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Topic: Analog clock III (Read 1620 times) |
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TimMann
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Analog clock III
« on: Aug 27th, 2002, 2:16am » |
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Nice problem. First note that the position of the minute hand is completely determined by the position of the hour hand, so to check whether we have a time that meets the conditions of the problem, we can note where the hour hand is, determine from that where the minute hand must be, move the hour hand to the minute hand's old position, and then check whether the minute hand's new position is the same as the hour hand's old position.
How to write this as an equation? Let's use modulo 1 arithmetic instead of modulo 12 -- that is, we'll write the initial position of the hour hand as a number t, 0 <= t < 1, which means a time >= noon and < midnight. As we move the hour hand from 0 to t, the minute hand moves twelve times as fast, so its initial position is (congruent modulo 1 to) 12t. Next, we move the hour hand ahead to 12t, which takes the minute hand to 144t.
Thus the initial position was a solution if t is congruent to 144t modulo 1. In other words, the part of 144t that follows the decimal point is equal to t. Another way to write this is t = 144t - k, where k is an integer; or simplifying, t = k/143. Since 0 <= t < 1, this equation has exactly 143 solutions -- those with k=0, 1, ..., 142.
The solutions where the hands coincide (that is, t is congruent to 12t modulo 1) are those where k is a multiple of 13: t = 0, t = 13/143 = 1/11, ..., t=130/143 = 10/11.
So yes, there are solutions where the hands don't coincide: there are 11 where they do coincide and 132 where they don't. (Note that the 132 counts each position of the hour hand as a separate solution. Another way to count would be to consider each pair of hand positions where making either hand the hour hand gives a valid time as just one solution; with that way of counting the total solutions would be 11 + 66.)
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| « Last Edit: Aug 31st, 2002, 1:05am by TimMann » |
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http://tim-mann.org/
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Eli
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 I don't understand how you got to the 13 period. Can you
explain a little bit more.
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Stephen Miller
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wouldn't the answers be as follows?
12:00
1:12
2:24
3:36
4:48
6:00
7:12
8:24
9:36
10:48
then back to 12:00
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electrical
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Re: Analog clock III
« Reply #3 on: Sep 30th, 2006, 10:23pm » |
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TimMann, that was masterful. Thanks.
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yevvi
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Re: Analog clock III
« Reply #4 on: Apr 11th, 2015, 2:43am » |
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Here are equations for positions of hour hand (h) and minute hand (m) that satisfy the problem (i divided the clock face into 60 units):
h = 5 * H + m/12
m = 5*M + h/12
where H, M are integers between 0 and 11. The first equation is to have the valid time. The 2nd is how hour and minute hands would move when they are reversed. The solution to these is:
h = (60/143)(12H + M)
m = (60/143)(12M + H)
where H, M range between 0 and 11, except we exclude case when H = M = 11 to keep h and m less than 60.
The case of H = M corresponds to when the hands point in the same direction.
So there are 143 positions total, 11 when the hands point in the same direction and 132 when they don't.
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