We have a regular analog clock with three hands - second, minute and hour. On a given day, how many times do:
(a) the minute and hour hands
(b) the minute and second hands
(c) the hour and second hands
(d) all three hands
meet?
One approach would be to just list out all the times they meet, but I'm looking for a more logical/mathematical approach.
Answer
I think there is a much simpler solution than all provided so far.
Consider the following three facts:
- hour hand will rotate 2 full times in a day
- minute hand will rotate 24 full times in a day
- second hand will rotate 1440 full times in a day
So then:
(a) the minute and hour hands, will meet exactly:
22 times: 24 - 2 (once every 24/22 hours)
(b) the minute and second hands, will meet exactly:
1416 times: 1440 - 24 (once every 24/1416 hours)
(c) the hour and second hands, will meet exactly:
1438 times: 1440 - 2 (once every 24/1438 hours)
(d) all three hands, will meet exactly:
twice: only at exactly 12:00:00 o'clock (noon and midnight)
Simply because:
The faster hand passes the slower hand by the number of laps it makes minus the number of laps the slower hand makes.
With the special case:
One hand lapping another won't necessarily coincide with the third hand being there. Try to find the common multiples of the three fractions 24/22, 24/1416, and 24/1438 and you will see there are only two at 24/1 and 24/2. (ie. 12 hours and 24 hours after the start).
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