Let us assume we have a perfect analog watch. That is, one whose hands move continuously at the same rate. So at 3:00, the big hand is pointing to 12, and the little hand points to 3. At 3:15, the big hand is a quarter of the way around, pointing to 3, and the little hand will be pointing one quarter of the way between 3 and 4.

Let x be the hours after 12 (so x = 3 at 3:15) and let y be the minutes after the hour (so y = 15 at 3:15). The amount around the minute hand will be around the clock is y/60. (The minute hand will be 15/60 = 1/4 around the clock at 3:15.)

The amount around the clock the hour hand will be is (x + y/60)/12. So at 3:15, which is 3.25 hours after 12:00, the hour hand will be 3.25/12 of the way around the clock. We need to solve for when (x + y/60)/12 = y/60, for integer values of x. Multiplying both sides by 60 gives us:

5(x + y/60) = y , so 5x + y/12 = y, or 5x = 11y/12 and finally, y = 60x/11 . The solutions are

x = 0, y = 0 (12:00)

x = 1, y = 60/11 ~ 5.4545 (1:05 and 27.27 seconds)

x = 2, y = 120/11 ~ 10.9090 (2:10 and 54.545 seconds)

x = 3, y = 180/11 ~ 16.3636 (3:16 and 21.818 seconds)

And so on through x = 10, y ~ 54.5454 (10:54 and 32.727 seconds). At x = 11, we are at 12:00 again.