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ccr
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Joined: September 18th, 2002, 2:55 am

Clock with equal length hands

July 11th, 2004, 4:09 pm

A favorite I heard a little while back ...On my clock, the hour hand and minute hand are of the same length. During a 12-hour period, how many times can I look at the clock and not be able to determine what time it is?(A quick caveat, if you quickly answer this question, be careful that you are not actually answering a different but similar brainteaser.)
 
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estcourt
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Clock with equal length hands

July 12th, 2004, 12:24 pm

11 times per hour so 132 times?
 
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ScilabGuru
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Clock with equal length hands

July 12th, 2004, 4:42 pm

I got 12 times per our so 144 times. My solution Being x from [0,1], the hour arrow always shows n+x while minute arrow shows 12x So to have swap of the arrows make sense we have to write that n + x = 12 y 12 x = m + y , where n is integer from 0 till 11, m is integer from 0 till 11 and x, y are real from [0,1] Solving y,m from x,n we have 123 solution for any n from 0 till 11. Indeed,y = (n +x) /12 must be from 0, till 1, is automatically satisified for all n = {0,...12} an x from [0,1]n = 0, -> x =12m /143, m = {0,.....1}n =1 -> x =12m+1 / 143 , m = {0,....1}........n=11, x =12m+11 /143 , m = {0,...1}So we have 12x12= 144
 
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estcourt
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Clock with equal length hands

July 13th, 2004, 7:10 am

I think the 12th time per hour you have is when the hands are overlapping but in this case there is no confusion over the time
 
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estcourt
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Clock with equal length hands

July 13th, 2004, 7:10 am

I think the 12th time per hour you have is when the hands are overlapping but in this case there is no confusion over the time
 
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longshort
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Clock with equal length hands

July 13th, 2004, 7:35 am

The only times you CAN determine what time it is is when the hands are overlapping. This happens once every 60 moves in an hour. So you will NOT be able to determine what time it is 59*12 = 708 times in a 12 hour period.
 
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rollyHeyHey
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Clock with equal length hands

July 13th, 2004, 10:09 am

LS: so if one hand is at 3 and the other is at 12 you couldn't figure out the time?
 
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longshort
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Clock with equal length hands

July 13th, 2004, 11:48 am

"LS: so if one hand is at 3 and the other is at 12 you couldn't figure out the time?"I see your point. You imply the it can only be 3 o'clock because if it was 0.15 one of the two hands would not be lying exactly on 12 but a bit forward?Yeah you are right....
 
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longshort
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Clock with equal length hands

July 13th, 2004, 11:52 am

Last edited by longshort on July 12th, 2004, 10:00 pm, edited 1 time in total.
 
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Aaron
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Clock with equal length hands

July 15th, 2004, 9:10 pm

I did it in a way that is equivalent to ScilabGuru's method. You can pick any two distinct hours in 12x11/2 = 66 ways. Call them H1 and H2. You cannot tell the difference between H1 plus 60*(H1 + 12*H2)/143 minutes and H2 plus 60*(12*H1 + H2)/143 minutes. Each of these 66 ways will occur twice every 12 hours for 132 instants at which you do not know the time.However, the answer is tricker than that, it depends on when you start the 12 hour period. Every 5 minutes and 2 14/143 seconds you get either one of the times or a time when the two hands exactly overlap. If you start the interval at one of these times, the 144th such time will occur exactly 12 hours later. If you start the interval at any other time, there will only be 143 occurances. The missing occurance could be one of the times when you can't tell the time, or an overlap. Therefore, some 12 hour intervals will have only 131 instants at which you do not know the time.
 
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genieman17
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Clock with equal length hands

July 16th, 2004, 5:57 pm

QuoteOriginally posted by: ccrA favorite I heard a little while back ...On my clock, the hour hand and minute hand are of the same length. During a 12-hour period, how many times can I look at the clock and not be able to determine what time it is?(A quick caveat, if you quickly answer this question, be careful that you are not actually answering a different but similar brainteaser.)Ok the answer to this question as it is asked is as many times as you want. There is no minimum or maximum limit to the number of times you may choose to randomly look at your watch within a 12 hour period.Assuming this hypothetical clock moves in a constant manner not like one of those ticking clocks that moves only when the minutes change: The only time you actually know what the time is, is when both hands are exactly (EXACTLY) at the same point. The number of times this happens is irrelevant because the span of time this occurs is infinitely small, since it happens for only a moment in time. Therefore, the rest of the time they are not exactly in the same place you cannot exactly know what time it is. Therefore for a total of 12 hours minus 12 (or 13?) moments in time, which is still equal to 12 hours, is the amount of time during which you do not know what time it actually is.
 
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Aaron
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Clock with equal length hands

July 16th, 2004, 7:30 pm

We went through this once before on the thread. The hands on a clock are related in that the distance of the minute hand from 12 is equal to 12 times the distance of the hour hand from the number behind it. Most of the time when you look at the clock you will be able to tell which hand is which because this relation will be satisfied only for one assignment. For example, if one hand is exactly on 12 and the other exactly on 6, it must be 6:00, not 12:30, because at 12:30 the hour hand is halfway between 12 and 1.If the hands move discretely with 60 positions around the clock face then you get intervals of uncertainty rather than instants. In your literal reading of the question that would allow you to look an infinite number of times at the same hand position and not know what time it was. But you'd still have 132 (or 131) ambiguous hand positions.
Last edited by Aaron on July 15th, 2004, 10:00 pm, edited 1 time in total.
 
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ccr
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Clock with equal length hands

July 17th, 2004, 4:33 pm

At the risk of answering my own question incorrectly...I don't see how you can have only 131 times.Consider if you start the clock at 12:03 (a "normal" non-overlap, non-special time). I think it is agreed that we get 132 times when we can't determine the time. Now start the clock at ~12:05:02+, (where the time looks the same as ~1:00:25+). To play devil's advocate, "Which time in the interval starting at 12:03, do we lose when we start the clock ahead about two minutes?"If 131 is correct, we have an odd number of times, which implies that we need three times to look the same. (We can't have just one time look the same, else we would know time it is.) The only way three times can look the same is if we have two of the three be the same, mod am/pm, which can only happen if we start at a special time, and consider the interval [0,12]. But I think this adds an instance, not subtracts one. I guess it would be possible to be 131, if: -You consider a 12-hour interval as (0,12)-We "start" at a special (but non-overlap) time-You are unaware what time we actually started.This way when it is ~1:00+, you wouldn't be able to rule out that it wasn't ~12:05+, even though both of those times are excluded from our interval. Sorry, I should have made the original question more precise. It was not intened to lead into "sweeping vs. ticking hands" and "does the interval contain the endpoints" debates.
 
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Aaron
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Clock with equal length hands

July 17th, 2004, 6:37 pm

I was wrong, but you are too I think. This turned out to be more entertaining (to me) than the original question.Define "overlap" times as when the hands exactly overlap, "uncertain" times as times you do not know the time and "special" time as either overlap or uncertain.Special times happen once every 5 minutes 2 14/143 seconds. Each 13th special time is an overlap time. 12 hours divided by 5 minutes 2 14/143 seconds is 143, so the only way to get 144 in a 12 hour interval is to have one on both endpoints. So I was correct up to this point, you will have either 143 or 144 special times in any 12 hour interval.My mistake was in assuming there were 12 overlap times. If that were true and your interval contained all of them, 143 special times means only 131 uncertain times.There are only 11 overlap times! Amazing, but true! Therefore you have either 133 or 132 uncertain times. You were right here, we can add a time but not exclude one.What is the missing overlap time? 11. There is no overlap time that starts with the hour 11. All the other hours have 11 uncertain times and one overlap for a total of 12; but 11 has only the 11 uncertain times, no overlap.What happened to 11's overlap time? It exists, but it's indistinguishable from 12's overlap time. 11:60 is the same as 12:00. It would be more elegant to start the minutes at 1 and end at 60, in which case it would be 12 that has no overlap time.
 
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estcourt
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Joined: June 9th, 2004, 7:35 am

Clock with equal length hands

July 19th, 2004, 3:13 pm

I thought about it slightly differently,we cannot tell the time when we can swap the hour and minute handswe always know the time (because we cannot confuse the hour) when both hands lie in the same sector (i.e. between the same two numbers on the dial so you can't change the hour)now it is is clear it is 132 times and no confusion over 12 o'clock
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