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Logic clockpuzzle proof 
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#1
Apr909, 05:07 AM

P: 4

Hi. So a classic logic puzzle goes like this:
At noon the hour, minute, and second hands coincide. In about one hour and five minutes the minute and hour hands will coincide again. What is the exact time (to the millisecond) when this occurs. (Assume that the clock hands move continuously.) If you like to solve this yourself I wrote the solution in a spoiler wrap, anything written under the spoiler wrap should also be considered a spoiler if you want to solve this:
Spoiler
The hour and minute hands coincide 11 times each 12 hour. There are 3600*12=43200 seconds in 12 hours. Since the hour hand and the minute hand moves with constant speed relatively to eachother, the time interval between each time they coincide is always the same. This time interval is 43200/11= 3927.273 seconds. So so since there is 3600 seconds per hour, and 5 minutes is 300 seconds, the clock is 13:05:27.273
Furthermore I would like to prove that the hour, minute and second hands only coincide two times per day, at noon and midnight: First off all, we already know that there are only 11 more possibilities where this can happen, in timesteps of 43200/11 seconds after noon/midnight. Therefore its sufficient to show that the times when the minute hand and second hand coincide does not fall in any of these timepoints. The minute and second hand coincide 59 times per hour. In timesteps of 3600/59 seconds after each full hour. If X*(43200/11)/(3600/59), where X [tex]\in[/tex] {1,10} does not equal a natural number, it proves that it only happens at noon and midnight. Using a spreadsheet I obtain: X [tex]\in[/tex] {1,10} , A=43200/11, B=3600/59 X  X*A  X*A/B  1  3927.273  64.364 2  7854.546  128.727 3  11781.819  193.091 4  15709.092  257.455 5  19636.365  321.818 6  23563.638  386.182 7  27490.911  450.546 8  31418.184  514.910 9  35345.457  579.273 10  39272.730  643.637 (11  43200  708 ) Can this be classified as a proof? Is it enough to show: (10*43200/11)/(3600/59) [tex]\neq[/tex] natural number , to prove this? Anyone have a more sophisticated/elegant way of proving it? That is if (10*43200/11)/(3600/59) [tex]\neq[/tex] natural number ,is not a proof, because if it is, I can't imagine any way to prove it more elegantly :) 


#2
Apr1009, 03:45 PM

P: 52

Hi kitz2.
I don't know if this is more elegant (since it's almost identical to your's) or even a proof, but you have that the time for n meetings of the second and minute hands since they last met is t_{1}=3600n/59 where n is a natural number, and the time for m meetings of the second and hour hands since they last met is t_{2}=43200m/719 where m is also a natural number. Assuming they meet at some time t=t_{0}, The three hands will subsequently meet whenever tt_{0}=t_{1}=t_{2}, so n=708m/719, and since 719 is prime and m is a natural number, n can only be a natural number if m=719k where again k is a natural number. Plugging that m in you get that tt_{0}=t_{2}=43200k seconds i.e they meet once every 12 hours. 


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