)/143. \end{eqnarray*} And we can substitute this back into (3) to get: \begin{eqnarray*} \beta & = 12(360 (12k+l))/143\\ & = 360(k+12l)/143. \end{eqnarray*} If $k=l$, then the hands lie one top of each other, and swapping the hands gives the same time, so we can still tell the time. However for each of the other $12 \times 11 = 132$ possible pairs of $k$ and $l$, the time is indistinguishable as by swapping over the hour and minute hands would give a different time. Therefore in the 12 hours from noon we can't tell the time on 132 occasions.

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