Clock Part II
In the solution for "clock": "if you just think about it, intuitively you'll see the minute hand passes the hour hand 11 times every 12 hours"
Uh, no. The hour hand goes around once in twelve hours. The minute hand goes around twelve times in twelve hours. So once the minute hand pulls away from the hour hand after 12:00, the difference in speed is 11 times. So the minute hand will catch up every time one-eleventh of 12 hours has elapsed. So the hands synch at 12:00, 1:05:27, 2:10:54, 3:16:22, 4:21:49, 5:27:16, 6:32:44, 7:38:11, 8:43:38, 9:49:06, and 10:54:33.
What do you mean the "difference in speed"? This should be a ratio, not a difference. The minute goes around the clock 12 times for every 1 time the hour hand goes around the clock. Therefore, the minute hand should pass the hour hand one twelfth of the time that it rotates around the clock face.
Okay, nevermind ... I see your point. My mistake was in assuming a ration since that would imply that the minute hand passed the hour hand at each hour.
Moreover, you solution indicates that there is no overlap during the 12th hour, which cannot be true as there is atleast one overlap at each hour.
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