The hour and minute hands of a clock are identical. How many moments are there in a day when it is not possible to tell from this clock what time it is?

You can off course do this problem analytically. But there is a very elegant solution that requires almost no usage of pen and paper. Before proceeding further, see if you can find such a solution.

First of all we note that we are not required to tell whether the time is AM or PM – that is not possible even with an ordinary clock. So we assume that we just have to tell the hour and the minute, without deciding whether its AM or PM.

Now, lets try to make more sense of the problem. What do we mean when we say we can’t tell the time? Visualize a bit. We can’t tell the time for exactly those positions of the hands, for which the positions are interchangeable. Sounds confusing?

Lets say one of the hands is exactly at 12, while the other is exactly at 6. Is there any problem in telling the time? No. We know that it is exactly 6:00, even though we can’t differentiate between the hands. What about one hand being somewhere between 4 and 5, and the other being exactly at 2? Obviously, the hand exactly at 2 cannot be the hour hand, since then the other hand should have been at 12.

So, we see that for certain positions, it can be easily determined which hand is the minute hand and which hand the hour hand. What are these positions? These are positions for which the hour and minute hands cannot take the place of each other. If one hand is at 12, and one at 6, we know that it is 6:00; there is no ambiguity, because the  hour hand could not have been at 12 with the minute hand being at 6. That is invalid. The only possibility is that the hour hand is the one at 6, while the other hand is the minute hand.

But, if you visualize carefully, there will be certain times for which the hour and minute hands will be interchangeable. If the hour hand is at position X, and the minute hand at Y, then there will come a time when the hour hand is at Y while the minute hand is at X. In such a case, we won’t be able to say whether the time is X:Y or Y:X. We have to precisely count these moments, the positions for which the hour and minute hands are interchangeable.

Having determined what we have to count, we start with the solution. Note that the minute hand moves 12 times as fast as the hour hand. If the hour hand moves from 12 to 1, the minute hand will have moved a complete circle.

Now, we add an imaginary third hand to the clock, one which moves 12 times as fast as the minute hand. Suppose now that it is 12:00, and all the three hands coincide. After some time,, let the hour hand and the third hand coincide. Call this position X, and let Y be the position of the minute hand at this instant. So it is X:Y. Note that at this instant, say T, the minute hand has traveled 12 times as far as the the hour hand, while the third hand has traveled 12 times as far as the minute hand.

Sometimes later, when the minute hand has traveled 12 times as far as it is at T, it will reach the position of the third hand, and also the hour hand, at time T, whereas, the hour hand will have traveled 12 times as far as it is at T, and it will reach the position of the minute hand. Therefore, the time at that instant is Y:X, which is ambiguous.

We see that all time instants where the hour hand and the fast hand coincide are ambiguous, so we just have to count such instants. The hour hand goes around 2 times a day, while the fast hand goes around 12 x 12 x 2 = 288 times a day. Therefore, these two hands are coincident 286 times. Out of these, there are some instants when the hour and the minute hands, and thus all three hands, coincide. There are precisely 22 such instants, because, the hour hand goes around 2 times a day, while the minute hand goes around 24 times a day. But these moments are not ambiguous, because even if the hour and the minute hands coincide, we can still tell the time.

Therefore, there are 286 – 22 = 264 such ambiguous moments.