Permalink Submitted by Anonymous on August 20, 2010

Although correct, I think the above proof doesn't give a good feel for what is going on. After all, this is just a clock -- a good solution ought to be obvious! A more intuitive approach came to me as I was considering the time 12:30. This time is unambiguous because if you treat the minute hand as an hour hand, its associated minute hand (which would point to 12) doesn't coincide with the hour hand. But, if you advance the time by one minute, the hour hand will move 1/12 minute, while the imaginary minute hand will move 12 minutes! It will overtake the hour hand, so at some point must coincide with it. From that key idea, the full proof was easy...

The hour and minute hands coincide at noon. The hour hand revolves once and the minute hand 12 times between noon and midnight; so, allowing for end cases, in that time the hour and minute hands coincide 12 - 1 = 11 times, including noon but excluding midnight.

Consider the minute hand as an hour hand, with an associated "imaginary" minute hand which we will refer to as the "third hand". The third hand revolves 12 times faster than its hour hand (the minute hand) and 12 * 12 = 144 times faster than the hour hand.

If the third hand does not coincide with the hour hand, then the time represented by the hour and minute hands is unambiguous. On the other hand, if the third hand does coincide with the hour hand, the time represented by the hour and minute hands cannot be distinguished from that represented by the minute and third hands -- unless the hour and minute hands coincide, in which case both times are the same and therefore unambiguous.

All three hands coincide at noon. The hour hand revolves once and the third hand 144 times between noon and midnight; so, allowing for end cases, in that time the hour and third hands coincide 144 - 1 = 143 times, including noon but excluding midnight.

As noted above, we must exclude the cases where the hour and minute hands coincide. Hence there are 143 - 11 = 132 ambiguous times on the clock, from noon until just before midnight, occurring in 66 pairs. (If we consider the period from noon to midnight inclusive, noon and midnight constitute an extra pair of ambiguous times.)

## A more intuitive approach

Although correct, I think the above proof doesn't give a good feel for what is going on. After all, this is just a clock -- a good solution ought to be obvious! A more intuitive approach came to me as I was considering the time 12:30. This time is unambiguous because if you treat the minute hand as an hour hand, its associated minute hand (which would point to 12) doesn't coincide with the hour hand. But, if you advance the time by one minute, the hour hand will move 1/12 minute, while the imaginary minute hand will move 12 minutes! It will overtake the hour hand, so at some point must coincide with it. From that key idea, the full proof was easy...

The hour and minute hands coincide at noon. The hour hand revolves once and the minute hand 12 times between noon and midnight; so, allowing for end cases, in that time the hour and minute hands coincide 12 - 1 = 11 times, including noon but excluding midnight.

Consider the minute hand as an hour hand, with an associated "imaginary" minute hand which we will refer to as the "third hand". The third hand revolves 12 times faster than its hour hand (the minute hand) and 12 * 12 = 144 times faster than the hour hand.

If the third hand does not coincide with the hour hand, then the time represented by the hour and minute hands is unambiguous. On the other hand, if the third hand does coincide with the hour hand, the time represented by the hour and minute hands cannot be distinguished from that represented by the minute and third hands -- unless the hour and minute hands coincide, in which case both times are the same and therefore unambiguous.

All three hands coincide at noon. The hour hand revolves once and the third hand 144 times between noon and midnight; so, allowing for end cases, in that time the hour and third hands coincide 144 - 1 = 143 times, including noon but excluding midnight.

As noted above, we must exclude the cases where the hour and minute hands coincide. Hence there are 143 - 11 = 132 ambiguous times on the clock, from noon until just before midnight, occurring in 66 pairs. (If we consider the period from noon to midnight

inclusive, noon and midnight constitute an extra pair of ambiguous times.)