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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    $\begingroup$ I'm a bit confused. So, the clock is stopped or it works? $\endgroup$ – rhsquared Nov 22 '18 at 9:22
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    $\begingroup$ @rhsquared There are two clocks :) $\endgroup$ – Anush Nov 22 '18 at 9:27
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    $\begingroup$ 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 ? $\endgroup$ – Ideogram Nov 23 '18 at 6:57
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    $\begingroup$ A stopped clock will tell the right time three times a day, if it's the day daylight saving ends ... $\endgroup$ – Julia Hayward Nov 23 '18 at 7:50
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    $\begingroup$ @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. $\endgroup$ – Bass Nov 23 '18 at 11:31

12 Answers 12


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.

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    $\begingroup$ What’s the general solution for the clock to be correct x times a day? $\endgroup$ – Anush Nov 22 '18 at 23:07
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    $\begingroup$ @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. $\endgroup$ – Bass Nov 23 '18 at 5:13
  • $\begingroup$ what happens if x=0? $\endgroup$ – JMP Nov 24 '18 at 13:40
  • $\begingroup$ @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.) $\endgroup$ – Bass Nov 24 '18 at 13:51
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    $\begingroup$ @Manuel, the English phrase "three times a day" means three times every day. If the clock were running backwards at one second each day, there would always be a couple of centuries between such days. $\endgroup$ – Bass Nov 26 '18 at 16:22

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.


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.

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    $\begingroup$ Nice and different. $\endgroup$ – Deduplicator Nov 22 '18 at 16:22

It should be right three times a day when

its stopped during Daylight Saving Time at 1:00


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)


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.


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.

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    $\begingroup$ Which are the hours it would be correct? $\endgroup$ – Anush Nov 22 '18 at 21:49

The clock needs to be moving at

2.5 times normal speed


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


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.

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    $\begingroup$ The same argument for digital clocks $\endgroup$ – gota Nov 23 '18 at 17:16

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 3 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)

Now, I initially started the fast clock at the same time as the perfect clock, meaning they were on an overtake from the start. What if I put the fast clock at the maximum disadvantage? If I start the fast clock at one second ahead of the perfect clock it's at a considerable disadvantage in the race and has a huge amount of catching up to do before it will register the same time as the perfect clock.

With a perfect clock at midnight and a faulty clock at midnight+1 second with a running rate of 3 times faster, the faulty clock will first catch up at 1 second past 6am, again at 1 second past noon, thirdly at 1 second past 6pm and the next time it registers the same time is at 1 second past midnight on the following day. If the fast clock goes any faster, 4 overtakes will occur in the day

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 or 3x 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


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.



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


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.

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    $\begingroup$ 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... $\endgroup$ – Chronocidal Nov 22 '18 at 9:38
  • $\begingroup$ @Chronocidal dang u right, i think my brain was working on a fast clock vs a stopped clock lol $\endgroup$ – AHKieran Nov 22 '18 at 9:54
  • $\begingroup$ You did manage to fool at least 2 more people though ;) $\endgroup$ – Hans Janssen Nov 22 '18 at 11:45

The speed has to be

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

  • $\begingroup$ 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 $\endgroup$ – gabbo1092 Nov 22 '18 at 14:27
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    $\begingroup$ This would be right at 12 too, wouldn't it? $\endgroup$ – jafe Nov 22 '18 at 14:27

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