# Ambiguous Clock

The hands of my alarm clock are indistinguishable. How many times throughout the day their positioning is such that one cannot figure out which is the hour hand and which is the minute hand?

*Remark: AM-PM is not important.*

**SOLUTION**

Imagine that you have a third hand which moves 12 times as fast as the minute hand. Then, at any time, if the hour hand moves to the location of the minute hand, the minute hand will move to the location of the imaginary hand. Therefore, our task is to find the number of times during the day when the hour hand and the imaginary hand are on top of each other, and the minute hand is not.

Since the imaginary hand moves 144 times faster than the hour hand, the two hands are on top of each other exactly 143 times between 12AM and 12PM. Out of these 143 times, 11 times all three arrows are on top of each other. Therefore, we have 2 × (143 – 11) = 264 times when we cannot figure out the exact time during the entire 24-hour cycle.

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2x(143-11)=264 times in a 24 hour period