A clock correct three times a day











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A stopped clock tells the right time twice a day. How fast must a
clock go for it to be right three times a day?











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  • 1




    I'm a bit confused. So, the clock is stopped or it works?
    – rhsquared
    2 days ago








  • 10




    @rhsquared There are two clocks :)
    – Anush
    2 days ago






  • 1




    Could one not argue that a clock that has stopped running at exactly 12.00 o'clock is also correct three times a day? Once at midnight, once at noon and again at midnight. Or, should we count these occurrences at midnight only as ½ ?
    – Ideogram
    yesterday






  • 7




    A stopped clock will tell the right time three times a day, if it's the day daylight saving ends ...
    – Julia Hayward
    yesterday






  • 1




    @Ideogram one definitely could argue that, but it's typically better not to. Whenever agreeing on conventions, one must have a purpose in mind; preserving simple addition is usually deemed quite important, so a midnight (00:00) is generally considered to belong to the next day. This results in a period of two days having as many midnights as day one and day two added together. If one counted the moment of midnight belonging to both days, this wouldn't hold.
    – Bass
    yesterday















up vote
44
down vote

favorite
6













A stopped clock tells the right time twice a day. How fast must a
clock go for it to be right three times a day?











share|improve this question


















  • 1




    I'm a bit confused. So, the clock is stopped or it works?
    – rhsquared
    2 days ago








  • 10




    @rhsquared There are two clocks :)
    – Anush
    2 days ago






  • 1




    Could one not argue that a clock that has stopped running at exactly 12.00 o'clock is also correct three times a day? Once at midnight, once at noon and again at midnight. Or, should we count these occurrences at midnight only as ½ ?
    – Ideogram
    yesterday






  • 7




    A stopped clock will tell the right time three times a day, if it's the day daylight saving ends ...
    – Julia Hayward
    yesterday






  • 1




    @Ideogram one definitely could argue that, but it's typically better not to. Whenever agreeing on conventions, one must have a purpose in mind; preserving simple addition is usually deemed quite important, so a midnight (00:00) is generally considered to belong to the next day. This results in a period of two days having as many midnights as day one and day two added together. If one counted the moment of midnight belonging to both days, this wouldn't hold.
    – Bass
    yesterday













up vote
44
down vote

favorite
6









up vote
44
down vote

favorite
6






6






A stopped clock tells the right time twice a day. How fast must a
clock go for it to be right three times a day?











share|improve this question














A stopped clock tells the right time twice a day. How fast must a
clock go for it to be right three times a day?








time






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share|improve this question




share|improve this question










asked 2 days ago









Anush

857420




857420








  • 1




    I'm a bit confused. So, the clock is stopped or it works?
    – rhsquared
    2 days ago








  • 10




    @rhsquared There are two clocks :)
    – Anush
    2 days ago






  • 1




    Could one not argue that a clock that has stopped running at exactly 12.00 o'clock is also correct three times a day? Once at midnight, once at noon and again at midnight. Or, should we count these occurrences at midnight only as ½ ?
    – Ideogram
    yesterday






  • 7




    A stopped clock will tell the right time three times a day, if it's the day daylight saving ends ...
    – Julia Hayward
    yesterday






  • 1




    @Ideogram one definitely could argue that, but it's typically better not to. Whenever agreeing on conventions, one must have a purpose in mind; preserving simple addition is usually deemed quite important, so a midnight (00:00) is generally considered to belong to the next day. This results in a period of two days having as many midnights as day one and day two added together. If one counted the moment of midnight belonging to both days, this wouldn't hold.
    – Bass
    yesterday














  • 1




    I'm a bit confused. So, the clock is stopped or it works?
    – rhsquared
    2 days ago








  • 10




    @rhsquared There are two clocks :)
    – Anush
    2 days ago






  • 1




    Could one not argue that a clock that has stopped running at exactly 12.00 o'clock is also correct three times a day? Once at midnight, once at noon and again at midnight. Or, should we count these occurrences at midnight only as ½ ?
    – Ideogram
    yesterday






  • 7




    A stopped clock will tell the right time three times a day, if it's the day daylight saving ends ...
    – Julia Hayward
    yesterday






  • 1




    @Ideogram one definitely could argue that, but it's typically better not to. Whenever agreeing on conventions, one must have a purpose in mind; preserving simple addition is usually deemed quite important, so a midnight (00:00) is generally considered to belong to the next day. This results in a period of two days having as many midnights as day one and day two added together. If one counted the moment of midnight belonging to both days, this wouldn't hold.
    – Bass
    yesterday








1




1




I'm a bit confused. So, the clock is stopped or it works?
– rhsquared
2 days ago






I'm a bit confused. So, the clock is stopped or it works?
– rhsquared
2 days ago






10




10




@rhsquared There are two clocks :)
– Anush
2 days ago




@rhsquared There are two clocks :)
– Anush
2 days ago




1




1




Could one not argue that a clock that has stopped running at exactly 12.00 o'clock is also correct three times a day? Once at midnight, once at noon and again at midnight. Or, should we count these occurrences at midnight only as ½ ?
– Ideogram
yesterday




Could one not argue that a clock that has stopped running at exactly 12.00 o'clock is also correct three times a day? Once at midnight, once at noon and again at midnight. Or, should we count these occurrences at midnight only as ½ ?
– Ideogram
yesterday




7




7




A stopped clock will tell the right time three times a day, if it's the day daylight saving ends ...
– Julia Hayward
yesterday




A stopped clock will tell the right time three times a day, if it's the day daylight saving ends ...
– Julia Hayward
yesterday




1




1




@Ideogram one definitely could argue that, but it's typically better not to. Whenever agreeing on conventions, one must have a purpose in mind; preserving simple addition is usually deemed quite important, so a midnight (00:00) is generally considered to belong to the next day. This results in a period of two days having as many midnights as day one and day two added together. If one counted the moment of midnight belonging to both days, this wouldn't hold.
– Bass
yesterday




@Ideogram one definitely could argue that, but it's typically better not to. Whenever agreeing on conventions, one must have a purpose in mind; preserving simple addition is usually deemed quite important, so a midnight (00:00) is generally considered to belong to the next day. This results in a period of two days having as many midnights as day one and day two added together. If one counted the moment of midnight belonging to both days, this wouldn't hold.
– Bass
yesterday










12 Answers
12






active

oldest

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up vote
54
down vote



accepted










Since a stopped clock already agrees with a correctly running clock twice per day,




let's just add a third time by having the clock run backwards once per day.




So the answer (that requires the smallest clock hand speed) is




half speed backwards.




Since the answer must be symmetrical (as long as the difference in speeds remains the same, it doesn't matter if our clock is getting ahead or behind the correctly running one), we can get the other answer by adding twice the speed difference:




$-0.5 + 2 times (1 - (-0.5)) = 2.5 $ times the regular speed.







share|improve this answer



















  • 1




    What’s the general solution for the clock to be correct x times a day?
    – Anush
    2 days ago






  • 2




    @Anush, the accumulated error must come to x full circles per day, and a regular clock runs at 2 full circles per day, so the the speed difference (rate of error accumulation) must be x/2 times the regular speed. So for example, for x=2, the error must accumulate at 1 (the unit is "hours per hour", or "circles per circle"), which gives the solutions of $1 pm 1$ times the regular speed.
    – Bass
    yesterday












  • what happens if x=0?
    – JonMark Perry
    14 hours ago










  • @JonMarkPerry Well, using the pattern from above, the error accumulation rate must then be exactly 0/2. (Any clock running at exactly the right speed, barring the one obvious exception, is never correct.)
    – Bass
    14 hours ago


















up vote
27
down vote













Suppose we have an accurate clock and the fast clock, and that they start off showing the same time.




Over the next period of 24 hours, we want the fast clock to equal the normal clock two more times. So it needs to overtake the real clock twice, and then catch up with it again at exactly the end of those 24 hours for the start of the next 24 hour period. So while the accurate clock goes around twice, the fast clock must go around $2+3=5$ times.


This means that the fast clock goes at $5/2=2.5$ times regular speed.




Here is an alternative method:




Three times a day means every $8$ hours. So in $8$ hours the fast clock must do the same as a regular clock plus one full revolution, i.e. $8+12=20$ hours. It therefore goes at $20/8=2.5$ times the regular speed.







share|improve this answer






























    up vote
    19
    down vote













    Alternative method:




    The clock is not moving its hands, but is instead moving around the earth through timezones. Each time "moves" around the earth to the west, so the clock travels east to meet them. In order to encounter each time 3 times in a 24-hour period, it has to travel half the distance around the earth. Actual speed would depend on the precise latitude, slower closer to the poles.




    The answer to how fast the clock must go:




    180 degrees eastward a day, or 7.5 degrees per hour.







    share|improve this answer










    New contributor




    Grollo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    • 2




      Nice and different.
      – Deduplicator
      2 days ago


















    up vote
    8
    down vote













    It should be right three times a day when




    its stopped during Daylight Saving Time at 1:00




    Reasoning:




    During the fall change, clocks move back one hour. If you had the clock set to 1:00, it would be right at 0100 hrs, 0200 hrs (when it switches back to 0100 hrs), then at 1300 hrs (1:00 pm)







    share|improve this answer




























      up vote
      4
      down vote













      The clock needs to be moving at




      2.5 times normal speed




      Reasoning:




      If you were to graph the stationary clock and normal time, then you would have a flat line, and 2 diagonal lines - these would cross at 2 points, which are the times where the clock is "correct".


      Since we cannot change normal time (barring DST or timezones) we need to increase the slope on the stopped (henceforth "wrong") clock until it intersects normal time thrice.


      This cannot happen until the Wrong Clock is moving at least as fast as the Correct Clock. At 1:1 the clock is either always right or always wrong. At 2:1 the clock will be "correct" at 12 hour intervals (e.g. Midnight, Noon and Midnight, or twice per day)


      As soon as your Wrong Clock is moving slightly faster than 2:1 then your "correct" periods will also be slightly more frequent than every 12 hours - or 3 times some days, and 2 times all the rest.


      To be right 3 times every day, you need to be right every 8 hours - which is a ratio of 5:2, or 2.5 time normal speed







      share|improve this answer




























        up vote
        3
        down vote














        A stopped clock tells the right time twice a day. How fast must a clock go for it to be right three times a day?




        It must travel at




        on average approximately 45–50mph, twice per day




        Here's how:




        Our clock is broken and has the time pointed at 4 o'clock. It starts in the city of Gary, Indiana, USA (EST, UTC-5).


        At 4am, our clock will be correct for the first time.


        At 4pm, our clock will be correct for the second time. Then we hop in our car with the clock.


        Using the I-90 E we can head westward to reach Chicago, Illinois in about 40 minutes (see Google Maps for the journey). 4pm is an off-peak time so travel ought to be fairly smooth. By the time we're in Chicago, we're in CST, UST-6, and the local time is about 3:40pm with good timing. Once it reaches 4pm again, our clock's been right for the third time today.


        Our job well done, we drive back to our home in Gary, Indiana and return our clock to where we kept it.







        share|improve this answer




























          up vote
          1
          down vote














          So if we are talking about an analog clock then it will come to same position three times a day if hours become 36 and to cover 36 hours in the routine 24 hours 1 second should become 1.5 second. $1.5*3600*24=129600$ seconds and since $36*3600=129600$ So speed of clock will increase so that 1 second will become equal to 1.5 second.







          share|improve this answer










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          Ihtisham Ali Farooq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          • Which are the hours it would be correct?
            – Anush
            2 days ago


















          up vote
          1
          down vote














          An analog clock with a second hand that jumps instantaneously every second is right 86400 times a day, so it's also right 3 times a day.







          share|improve this answer

















          • 1




            The same argument for digital clocks
            – gota
            yesterday


















          up vote
          0
          down vote













          Assumption 1: The regular 12 hour clock-face.

          Assumption 2: Fast here does not mean an offset but that the clock is actually running at a higher speed.




          It should rotate at 60 (false) hours per day (instead of 24 hours per
          day).




          Assuming we start at 00:00 on day 1 (actual time) and set the false clock at 00:00 and switch it ON. -- FIRST TIME it is RIGHT



          By the time it is 08:00 (AM actual time) which 1/3 of the day past, the false time would show as 08:00 too as it would finish 1/3 of 60 hours which is 20 hours or 12 hours false time (one complete rotation from 00:00) + 8 hours and it would point to 08:00 on the clock. -- SECOND TIME it is RIGHT



          By the time it is 04:00 (PM actual time) which 2/3 of the day past, the false time would show as 04:00 too as it would finish 2/3 of 60 hours which is 40 hours or 36 hours false time (3 complete rotations from 00:00) + 4 hours and it would point to 04:00 on the clock. -- THIRD TIME it is RIGHT



          Using the same math, it would be RIGHT again at 00:00 (FIRST TIME of next day) and so on.






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          New contributor




          alwayslearning is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            up vote
            0
            down vote













            There is a slight ambiguity/lack of precision in the question wording which (in my interpretation) means there isn't a definitive answer as other answers suggest.



            Up front, my answer is:




            A clock that is more than 2 times faster and less than or equal to 2.5 times faster than a perfect clock, can be observed to be correct 3 times in one day




            First I'd like to introduce a concept that makes the rest of this answer easier to explain



            Suppose I have a perfect clock, that is accurate and on time. Suppose I also have a broken clock that I position so it shows a time one minute before the current time on the perfect clock, and then I move the minute hand of the broken clock so it shows a time one minute after the perfect clock. At some point during that motion, the broken clock was showing a time that was accurate. It occurred just (infinitesimally) prior to the exact moment the broken clock time moved past the perfect clock time.



            Hence, when a clock is faulty (running fast) and it "overtakes" a perfect clock, for the briefest of moments the time the faulty clock is observed to be perfectly accurate.



            We must hence work out how fast a clock must run to overtake a perfect clock 3 times in a 24 hour period






            • If a faulty clock runs twice as fast as perfect clock, and we set them going both showing 12:00, they are in sync. This is the first time they are in sync. Let's also say it's Monday

            • 6 hours after we started, the faulty fast clock is back at the start position, but the perfect clock is only half way round

            • 12 hours after we started, the faulty fast clock will overtake the perfect clock and it's the second time it has happened

            • 24 hours after we started, the fast clock again overtakes the perfect clock. It does so on the exact stroke of midnight but this means it's a new day (i.e. it's now Tuesday)

            • Our fast clock only overtook the perfect clock twice - at the outset, and at noon




            This means...




            ... 2 times normal speed isn't enough to get the fast clock to overtake the perfect clock 3 times




            However, if a clock runs...




            ... just fractionally faster than 2 times normal speed, then it will overtake a perfect clock 3 times in one day. If we have a clock that is a double speed plus one second faster (i.e. it counts (86400 * 2) + 1 seconds in a single day) then from being synced to a perfect clock on midnight monday (first overtake) it will overtake the perfect clock at around 11:59:59.5 (half a second to noon) and again at around 23:59:59 (just before it becomes Tuesday)




            Hence the answer given above..



            But before I finish:



            A Special Note




            2.5 times is a special speed factor; it's the only speed factor where the faulty fast clock will overtake (=show the correct time) three times a day, day in day out without being manipulated further. Any speed factor slower than this will, after some variable number of days depending on the speed factor, result in a clock that has too much catching up to do at the start of the day and won't run fast enough to overtake the perfect clock 3 times in the day. How often these "can't overtake 3 times" days occur is a function of how slow the clock is. Values near 2x speed will take a long time to get back to a state where they can have a "3 overtakes" day. Values near 2.5x speed will have many days in a row where they perform 3 overtakes, only occasionally dipping to a 2-overtake day




            The question didn't call for every day (for the rest of the clock's life) being a 3-overtake day, only that a faulty clock should show the correct time 3 times in a day, hence why I feel the answer has to be..




            ..a range







            share|improve this answer




























              up vote
              -1
              down vote













              Answer:




              It must move 3 seconds for every 2 seconds that pass, so 1.5 times as fast as a normal clock.




              Reason:




              In a 24 hour period, 36 hours will have transpired on this faster clock, meaning it will have made 3 rotations within the day. At some point in each of those rotations, the correct time will have been passed, meaning it is right 3 times a day.







              share|improve this answer

















              • 3




                You may want to check your premise: Using your example, if the clock shows the correct time at Midnight, it will not show the correct time again until the following Midnight...
                – Chronocidal
                2 days ago










              • @Chronocidal dang u right, i think my brain was working on a fast clock vs a stopped clock lol
                – AHKieran
                2 days ago










              • You did manage to fool at least 2 more people though ;)
                – Geliormth
                2 days ago


















              up vote
              -2
              down vote













              The speed has to be




              -1 so the clock will be at the same spot at 00, 06 and 18







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              • Welcome to Puzzling SE. Could you add more explanation to what you mean by a speed of -1? Also if you'd like to learn more about this site and earn your first badge, check out the tour at this link: puzzling.stackexchange.com/tour
                – gabbo1092
                2 days ago






              • 2




                This would be right at 12 too, wouldn't it?
                – jafe
                2 days ago










              protected by JonMark Perry 20 hours ago



              Thank you for your interest in this question.
              Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).



              Would you like to answer one of these unanswered questions instead?














              12 Answers
              12






              active

              oldest

              votes








              12 Answers
              12






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              54
              down vote



              accepted










              Since a stopped clock already agrees with a correctly running clock twice per day,




              let's just add a third time by having the clock run backwards once per day.




              So the answer (that requires the smallest clock hand speed) is




              half speed backwards.




              Since the answer must be symmetrical (as long as the difference in speeds remains the same, it doesn't matter if our clock is getting ahead or behind the correctly running one), we can get the other answer by adding twice the speed difference:




              $-0.5 + 2 times (1 - (-0.5)) = 2.5 $ times the regular speed.







              share|improve this answer



















              • 1




                What’s the general solution for the clock to be correct x times a day?
                – Anush
                2 days ago






              • 2




                @Anush, the accumulated error must come to x full circles per day, and a regular clock runs at 2 full circles per day, so the the speed difference (rate of error accumulation) must be x/2 times the regular speed. So for example, for x=2, the error must accumulate at 1 (the unit is "hours per hour", or "circles per circle"), which gives the solutions of $1 pm 1$ times the regular speed.
                – Bass
                yesterday












              • what happens if x=0?
                – JonMark Perry
                14 hours ago










              • @JonMarkPerry Well, using the pattern from above, the error accumulation rate must then be exactly 0/2. (Any clock running at exactly the right speed, barring the one obvious exception, is never correct.)
                – Bass
                14 hours ago















              up vote
              54
              down vote



              accepted










              Since a stopped clock already agrees with a correctly running clock twice per day,




              let's just add a third time by having the clock run backwards once per day.




              So the answer (that requires the smallest clock hand speed) is




              half speed backwards.




              Since the answer must be symmetrical (as long as the difference in speeds remains the same, it doesn't matter if our clock is getting ahead or behind the correctly running one), we can get the other answer by adding twice the speed difference:




              $-0.5 + 2 times (1 - (-0.5)) = 2.5 $ times the regular speed.







              share|improve this answer



















              • 1




                What’s the general solution for the clock to be correct x times a day?
                – Anush
                2 days ago






              • 2




                @Anush, the accumulated error must come to x full circles per day, and a regular clock runs at 2 full circles per day, so the the speed difference (rate of error accumulation) must be x/2 times the regular speed. So for example, for x=2, the error must accumulate at 1 (the unit is "hours per hour", or "circles per circle"), which gives the solutions of $1 pm 1$ times the regular speed.
                – Bass
                yesterday












              • what happens if x=0?
                – JonMark Perry
                14 hours ago










              • @JonMarkPerry Well, using the pattern from above, the error accumulation rate must then be exactly 0/2. (Any clock running at exactly the right speed, barring the one obvious exception, is never correct.)
                – Bass
                14 hours ago













              up vote
              54
              down vote



              accepted







              up vote
              54
              down vote



              accepted






              Since a stopped clock already agrees with a correctly running clock twice per day,




              let's just add a third time by having the clock run backwards once per day.




              So the answer (that requires the smallest clock hand speed) is




              half speed backwards.




              Since the answer must be symmetrical (as long as the difference in speeds remains the same, it doesn't matter if our clock is getting ahead or behind the correctly running one), we can get the other answer by adding twice the speed difference:




              $-0.5 + 2 times (1 - (-0.5)) = 2.5 $ times the regular speed.







              share|improve this answer














              Since a stopped clock already agrees with a correctly running clock twice per day,




              let's just add a third time by having the clock run backwards once per day.




              So the answer (that requires the smallest clock hand speed) is




              half speed backwards.




              Since the answer must be symmetrical (as long as the difference in speeds remains the same, it doesn't matter if our clock is getting ahead or behind the correctly running one), we can get the other answer by adding twice the speed difference:




              $-0.5 + 2 times (1 - (-0.5)) = 2.5 $ times the regular speed.








              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 2 days ago

























              answered 2 days ago









              Bass

              26.4k464165




              26.4k464165








              • 1




                What’s the general solution for the clock to be correct x times a day?
                – Anush
                2 days ago






              • 2




                @Anush, the accumulated error must come to x full circles per day, and a regular clock runs at 2 full circles per day, so the the speed difference (rate of error accumulation) must be x/2 times the regular speed. So for example, for x=2, the error must accumulate at 1 (the unit is "hours per hour", or "circles per circle"), which gives the solutions of $1 pm 1$ times the regular speed.
                – Bass
                yesterday












              • what happens if x=0?
                – JonMark Perry
                14 hours ago










              • @JonMarkPerry Well, using the pattern from above, the error accumulation rate must then be exactly 0/2. (Any clock running at exactly the right speed, barring the one obvious exception, is never correct.)
                – Bass
                14 hours ago














              • 1




                What’s the general solution for the clock to be correct x times a day?
                – Anush
                2 days ago






              • 2




                @Anush, the accumulated error must come to x full circles per day, and a regular clock runs at 2 full circles per day, so the the speed difference (rate of error accumulation) must be x/2 times the regular speed. So for example, for x=2, the error must accumulate at 1 (the unit is "hours per hour", or "circles per circle"), which gives the solutions of $1 pm 1$ times the regular speed.
                – Bass
                yesterday












              • what happens if x=0?
                – JonMark Perry
                14 hours ago










              • @JonMarkPerry Well, using the pattern from above, the error accumulation rate must then be exactly 0/2. (Any clock running at exactly the right speed, barring the one obvious exception, is never correct.)
                – Bass
                14 hours ago








              1




              1




              What’s the general solution for the clock to be correct x times a day?
              – Anush
              2 days ago




              What’s the general solution for the clock to be correct x times a day?
              – Anush
              2 days ago




              2




              2




              @Anush, the accumulated error must come to x full circles per day, and a regular clock runs at 2 full circles per day, so the the speed difference (rate of error accumulation) must be x/2 times the regular speed. So for example, for x=2, the error must accumulate at 1 (the unit is "hours per hour", or "circles per circle"), which gives the solutions of $1 pm 1$ times the regular speed.
              – Bass
              yesterday






              @Anush, the accumulated error must come to x full circles per day, and a regular clock runs at 2 full circles per day, so the the speed difference (rate of error accumulation) must be x/2 times the regular speed. So for example, for x=2, the error must accumulate at 1 (the unit is "hours per hour", or "circles per circle"), which gives the solutions of $1 pm 1$ times the regular speed.
              – Bass
              yesterday














              what happens if x=0?
              – JonMark Perry
              14 hours ago




              what happens if x=0?
              – JonMark Perry
              14 hours ago












              @JonMarkPerry Well, using the pattern from above, the error accumulation rate must then be exactly 0/2. (Any clock running at exactly the right speed, barring the one obvious exception, is never correct.)
              – Bass
              14 hours ago




              @JonMarkPerry Well, using the pattern from above, the error accumulation rate must then be exactly 0/2. (Any clock running at exactly the right speed, barring the one obvious exception, is never correct.)
              – Bass
              14 hours ago










              up vote
              27
              down vote













              Suppose we have an accurate clock and the fast clock, and that they start off showing the same time.




              Over the next period of 24 hours, we want the fast clock to equal the normal clock two more times. So it needs to overtake the real clock twice, and then catch up with it again at exactly the end of those 24 hours for the start of the next 24 hour period. So while the accurate clock goes around twice, the fast clock must go around $2+3=5$ times.


              This means that the fast clock goes at $5/2=2.5$ times regular speed.




              Here is an alternative method:




              Three times a day means every $8$ hours. So in $8$ hours the fast clock must do the same as a regular clock plus one full revolution, i.e. $8+12=20$ hours. It therefore goes at $20/8=2.5$ times the regular speed.







              share|improve this answer



























                up vote
                27
                down vote













                Suppose we have an accurate clock and the fast clock, and that they start off showing the same time.




                Over the next period of 24 hours, we want the fast clock to equal the normal clock two more times. So it needs to overtake the real clock twice, and then catch up with it again at exactly the end of those 24 hours for the start of the next 24 hour period. So while the accurate clock goes around twice, the fast clock must go around $2+3=5$ times.


                This means that the fast clock goes at $5/2=2.5$ times regular speed.




                Here is an alternative method:




                Three times a day means every $8$ hours. So in $8$ hours the fast clock must do the same as a regular clock plus one full revolution, i.e. $8+12=20$ hours. It therefore goes at $20/8=2.5$ times the regular speed.







                share|improve this answer

























                  up vote
                  27
                  down vote










                  up vote
                  27
                  down vote









                  Suppose we have an accurate clock and the fast clock, and that they start off showing the same time.




                  Over the next period of 24 hours, we want the fast clock to equal the normal clock two more times. So it needs to overtake the real clock twice, and then catch up with it again at exactly the end of those 24 hours for the start of the next 24 hour period. So while the accurate clock goes around twice, the fast clock must go around $2+3=5$ times.


                  This means that the fast clock goes at $5/2=2.5$ times regular speed.




                  Here is an alternative method:




                  Three times a day means every $8$ hours. So in $8$ hours the fast clock must do the same as a regular clock plus one full revolution, i.e. $8+12=20$ hours. It therefore goes at $20/8=2.5$ times the regular speed.







                  share|improve this answer














                  Suppose we have an accurate clock and the fast clock, and that they start off showing the same time.




                  Over the next period of 24 hours, we want the fast clock to equal the normal clock two more times. So it needs to overtake the real clock twice, and then catch up with it again at exactly the end of those 24 hours for the start of the next 24 hour period. So while the accurate clock goes around twice, the fast clock must go around $2+3=5$ times.


                  This means that the fast clock goes at $5/2=2.5$ times regular speed.




                  Here is an alternative method:




                  Three times a day means every $8$ hours. So in $8$ hours the fast clock must do the same as a regular clock plus one full revolution, i.e. $8+12=20$ hours. It therefore goes at $20/8=2.5$ times the regular speed.








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  Jaap Scherphuis

                  14k12462




                  14k12462






















                      up vote
                      19
                      down vote













                      Alternative method:




                      The clock is not moving its hands, but is instead moving around the earth through timezones. Each time "moves" around the earth to the west, so the clock travels east to meet them. In order to encounter each time 3 times in a 24-hour period, it has to travel half the distance around the earth. Actual speed would depend on the precise latitude, slower closer to the poles.




                      The answer to how fast the clock must go:




                      180 degrees eastward a day, or 7.5 degrees per hour.







                      share|improve this answer










                      New contributor




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                      • 2




                        Nice and different.
                        – Deduplicator
                        2 days ago















                      up vote
                      19
                      down vote













                      Alternative method:




                      The clock is not moving its hands, but is instead moving around the earth through timezones. Each time "moves" around the earth to the west, so the clock travels east to meet them. In order to encounter each time 3 times in a 24-hour period, it has to travel half the distance around the earth. Actual speed would depend on the precise latitude, slower closer to the poles.




                      The answer to how fast the clock must go:




                      180 degrees eastward a day, or 7.5 degrees per hour.







                      share|improve this answer










                      New contributor




                      Grollo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                      • 2




                        Nice and different.
                        – Deduplicator
                        2 days ago













                      up vote
                      19
                      down vote










                      up vote
                      19
                      down vote









                      Alternative method:




                      The clock is not moving its hands, but is instead moving around the earth through timezones. Each time "moves" around the earth to the west, so the clock travels east to meet them. In order to encounter each time 3 times in a 24-hour period, it has to travel half the distance around the earth. Actual speed would depend on the precise latitude, slower closer to the poles.




                      The answer to how fast the clock must go:




                      180 degrees eastward a day, or 7.5 degrees per hour.







                      share|improve this answer










                      New contributor




                      Grollo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                      Check out our Code of Conduct.









                      Alternative method:




                      The clock is not moving its hands, but is instead moving around the earth through timezones. Each time "moves" around the earth to the west, so the clock travels east to meet them. In order to encounter each time 3 times in a 24-hour period, it has to travel half the distance around the earth. Actual speed would depend on the precise latitude, slower closer to the poles.




                      The answer to how fast the clock must go:




                      180 degrees eastward a day, or 7.5 degrees per hour.








                      share|improve this answer










                      New contributor




                      Grollo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                      share|improve this answer



                      share|improve this answer








                      edited 2 days ago





















                      New contributor




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                      answered 2 days ago









                      Grollo

                      2915




                      2915




                      New contributor




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                      New contributor





                      Grollo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                      • 2




                        Nice and different.
                        – Deduplicator
                        2 days ago














                      • 2




                        Nice and different.
                        – Deduplicator
                        2 days ago








                      2




                      2




                      Nice and different.
                      – Deduplicator
                      2 days ago




                      Nice and different.
                      – Deduplicator
                      2 days ago










                      up vote
                      8
                      down vote













                      It should be right three times a day when




                      its stopped during Daylight Saving Time at 1:00




                      Reasoning:




                      During the fall change, clocks move back one hour. If you had the clock set to 1:00, it would be right at 0100 hrs, 0200 hrs (when it switches back to 0100 hrs), then at 1300 hrs (1:00 pm)







                      share|improve this answer

























                        up vote
                        8
                        down vote













                        It should be right three times a day when




                        its stopped during Daylight Saving Time at 1:00




                        Reasoning:




                        During the fall change, clocks move back one hour. If you had the clock set to 1:00, it would be right at 0100 hrs, 0200 hrs (when it switches back to 0100 hrs), then at 1300 hrs (1:00 pm)







                        share|improve this answer























                          up vote
                          8
                          down vote










                          up vote
                          8
                          down vote









                          It should be right three times a day when




                          its stopped during Daylight Saving Time at 1:00




                          Reasoning:




                          During the fall change, clocks move back one hour. If you had the clock set to 1:00, it would be right at 0100 hrs, 0200 hrs (when it switches back to 0100 hrs), then at 1300 hrs (1:00 pm)







                          share|improve this answer












                          It should be right three times a day when




                          its stopped during Daylight Saving Time at 1:00




                          Reasoning:




                          During the fall change, clocks move back one hour. If you had the clock set to 1:00, it would be right at 0100 hrs, 0200 hrs (when it switches back to 0100 hrs), then at 1300 hrs (1:00 pm)








                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 2 days ago









                          Canadian Luke

                          1928




                          1928






















                              up vote
                              4
                              down vote













                              The clock needs to be moving at




                              2.5 times normal speed




                              Reasoning:




                              If you were to graph the stationary clock and normal time, then you would have a flat line, and 2 diagonal lines - these would cross at 2 points, which are the times where the clock is "correct".


                              Since we cannot change normal time (barring DST or timezones) we need to increase the slope on the stopped (henceforth "wrong") clock until it intersects normal time thrice.


                              This cannot happen until the Wrong Clock is moving at least as fast as the Correct Clock. At 1:1 the clock is either always right or always wrong. At 2:1 the clock will be "correct" at 12 hour intervals (e.g. Midnight, Noon and Midnight, or twice per day)


                              As soon as your Wrong Clock is moving slightly faster than 2:1 then your "correct" periods will also be slightly more frequent than every 12 hours - or 3 times some days, and 2 times all the rest.


                              To be right 3 times every day, you need to be right every 8 hours - which is a ratio of 5:2, or 2.5 time normal speed







                              share|improve this answer

























                                up vote
                                4
                                down vote













                                The clock needs to be moving at




                                2.5 times normal speed




                                Reasoning:




                                If you were to graph the stationary clock and normal time, then you would have a flat line, and 2 diagonal lines - these would cross at 2 points, which are the times where the clock is "correct".


                                Since we cannot change normal time (barring DST or timezones) we need to increase the slope on the stopped (henceforth "wrong") clock until it intersects normal time thrice.


                                This cannot happen until the Wrong Clock is moving at least as fast as the Correct Clock. At 1:1 the clock is either always right or always wrong. At 2:1 the clock will be "correct" at 12 hour intervals (e.g. Midnight, Noon and Midnight, or twice per day)


                                As soon as your Wrong Clock is moving slightly faster than 2:1 then your "correct" periods will also be slightly more frequent than every 12 hours - or 3 times some days, and 2 times all the rest.


                                To be right 3 times every day, you need to be right every 8 hours - which is a ratio of 5:2, or 2.5 time normal speed







                                share|improve this answer























                                  up vote
                                  4
                                  down vote










                                  up vote
                                  4
                                  down vote









                                  The clock needs to be moving at




                                  2.5 times normal speed




                                  Reasoning:




                                  If you were to graph the stationary clock and normal time, then you would have a flat line, and 2 diagonal lines - these would cross at 2 points, which are the times where the clock is "correct".


                                  Since we cannot change normal time (barring DST or timezones) we need to increase the slope on the stopped (henceforth "wrong") clock until it intersects normal time thrice.


                                  This cannot happen until the Wrong Clock is moving at least as fast as the Correct Clock. At 1:1 the clock is either always right or always wrong. At 2:1 the clock will be "correct" at 12 hour intervals (e.g. Midnight, Noon and Midnight, or twice per day)


                                  As soon as your Wrong Clock is moving slightly faster than 2:1 then your "correct" periods will also be slightly more frequent than every 12 hours - or 3 times some days, and 2 times all the rest.


                                  To be right 3 times every day, you need to be right every 8 hours - which is a ratio of 5:2, or 2.5 time normal speed







                                  share|improve this answer












                                  The clock needs to be moving at




                                  2.5 times normal speed




                                  Reasoning:




                                  If you were to graph the stationary clock and normal time, then you would have a flat line, and 2 diagonal lines - these would cross at 2 points, which are the times where the clock is "correct".


                                  Since we cannot change normal time (barring DST or timezones) we need to increase the slope on the stopped (henceforth "wrong") clock until it intersects normal time thrice.


                                  This cannot happen until the Wrong Clock is moving at least as fast as the Correct Clock. At 1:1 the clock is either always right or always wrong. At 2:1 the clock will be "correct" at 12 hour intervals (e.g. Midnight, Noon and Midnight, or twice per day)


                                  As soon as your Wrong Clock is moving slightly faster than 2:1 then your "correct" periods will also be slightly more frequent than every 12 hours - or 3 times some days, and 2 times all the rest.


                                  To be right 3 times every day, you need to be right every 8 hours - which is a ratio of 5:2, or 2.5 time normal speed








                                  share|improve this answer












                                  share|improve this answer



                                  share|improve this answer










                                  answered 2 days ago









                                  Chronocidal

                                  674111




                                  674111






















                                      up vote
                                      3
                                      down vote














                                      A stopped clock tells the right time twice a day. How fast must a clock go for it to be right three times a day?




                                      It must travel at




                                      on average approximately 45–50mph, twice per day




                                      Here's how:




                                      Our clock is broken and has the time pointed at 4 o'clock. It starts in the city of Gary, Indiana, USA (EST, UTC-5).


                                      At 4am, our clock will be correct for the first time.


                                      At 4pm, our clock will be correct for the second time. Then we hop in our car with the clock.


                                      Using the I-90 E we can head westward to reach Chicago, Illinois in about 40 minutes (see Google Maps for the journey). 4pm is an off-peak time so travel ought to be fairly smooth. By the time we're in Chicago, we're in CST, UST-6, and the local time is about 3:40pm with good timing. Once it reaches 4pm again, our clock's been right for the third time today.


                                      Our job well done, we drive back to our home in Gary, Indiana and return our clock to where we kept it.







                                      share|improve this answer

























                                        up vote
                                        3
                                        down vote














                                        A stopped clock tells the right time twice a day. How fast must a clock go for it to be right three times a day?




                                        It must travel at




                                        on average approximately 45–50mph, twice per day




                                        Here's how:




                                        Our clock is broken and has the time pointed at 4 o'clock. It starts in the city of Gary, Indiana, USA (EST, UTC-5).


                                        At 4am, our clock will be correct for the first time.


                                        At 4pm, our clock will be correct for the second time. Then we hop in our car with the clock.


                                        Using the I-90 E we can head westward to reach Chicago, Illinois in about 40 minutes (see Google Maps for the journey). 4pm is an off-peak time so travel ought to be fairly smooth. By the time we're in Chicago, we're in CST, UST-6, and the local time is about 3:40pm with good timing. Once it reaches 4pm again, our clock's been right for the third time today.


                                        Our job well done, we drive back to our home in Gary, Indiana and return our clock to where we kept it.







                                        share|improve this answer























                                          up vote
                                          3
                                          down vote










                                          up vote
                                          3
                                          down vote










                                          A stopped clock tells the right time twice a day. How fast must a clock go for it to be right three times a day?




                                          It must travel at




                                          on average approximately 45–50mph, twice per day




                                          Here's how:




                                          Our clock is broken and has the time pointed at 4 o'clock. It starts in the city of Gary, Indiana, USA (EST, UTC-5).


                                          At 4am, our clock will be correct for the first time.


                                          At 4pm, our clock will be correct for the second time. Then we hop in our car with the clock.


                                          Using the I-90 E we can head westward to reach Chicago, Illinois in about 40 minutes (see Google Maps for the journey). 4pm is an off-peak time so travel ought to be fairly smooth. By the time we're in Chicago, we're in CST, UST-6, and the local time is about 3:40pm with good timing. Once it reaches 4pm again, our clock's been right for the third time today.


                                          Our job well done, we drive back to our home in Gary, Indiana and return our clock to where we kept it.







                                          share|improve this answer













                                          A stopped clock tells the right time twice a day. How fast must a clock go for it to be right three times a day?




                                          It must travel at




                                          on average approximately 45–50mph, twice per day




                                          Here's how:




                                          Our clock is broken and has the time pointed at 4 o'clock. It starts in the city of Gary, Indiana, USA (EST, UTC-5).


                                          At 4am, our clock will be correct for the first time.


                                          At 4pm, our clock will be correct for the second time. Then we hop in our car with the clock.


                                          Using the I-90 E we can head westward to reach Chicago, Illinois in about 40 minutes (see Google Maps for the journey). 4pm is an off-peak time so travel ought to be fairly smooth. By the time we're in Chicago, we're in CST, UST-6, and the local time is about 3:40pm with good timing. Once it reaches 4pm again, our clock's been right for the third time today.


                                          Our job well done, we drive back to our home in Gary, Indiana and return our clock to where we kept it.








                                          share|improve this answer












                                          share|improve this answer



                                          share|improve this answer










                                          answered 2 days ago









                                          doppelgreener

                                          246212




                                          246212






















                                              up vote
                                              1
                                              down vote














                                              So if we are talking about an analog clock then it will come to same position three times a day if hours become 36 and to cover 36 hours in the routine 24 hours 1 second should become 1.5 second. $1.5*3600*24=129600$ seconds and since $36*3600=129600$ So speed of clock will increase so that 1 second will become equal to 1.5 second.







                                              share|improve this answer










                                              New contributor




                                              Ihtisham Ali Farooq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.


















                                              • Which are the hours it would be correct?
                                                – Anush
                                                2 days ago















                                              up vote
                                              1
                                              down vote














                                              So if we are talking about an analog clock then it will come to same position three times a day if hours become 36 and to cover 36 hours in the routine 24 hours 1 second should become 1.5 second. $1.5*3600*24=129600$ seconds and since $36*3600=129600$ So speed of clock will increase so that 1 second will become equal to 1.5 second.







                                              share|improve this answer










                                              New contributor




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                                              Check out our Code of Conduct.


















                                              • Which are the hours it would be correct?
                                                – Anush
                                                2 days ago













                                              up vote
                                              1
                                              down vote










                                              up vote
                                              1
                                              down vote










                                              So if we are talking about an analog clock then it will come to same position three times a day if hours become 36 and to cover 36 hours in the routine 24 hours 1 second should become 1.5 second. $1.5*3600*24=129600$ seconds and since $36*3600=129600$ So speed of clock will increase so that 1 second will become equal to 1.5 second.







                                              share|improve this answer










                                              New contributor




                                              Ihtisham Ali Farooq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.










                                              So if we are talking about an analog clock then it will come to same position three times a day if hours become 36 and to cover 36 hours in the routine 24 hours 1 second should become 1.5 second. $1.5*3600*24=129600$ seconds and since $36*3600=129600$ So speed of clock will increase so that 1 second will become equal to 1.5 second.








                                              share|improve this answer










                                              New contributor




                                              Ihtisham Ali Farooq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.









                                              share|improve this answer



                                              share|improve this answer








                                              edited 2 days ago









                                              gabbo1092

                                              4,539635




                                              4,539635






                                              New contributor




                                              Ihtisham Ali Farooq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                              answered 2 days ago









                                              Ihtisham Ali Farooq

                                              284




                                              284




                                              New contributor




                                              Ihtisham Ali Farooq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                                              New contributor





                                              Ihtisham Ali Farooq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.






                                              Ihtisham Ali Farooq is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.












                                              • Which are the hours it would be correct?
                                                – Anush
                                                2 days ago


















                                              • Which are the hours it would be correct?
                                                – Anush
                                                2 days ago
















                                              Which are the hours it would be correct?
                                              – Anush
                                              2 days ago




                                              Which are the hours it would be correct?
                                              – Anush
                                              2 days ago










                                              up vote
                                              1
                                              down vote














                                              An analog clock with a second hand that jumps instantaneously every second is right 86400 times a day, so it's also right 3 times a day.







                                              share|improve this answer

















                                              • 1




                                                The same argument for digital clocks
                                                – gota
                                                yesterday















                                              up vote
                                              1
                                              down vote














                                              An analog clock with a second hand that jumps instantaneously every second is right 86400 times a day, so it's also right 3 times a day.







                                              share|improve this answer

















                                              • 1




                                                The same argument for digital clocks
                                                – gota
                                                yesterday













                                              up vote
                                              1
                                              down vote










                                              up vote
                                              1
                                              down vote










                                              An analog clock with a second hand that jumps instantaneously every second is right 86400 times a day, so it's also right 3 times a day.







                                              share|improve this answer













                                              An analog clock with a second hand that jumps instantaneously every second is right 86400 times a day, so it's also right 3 times a day.








                                              share|improve this answer












                                              share|improve this answer



                                              share|improve this answer










                                              answered yesterday









                                              Peter

                                              177110




                                              177110








                                              • 1




                                                The same argument for digital clocks
                                                – gota
                                                yesterday














                                              • 1




                                                The same argument for digital clocks
                                                – gota
                                                yesterday








                                              1




                                              1




                                              The same argument for digital clocks
                                              – gota
                                              yesterday




                                              The same argument for digital clocks
                                              – gota
                                              yesterday










                                              up vote
                                              0
                                              down vote













                                              Assumption 1: The regular 12 hour clock-face.

                                              Assumption 2: Fast here does not mean an offset but that the clock is actually running at a higher speed.




                                              It should rotate at 60 (false) hours per day (instead of 24 hours per
                                              day).




                                              Assuming we start at 00:00 on day 1 (actual time) and set the false clock at 00:00 and switch it ON. -- FIRST TIME it is RIGHT



                                              By the time it is 08:00 (AM actual time) which 1/3 of the day past, the false time would show as 08:00 too as it would finish 1/3 of 60 hours which is 20 hours or 12 hours false time (one complete rotation from 00:00) + 8 hours and it would point to 08:00 on the clock. -- SECOND TIME it is RIGHT



                                              By the time it is 04:00 (PM actual time) which 2/3 of the day past, the false time would show as 04:00 too as it would finish 2/3 of 60 hours which is 40 hours or 36 hours false time (3 complete rotations from 00:00) + 4 hours and it would point to 04:00 on the clock. -- THIRD TIME it is RIGHT



                                              Using the same math, it would be RIGHT again at 00:00 (FIRST TIME of next day) and so on.






                                              share|improve this answer










                                              New contributor




                                              alwayslearning is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                              Check out our Code of Conduct.






















                                                up vote
                                                0
                                                down vote













                                                Assumption 1: The regular 12 hour clock-face.

                                                Assumption 2: Fast here does not mean an offset but that the clock is actually running at a higher speed.




                                                It should rotate at 60 (false) hours per day (instead of 24 hours per
                                                day).




                                                Assuming we start at 00:00 on day 1 (actual time) and set the false clock at 00:00 and switch it ON. -- FIRST TIME it is RIGHT



                                                By the time it is 08:00 (AM actual time) which 1/3 of the day past, the false time would show as 08:00 too as it would finish 1/3 of 60 hours which is 20 hours or 12 hours false time (one complete rotation from 00:00) + 8 hours and it would point to 08:00 on the clock. -- SECOND TIME it is RIGHT



                                                By the time it is 04:00 (PM actual time) which 2/3 of the day past, the false time would show as 04:00 too as it would finish 2/3 of 60 hours which is 40 hours or 36 hours false time (3 complete rotations from 00:00) + 4 hours and it would point to 04:00 on the clock. -- THIRD TIME it is RIGHT



                                                Using the same math, it would be RIGHT again at 00:00 (FIRST TIME of next day) and so on.






                                                share|improve this answer










                                                New contributor




                                                alwayslearning is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                Check out our Code of Conduct.




















                                                  up vote
                                                  0
                                                  down vote










                                                  up vote
                                                  0
                                                  down vote









                                                  Assumption 1: The regular 12 hour clock-face.

                                                  Assumption 2: Fast here does not mean an offset but that the clock is actually running at a higher speed.




                                                  It should rotate at 60 (false) hours per day (instead of 24 hours per
                                                  day).




                                                  Assuming we start at 00:00 on day 1 (actual time) and set the false clock at 00:00 and switch it ON. -- FIRST TIME it is RIGHT



                                                  By the time it is 08:00 (AM actual time) which 1/3 of the day past, the false time would show as 08:00 too as it would finish 1/3 of 60 hours which is 20 hours or 12 hours false time (one complete rotation from 00:00) + 8 hours and it would point to 08:00 on the clock. -- SECOND TIME it is RIGHT



                                                  By the time it is 04:00 (PM actual time) which 2/3 of the day past, the false time would show as 04:00 too as it would finish 2/3 of 60 hours which is 40 hours or 36 hours false time (3 complete rotations from 00:00) + 4 hours and it would point to 04:00 on the clock. -- THIRD TIME it is RIGHT



                                                  Using the same math, it would be RIGHT again at 00:00 (FIRST TIME of next day) and so on.






                                                  share|improve this answer










                                                  New contributor




                                                  alwayslearning is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                  Check out our Code of Conduct.









                                                  Assumption 1: The regular 12 hour clock-face.

                                                  Assumption 2: Fast here does not mean an offset but that the clock is actually running at a higher speed.




                                                  It should rotate at 60 (false) hours per day (instead of 24 hours per
                                                  day).




                                                  Assuming we start at 00:00 on day 1 (actual time) and set the false clock at 00:00 and switch it ON. -- FIRST TIME it is RIGHT



                                                  By the time it is 08:00 (AM actual time) which 1/3 of the day past, the false time would show as 08:00 too as it would finish 1/3 of 60 hours which is 20 hours or 12 hours false time (one complete rotation from 00:00) + 8 hours and it would point to 08:00 on the clock. -- SECOND TIME it is RIGHT



                                                  By the time it is 04:00 (PM actual time) which 2/3 of the day past, the false time would show as 04:00 too as it would finish 2/3 of 60 hours which is 40 hours or 36 hours false time (3 complete rotations from 00:00) + 4 hours and it would point to 04:00 on the clock. -- THIRD TIME it is RIGHT



                                                  Using the same math, it would be RIGHT again at 00:00 (FIRST TIME of next day) and so on.







                                                  share|improve this answer










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                                                  share|improve this answer








                                                  edited 2 days ago





















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                                                  answered 2 days ago









                                                  alwayslearning

                                                  21714




                                                  21714




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                                                      up vote
                                                      0
                                                      down vote













                                                      There is a slight ambiguity/lack of precision in the question wording which (in my interpretation) means there isn't a definitive answer as other answers suggest.



                                                      Up front, my answer is:




                                                      A clock that is more than 2 times faster and less than or equal to 2.5 times faster than a perfect clock, can be observed to be correct 3 times in one day




                                                      First I'd like to introduce a concept that makes the rest of this answer easier to explain



                                                      Suppose I have a perfect clock, that is accurate and on time. Suppose I also have a broken clock that I position so it shows a time one minute before the current time on the perfect clock, and then I move the minute hand of the broken clock so it shows a time one minute after the perfect clock. At some point during that motion, the broken clock was showing a time that was accurate. It occurred just (infinitesimally) prior to the exact moment the broken clock time moved past the perfect clock time.



                                                      Hence, when a clock is faulty (running fast) and it "overtakes" a perfect clock, for the briefest of moments the time the faulty clock is observed to be perfectly accurate.



                                                      We must hence work out how fast a clock must run to overtake a perfect clock 3 times in a 24 hour period






                                                      • If a faulty clock runs twice as fast as perfect clock, and we set them going both showing 12:00, they are in sync. This is the first time they are in sync. Let's also say it's Monday

                                                      • 6 hours after we started, the faulty fast clock is back at the start position, but the perfect clock is only half way round

                                                      • 12 hours after we started, the faulty fast clock will overtake the perfect clock and it's the second time it has happened

                                                      • 24 hours after we started, the fast clock again overtakes the perfect clock. It does so on the exact stroke of midnight but this means it's a new day (i.e. it's now Tuesday)

                                                      • Our fast clock only overtook the perfect clock twice - at the outset, and at noon




                                                      This means...




                                                      ... 2 times normal speed isn't enough to get the fast clock to overtake the perfect clock 3 times




                                                      However, if a clock runs...




                                                      ... just fractionally faster than 2 times normal speed, then it will overtake a perfect clock 3 times in one day. If we have a clock that is a double speed plus one second faster (i.e. it counts (86400 * 2) + 1 seconds in a single day) then from being synced to a perfect clock on midnight monday (first overtake) it will overtake the perfect clock at around 11:59:59.5 (half a second to noon) and again at around 23:59:59 (just before it becomes Tuesday)




                                                      Hence the answer given above..



                                                      But before I finish:



                                                      A Special Note




                                                      2.5 times is a special speed factor; it's the only speed factor where the faulty fast clock will overtake (=show the correct time) three times a day, day in day out without being manipulated further. Any speed factor slower than this will, after some variable number of days depending on the speed factor, result in a clock that has too much catching up to do at the start of the day and won't run fast enough to overtake the perfect clock 3 times in the day. How often these "can't overtake 3 times" days occur is a function of how slow the clock is. Values near 2x speed will take a long time to get back to a state where they can have a "3 overtakes" day. Values near 2.5x speed will have many days in a row where they perform 3 overtakes, only occasionally dipping to a 2-overtake day




                                                      The question didn't call for every day (for the rest of the clock's life) being a 3-overtake day, only that a faulty clock should show the correct time 3 times in a day, hence why I feel the answer has to be..




                                                      ..a range







                                                      share|improve this answer

























                                                        up vote
                                                        0
                                                        down vote













                                                        There is a slight ambiguity/lack of precision in the question wording which (in my interpretation) means there isn't a definitive answer as other answers suggest.



                                                        Up front, my answer is:




                                                        A clock that is more than 2 times faster and less than or equal to 2.5 times faster than a perfect clock, can be observed to be correct 3 times in one day




                                                        First I'd like to introduce a concept that makes the rest of this answer easier to explain



                                                        Suppose I have a perfect clock, that is accurate and on time. Suppose I also have a broken clock that I position so it shows a time one minute before the current time on the perfect clock, and then I move the minute hand of the broken clock so it shows a time one minute after the perfect clock. At some point during that motion, the broken clock was showing a time that was accurate. It occurred just (infinitesimally) prior to the exact moment the broken clock time moved past the perfect clock time.



                                                        Hence, when a clock is faulty (running fast) and it "overtakes" a perfect clock, for the briefest of moments the time the faulty clock is observed to be perfectly accurate.



                                                        We must hence work out how fast a clock must run to overtake a perfect clock 3 times in a 24 hour period






                                                        • If a faulty clock runs twice as fast as perfect clock, and we set them going both showing 12:00, they are in sync. This is the first time they are in sync. Let's also say it's Monday

                                                        • 6 hours after we started, the faulty fast clock is back at the start position, but the perfect clock is only half way round

                                                        • 12 hours after we started, the faulty fast clock will overtake the perfect clock and it's the second time it has happened

                                                        • 24 hours after we started, the fast clock again overtakes the perfect clock. It does so on the exact stroke of midnight but this means it's a new day (i.e. it's now Tuesday)

                                                        • Our fast clock only overtook the perfect clock twice - at the outset, and at noon




                                                        This means...




                                                        ... 2 times normal speed isn't enough to get the fast clock to overtake the perfect clock 3 times




                                                        However, if a clock runs...




                                                        ... just fractionally faster than 2 times normal speed, then it will overtake a perfect clock 3 times in one day. If we have a clock that is a double speed plus one second faster (i.e. it counts (86400 * 2) + 1 seconds in a single day) then from being synced to a perfect clock on midnight monday (first overtake) it will overtake the perfect clock at around 11:59:59.5 (half a second to noon) and again at around 23:59:59 (just before it becomes Tuesday)




                                                        Hence the answer given above..



                                                        But before I finish:



                                                        A Special Note




                                                        2.5 times is a special speed factor; it's the only speed factor where the faulty fast clock will overtake (=show the correct time) three times a day, day in day out without being manipulated further. Any speed factor slower than this will, after some variable number of days depending on the speed factor, result in a clock that has too much catching up to do at the start of the day and won't run fast enough to overtake the perfect clock 3 times in the day. How often these "can't overtake 3 times" days occur is a function of how slow the clock is. Values near 2x speed will take a long time to get back to a state where they can have a "3 overtakes" day. Values near 2.5x speed will have many days in a row where they perform 3 overtakes, only occasionally dipping to a 2-overtake day




                                                        The question didn't call for every day (for the rest of the clock's life) being a 3-overtake day, only that a faulty clock should show the correct time 3 times in a day, hence why I feel the answer has to be..




                                                        ..a range







                                                        share|improve this answer























                                                          up vote
                                                          0
                                                          down vote










                                                          up vote
                                                          0
                                                          down vote









                                                          There is a slight ambiguity/lack of precision in the question wording which (in my interpretation) means there isn't a definitive answer as other answers suggest.



                                                          Up front, my answer is:




                                                          A clock that is more than 2 times faster and less than or equal to 2.5 times faster than a perfect clock, can be observed to be correct 3 times in one day




                                                          First I'd like to introduce a concept that makes the rest of this answer easier to explain



                                                          Suppose I have a perfect clock, that is accurate and on time. Suppose I also have a broken clock that I position so it shows a time one minute before the current time on the perfect clock, and then I move the minute hand of the broken clock so it shows a time one minute after the perfect clock. At some point during that motion, the broken clock was showing a time that was accurate. It occurred just (infinitesimally) prior to the exact moment the broken clock time moved past the perfect clock time.



                                                          Hence, when a clock is faulty (running fast) and it "overtakes" a perfect clock, for the briefest of moments the time the faulty clock is observed to be perfectly accurate.



                                                          We must hence work out how fast a clock must run to overtake a perfect clock 3 times in a 24 hour period






                                                          • If a faulty clock runs twice as fast as perfect clock, and we set them going both showing 12:00, they are in sync. This is the first time they are in sync. Let's also say it's Monday

                                                          • 6 hours after we started, the faulty fast clock is back at the start position, but the perfect clock is only half way round

                                                          • 12 hours after we started, the faulty fast clock will overtake the perfect clock and it's the second time it has happened

                                                          • 24 hours after we started, the fast clock again overtakes the perfect clock. It does so on the exact stroke of midnight but this means it's a new day (i.e. it's now Tuesday)

                                                          • Our fast clock only overtook the perfect clock twice - at the outset, and at noon




                                                          This means...




                                                          ... 2 times normal speed isn't enough to get the fast clock to overtake the perfect clock 3 times




                                                          However, if a clock runs...




                                                          ... just fractionally faster than 2 times normal speed, then it will overtake a perfect clock 3 times in one day. If we have a clock that is a double speed plus one second faster (i.e. it counts (86400 * 2) + 1 seconds in a single day) then from being synced to a perfect clock on midnight monday (first overtake) it will overtake the perfect clock at around 11:59:59.5 (half a second to noon) and again at around 23:59:59 (just before it becomes Tuesday)




                                                          Hence the answer given above..



                                                          But before I finish:



                                                          A Special Note




                                                          2.5 times is a special speed factor; it's the only speed factor where the faulty fast clock will overtake (=show the correct time) three times a day, day in day out without being manipulated further. Any speed factor slower than this will, after some variable number of days depending on the speed factor, result in a clock that has too much catching up to do at the start of the day and won't run fast enough to overtake the perfect clock 3 times in the day. How often these "can't overtake 3 times" days occur is a function of how slow the clock is. Values near 2x speed will take a long time to get back to a state where they can have a "3 overtakes" day. Values near 2.5x speed will have many days in a row where they perform 3 overtakes, only occasionally dipping to a 2-overtake day




                                                          The question didn't call for every day (for the rest of the clock's life) being a 3-overtake day, only that a faulty clock should show the correct time 3 times in a day, hence why I feel the answer has to be..




                                                          ..a range







                                                          share|improve this answer












                                                          There is a slight ambiguity/lack of precision in the question wording which (in my interpretation) means there isn't a definitive answer as other answers suggest.



                                                          Up front, my answer is:




                                                          A clock that is more than 2 times faster and less than or equal to 2.5 times faster than a perfect clock, can be observed to be correct 3 times in one day




                                                          First I'd like to introduce a concept that makes the rest of this answer easier to explain



                                                          Suppose I have a perfect clock, that is accurate and on time. Suppose I also have a broken clock that I position so it shows a time one minute before the current time on the perfect clock, and then I move the minute hand of the broken clock so it shows a time one minute after the perfect clock. At some point during that motion, the broken clock was showing a time that was accurate. It occurred just (infinitesimally) prior to the exact moment the broken clock time moved past the perfect clock time.



                                                          Hence, when a clock is faulty (running fast) and it "overtakes" a perfect clock, for the briefest of moments the time the faulty clock is observed to be perfectly accurate.



                                                          We must hence work out how fast a clock must run to overtake a perfect clock 3 times in a 24 hour period






                                                          • If a faulty clock runs twice as fast as perfect clock, and we set them going both showing 12:00, they are in sync. This is the first time they are in sync. Let's also say it's Monday

                                                          • 6 hours after we started, the faulty fast clock is back at the start position, but the perfect clock is only half way round

                                                          • 12 hours after we started, the faulty fast clock will overtake the perfect clock and it's the second time it has happened

                                                          • 24 hours after we started, the fast clock again overtakes the perfect clock. It does so on the exact stroke of midnight but this means it's a new day (i.e. it's now Tuesday)

                                                          • Our fast clock only overtook the perfect clock twice - at the outset, and at noon




                                                          This means...




                                                          ... 2 times normal speed isn't enough to get the fast clock to overtake the perfect clock 3 times




                                                          However, if a clock runs...




                                                          ... just fractionally faster than 2 times normal speed, then it will overtake a perfect clock 3 times in one day. If we have a clock that is a double speed plus one second faster (i.e. it counts (86400 * 2) + 1 seconds in a single day) then from being synced to a perfect clock on midnight monday (first overtake) it will overtake the perfect clock at around 11:59:59.5 (half a second to noon) and again at around 23:59:59 (just before it becomes Tuesday)




                                                          Hence the answer given above..



                                                          But before I finish:



                                                          A Special Note




                                                          2.5 times is a special speed factor; it's the only speed factor where the faulty fast clock will overtake (=show the correct time) three times a day, day in day out without being manipulated further. Any speed factor slower than this will, after some variable number of days depending on the speed factor, result in a clock that has too much catching up to do at the start of the day and won't run fast enough to overtake the perfect clock 3 times in the day. How often these "can't overtake 3 times" days occur is a function of how slow the clock is. Values near 2x speed will take a long time to get back to a state where they can have a "3 overtakes" day. Values near 2.5x speed will have many days in a row where they perform 3 overtakes, only occasionally dipping to a 2-overtake day




                                                          The question didn't call for every day (for the rest of the clock's life) being a 3-overtake day, only that a faulty clock should show the correct time 3 times in a day, hence why I feel the answer has to be..




                                                          ..a range








                                                          share|improve this answer












                                                          share|improve this answer



                                                          share|improve this answer










                                                          answered 5 hours ago









                                                          Caius Jard

                                                          837153




                                                          837153






















                                                              up vote
                                                              -1
                                                              down vote













                                                              Answer:




                                                              It must move 3 seconds for every 2 seconds that pass, so 1.5 times as fast as a normal clock.




                                                              Reason:




                                                              In a 24 hour period, 36 hours will have transpired on this faster clock, meaning it will have made 3 rotations within the day. At some point in each of those rotations, the correct time will have been passed, meaning it is right 3 times a day.







                                                              share|improve this answer

















                                                              • 3




                                                                You may want to check your premise: Using your example, if the clock shows the correct time at Midnight, it will not show the correct time again until the following Midnight...
                                                                – Chronocidal
                                                                2 days ago










                                                              • @Chronocidal dang u right, i think my brain was working on a fast clock vs a stopped clock lol
                                                                – AHKieran
                                                                2 days ago










                                                              • You did manage to fool at least 2 more people though ;)
                                                                – Geliormth
                                                                2 days ago















                                                              up vote
                                                              -1
                                                              down vote













                                                              Answer:




                                                              It must move 3 seconds for every 2 seconds that pass, so 1.5 times as fast as a normal clock.




                                                              Reason:




                                                              In a 24 hour period, 36 hours will have transpired on this faster clock, meaning it will have made 3 rotations within the day. At some point in each of those rotations, the correct time will have been passed, meaning it is right 3 times a day.







                                                              share|improve this answer

















                                                              • 3




                                                                You may want to check your premise: Using your example, if the clock shows the correct time at Midnight, it will not show the correct time again until the following Midnight...
                                                                – Chronocidal
                                                                2 days ago










                                                              • @Chronocidal dang u right, i think my brain was working on a fast clock vs a stopped clock lol
                                                                – AHKieran
                                                                2 days ago










                                                              • You did manage to fool at least 2 more people though ;)
                                                                – Geliormth
                                                                2 days ago













                                                              up vote
                                                              -1
                                                              down vote










                                                              up vote
                                                              -1
                                                              down vote









                                                              Answer:




                                                              It must move 3 seconds for every 2 seconds that pass, so 1.5 times as fast as a normal clock.




                                                              Reason:




                                                              In a 24 hour period, 36 hours will have transpired on this faster clock, meaning it will have made 3 rotations within the day. At some point in each of those rotations, the correct time will have been passed, meaning it is right 3 times a day.







                                                              share|improve this answer












                                                              Answer:




                                                              It must move 3 seconds for every 2 seconds that pass, so 1.5 times as fast as a normal clock.




                                                              Reason:




                                                              In a 24 hour period, 36 hours will have transpired on this faster clock, meaning it will have made 3 rotations within the day. At some point in each of those rotations, the correct time will have been passed, meaning it is right 3 times a day.








                                                              share|improve this answer












                                                              share|improve this answer



                                                              share|improve this answer










                                                              answered 2 days ago









                                                              AHKieran

                                                              3,638632




                                                              3,638632








                                                              • 3




                                                                You may want to check your premise: Using your example, if the clock shows the correct time at Midnight, it will not show the correct time again until the following Midnight...
                                                                – Chronocidal
                                                                2 days ago










                                                              • @Chronocidal dang u right, i think my brain was working on a fast clock vs a stopped clock lol
                                                                – AHKieran
                                                                2 days ago










                                                              • You did manage to fool at least 2 more people though ;)
                                                                – Geliormth
                                                                2 days ago














                                                              • 3




                                                                You may want to check your premise: Using your example, if the clock shows the correct time at Midnight, it will not show the correct time again until the following Midnight...
                                                                – Chronocidal
                                                                2 days ago










                                                              • @Chronocidal dang u right, i think my brain was working on a fast clock vs a stopped clock lol
                                                                – AHKieran
                                                                2 days ago










                                                              • You did manage to fool at least 2 more people though ;)
                                                                – Geliormth
                                                                2 days ago








                                                              3




                                                              3




                                                              You may want to check your premise: Using your example, if the clock shows the correct time at Midnight, it will not show the correct time again until the following Midnight...
                                                              – Chronocidal
                                                              2 days ago




                                                              You may want to check your premise: Using your example, if the clock shows the correct time at Midnight, it will not show the correct time again until the following Midnight...
                                                              – Chronocidal
                                                              2 days ago












                                                              @Chronocidal dang u right, i think my brain was working on a fast clock vs a stopped clock lol
                                                              – AHKieran
                                                              2 days ago




                                                              @Chronocidal dang u right, i think my brain was working on a fast clock vs a stopped clock lol
                                                              – AHKieran
                                                              2 days ago












                                                              You did manage to fool at least 2 more people though ;)
                                                              – Geliormth
                                                              2 days ago




                                                              You did manage to fool at least 2 more people though ;)
                                                              – Geliormth
                                                              2 days ago










                                                              up vote
                                                              -2
                                                              down vote













                                                              The speed has to be




                                                              -1 so the clock will be at the same spot at 00, 06 and 18







                                                              share|improve this answer










                                                              New contributor




                                                              user54121 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                              Check out our Code of Conduct.


















                                                              • Welcome to Puzzling SE. Could you add more explanation to what you mean by a speed of -1? Also if you'd like to learn more about this site and earn your first badge, check out the tour at this link: puzzling.stackexchange.com/tour
                                                                – gabbo1092
                                                                2 days ago






                                                              • 2




                                                                This would be right at 12 too, wouldn't it?
                                                                – jafe
                                                                2 days ago















                                                              up vote
                                                              -2
                                                              down vote













                                                              The speed has to be




                                                              -1 so the clock will be at the same spot at 00, 06 and 18







                                                              share|improve this answer










                                                              New contributor




                                                              user54121 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                              Check out our Code of Conduct.


















                                                              • Welcome to Puzzling SE. Could you add more explanation to what you mean by a speed of -1? Also if you'd like to learn more about this site and earn your first badge, check out the tour at this link: puzzling.stackexchange.com/tour
                                                                – gabbo1092
                                                                2 days ago






                                                              • 2




                                                                This would be right at 12 too, wouldn't it?
                                                                – jafe
                                                                2 days ago













                                                              up vote
                                                              -2
                                                              down vote










                                                              up vote
                                                              -2
                                                              down vote









                                                              The speed has to be




                                                              -1 so the clock will be at the same spot at 00, 06 and 18







                                                              share|improve this answer










                                                              New contributor




                                                              user54121 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                              Check out our Code of Conduct.









                                                              The speed has to be




                                                              -1 so the clock will be at the same spot at 00, 06 and 18








                                                              share|improve this answer










                                                              New contributor




                                                              user54121 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                              Check out our Code of Conduct.









                                                              share|improve this answer



                                                              share|improve this answer








                                                              edited 2 days ago









                                                              gabbo1092

                                                              4,539635




                                                              4,539635






                                                              New contributor




                                                              user54121 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                              Check out our Code of Conduct.









                                                              answered 2 days ago









                                                              user54121

                                                              1




                                                              1




                                                              New contributor




                                                              user54121 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                              Check out our Code of Conduct.





                                                              New contributor





                                                              user54121 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                              Check out our Code of Conduct.






                                                              user54121 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                                                              Check out our Code of Conduct.












                                                              • Welcome to Puzzling SE. Could you add more explanation to what you mean by a speed of -1? Also if you'd like to learn more about this site and earn your first badge, check out the tour at this link: puzzling.stackexchange.com/tour
                                                                – gabbo1092
                                                                2 days ago






                                                              • 2




                                                                This would be right at 12 too, wouldn't it?
                                                                – jafe
                                                                2 days ago


















                                                              • Welcome to Puzzling SE. Could you add more explanation to what you mean by a speed of -1? Also if you'd like to learn more about this site and earn your first badge, check out the tour at this link: puzzling.stackexchange.com/tour
                                                                – gabbo1092
                                                                2 days ago






                                                              • 2




                                                                This would be right at 12 too, wouldn't it?
                                                                – jafe
                                                                2 days ago
















                                                              Welcome to Puzzling SE. Could you add more explanation to what you mean by a speed of -1? Also if you'd like to learn more about this site and earn your first badge, check out the tour at this link: puzzling.stackexchange.com/tour
                                                              – gabbo1092
                                                              2 days ago




                                                              Welcome to Puzzling SE. Could you add more explanation to what you mean by a speed of -1? Also if you'd like to learn more about this site and earn your first badge, check out the tour at this link: puzzling.stackexchange.com/tour
                                                              – gabbo1092
                                                              2 days ago




                                                              2




                                                              2




                                                              This would be right at 12 too, wouldn't it?
                                                              – jafe
                                                              2 days ago




                                                              This would be right at 12 too, wouldn't it?
                                                              – jafe
                                                              2 days ago





                                                              protected by JonMark Perry 20 hours ago



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