When both hands are together and moving CW,

it takes 1 hour 5 minutes + epsilon to meet again, when HH stops, MH continues CW.

One hour later, MH meets again and reverses direction (to CCW), while HH starts again.

55 minutes - epsilon later, they meet again, HH stops, MH continues CCW

one hour later, MH meets again and reverses direction to CW, while HH starts again.

In other words, in exactly 4 hours, the two hands moved exactly one sixth of the circle (that is, they are both pointing exactly at the number 2 more than last time.

So it would take 24 hours for the hands to get together at 12 with both hands moving CW. But only 23 hours for them to get together at 12, because that time, MH is still going CCW, and HH stops.

So, from a standing start--both hands at 12, both moving CW, it takes 23 hours to get back to both hands at 12 , 24 hours to do so with both hands moving CW. I agree with Rob's numbers.

BobbyGo may have envisioned the full cycle, but missed that last backward run.

If we are to tell how long from when the clock has HH past 10, MH at 12, we notice several things:

* HH must be moving--it only stops at a little after the odd hours, and exactly on the even hours; it's moving at other times

* MH must be moving CW--same condition

* Therefore it will be (3 hours - 22 minutes) before the hands are together at 12 and 12 (and that will be the 23 hour situation).