I remember that some old analog clocks don't have continuous movement on the minute hand, instead it skips forward in one minute increments. In fact the ratchet mechanism involved causes the hand to sweep back a few degrees for a moment before it snaps forward to the next minute. In this case you could have the hands overlap upto 3 times in an hour, depending on where the hour hand is compared to the distance of the back sweep. The snarky solution to puzzle is that the hands are always overlapped when you consider the center point.
Arguably, at the center point it's not the hands that overlap, but the _bases_ of the hands that overlap. Just as the base of a statue is not the statue, the base of a hand is not the hand.
There's also 24h analog clocks, just to make things more annoying. And clocks with a second hand in addition to minute and hour. There's probably one that has fully seperate displays for each hand, as well.
@@StephenByersJ Yeah but it still had specific definitions like what overlaps at 12 count and which don't which is why most probably got it wrong by just 1 or 2.
Not so fast. An electronic sensor system would get a different count than a mathematician because on a physical clock the two hands will overlap even when they are not pointing at the same point simply because the physical hands are too thick.
The overlaps would be longer, but the amount of them would remain the same except for the very last time, where the overlap would begin right before the deadline is hit.
The fact that you can form an infinite series, is an amazing illustration of how mathematicians defined real numbers, it's a limit of rational numbers, even though the minute hand eventually catches up, but that time stamp is never a precise one
The (limit of the) infinite sum is still rational though. The time stamp is "never precise" merely because we haven't divided the hour in a multiple of 11 minutes. So I'd say that this particular infinite series has nothing to do with the definition of real numbers.
..hum the precice time is for n[1 to 22] = n(720)/11 minuits wich is about 65.45 minuits, or excatly 65 minuits, 27 seconds and (9090909090898153381683397656577 / 33333333333293229066172458074116) hundreds of a second (aprox 27.3) times n, where n is number of hours ..can also be written as n60(12/11) or n60 * (1 + 1/11) ...n is 11 for each rotation of the minuit hand n*2 rotations = 22
@@Patrik6920 The interval between two consecutive overlaps is exactly 65 minutes and (27 + 3/11) seconds. (note: 27 + 3/11 = 27.27272727... ) And not "minuit", because _minuit_ is French and means "midnight".
I used to think of this problem as a child. It is like the Achilles-tortoise parable. Think of 1 o'clock. the hour hand is the tortoise and Achilles is the minute hand. The minute hand will pass the hour hand even though it subdivides into an infinite number of increments. As in your last solution.
ah yeah, the old Achilles-tortoise parable that we all thought of as a child. Old news, everyone knows it as a way of telling time. I often say It's about Achilles past Tortoise, and then leave it to the stranger on the street to subdivide it into an infinite number of increments.
This is analogous to a general fact in astronomy, where there is 1 more sidereal days or months than synodic days or months per year (given the rotation of the planet around its axis is the same as around the sun, which is not true for Venus). Here a "synodic hour" is an overlap and a "sidereal hour" is a normal hour. Thus 11 "synodic hours" per one "clock year" or 11 overlaps for half a day.
Exactly what I was thinking. Applies to number of rotations of the Earth about its axis per year and number of rotations of the Moon about the Earth per year. If rotations are considered relative to the fixed stars, you get one more rotation in each case than the way we usually count, based on the position of the sun.
1:50 If you're initially assuming once for every hour of the day then that would be 24, one each for the 24 hours from 0 to 23 using 24 hour clock, so one of the midnights is already not included, otherwise you'd start with 25 hours in a day (0-23 and another 0) and subtract 1 to get 24.
@@9adam4 Yes, we know. The point here is that Presh's intuitive reasoning to arrive at 23 ("Once every hour makes 24, but then I subtract 1 in order to not double count the end points") doesn't make sense.
Following the idea of one of the comments. In 24 hours, the minute hand runs 24 circles, the hour hand runs 2 circles, so the minute hand overpasses the hour hand 24-2 = 22 times, or in other words, they overlap 22 times. The interval between two overlaps is apparently a constant, so the interval is 24/22 hours (h), or 12/11 (h), which is 1 hour 5 minutes 27 and 3/11 seconds, and the kth overlap simply corresponds to the time (12/11)k (h), k = 1, 2, 3, ..., 22. When k = 11, it is the noon time; when k = 22, it is the end of the day, or 24 (h). k can take the value 23 or more, but they represent overlaps of the next day.
Got it right the first time. I went empirically through the meeting points and no later than near the bottom of the clock I realized we can't have the minutes hand at the half hour mark and the hour hand at the full 6th hour mark. Then I sketched it up on a sheet of paper and saw that the meeting point happens at 12:00 as well as intermediately between every two hours except the first and the last. This puzzle reminds me of the fact that even though we colloquially know that Earth rotates 365 days per year, making 365 days, it actually rotates 366 times per year if you look at it from a fixed reference frame outside of the solar system. Our reference frame normally is fixated on the sun, so it rotates one turn per year, adding one turn to the 365 turns we observe relative to the sun.
More precisely, 1 Orbit contains 365.242374 days, which would mean the earth rotates 366.242374 times. We compensate with leap years. To approximate 365¼ we have a leap year every fourth year (so 365.25), to approximate the slight error in that (≈0.01) we skip it if the year is also divisible by 100 (so 365.24). To further compensate for the ≈0.002 error in that, we intentionally fail to skip it on years divisible by 400... fun stuff
Btw, a reference frame fixated on the sun is not "a fixed reference frame outside the solar system"... the sun is very much in the solar system (where you got your ≈366 rotations). It would be the same outside the solar system too, until you get far enough away to consider the rotation of the galaxy and such.
@@parodoxis I referred to the fact that normally in our earthly life, our reference frame for counting days is based on the vector from Earth to Sun, and that vector rotates once per year around the Sun, as does Earth.
@@MrSaemichlaus right, and so did I, that's rotation # 366. No confusion there, did you instead mean to clarify your reference to using a frame "outside our solar system"? That's the part I didn't get. Either way, no worries.
This problem has a special place in my heart and still irks me to this day. Not sure where I first met it in HS, but the same as presented here. It didn't specify how many hands (I thought Hour / Minute / Second) as the question also specified find to the nearest second. So I went about solving it as though all 3 hands had to overlap =____=, got it wrong even though i thought it was kind of straight forward. per min: hour hand (0.5 degrees) | min hand (6 degrees) | sec hand (360 degrees) relative speed (hour v min) 5.5 / min >> 360/5.5 = 720/11 to overlap (720 mins is 12 hrs, so every 12 hours there are 11 overlap) relative speed (hour v sec) 0.5 / min >> 360/0.5 = 720 to overlap ... so once everyone 12 hours... so all three overlap at exactly noon and midnight, twice. aside: Saw this question again at at some years later (prob interview, don't remember), and remembering that it was poorly worded, i calculated the total overlaps (second v min, second v hour, min v hour) and got it wrong again =_____=
How do you know you got it wrong? How do I know you got it wrong. You don't state your answer to the question and you don't state what you believe to be the correct answer.
Writing this before I watch the video. If you just want the number of times, you can quickly recognize that you only need to consider the first 12 hours because the next 12 hours are duplicated positions. Then, your logic of "subtract 1 for the minute hand passing the hour hand" does indeed give you the correct number of 11 times in a half day. Thus, there are 22 overlaps in a day. If you want the actual times, use basic modular arithmetic. Using the hour ticks as the unit of circular distance and denoting time passed T in units of hours, the hour hand travels at a rate of T and the minute hand travels at a rate of 12T (so example, when T = 1.5, then 1.5 hours has passed, the hour hand traveled 1.5 units and the minute hand traveled 18 units). The hour and minute hand are lined up when 12T = T mod 12, or in other words, when 11T = 0 mod 12. So just start enumerating the instances: 11T = 0 -> T = 0. So at 00:00, they line up. 11T = 12 -> T = 12/11. So when 12/11 hours passed, they line up again (the time would be 01:05.27) At this point, you can keep going or just recognize that you can get the rest of the answers by adding an additional time delta of 01:05.27. Keep on finding solutions for as long as T < 24 and you'll see that there are only 22 values of T which can satisfy the modular equation.
Another way, once you discover they overlap 22 times in a day, is to calculate the total seconds in the day, 24*60*60 = 86,400 seconds, divide by the 22 overlaps, and you get ~3927.2 seconds, which equates to 1 hour, 5 minutes, 27 seconds. Just keep adding them to get the times, to the nearest second, that the hands overlap.
I answered 23 based on the clock hands having non-zero width. That would account for the last (albeit not total) overlap starting before midnight. It should have said "total" or"full" overlap to make it clear the endpoint was not included. When doing a lap around a circle you could use a half-open interval, but most would think that both the start and the stop points are included, hence a closed interval.
You're right. If you start the clock at midnight then the hands are overlapping. If you stop it the following midnight then they are also overlapping. You can't just count the first time and ignore the second. It's an arbitrary decision to assign midnight as the start of a day, and not the end of the previous day. That's a human construct, not a mathematical calculation of how many times the hands overlap. Either ignore both ends and only count the overlaps occurring in the duration of the day, or count both ends. You can't have half an overlap, and you can't arbitrarily assign it to a certain day.
So many people are doing this problem wrong. Step one: create a pre-planning meeting to establish a strategy for acquiring a random sample of different clocks. … step fourteen: determine which of the squired clocks comply with the ethical sourcing standards we were supposed to be using, discarding the rest …. step one hundred-four: apply for an out-of-budget request for an additional $1.2 million for the overtime required to prepare for the pre-meeting for the project ….step fifteen-thousand, six-hundred, forty-one: submit the final version of the 4,231-page environmental impact report to the internal review committee, now in the correct font.
Think of clock hand positions as complex numbers on the unit circle. If z is the position of the hour hand, the minute hand is at z^12. (Clock is on its side.). They meet when z = z^12. This resolves to z(z^11-1) = 0. Since z=0 is not the kind of solution we want, there are 11 distinct solutions on the unit circle. Thus 22 for each 24 hour period.
@@OLJeffo Justify doing what you propose in your original comment. The argument is implied in the fact that the correct answer is 23 so either your math is wrong or your approach is wrong. That's how things work.
@@peterbaruxis2511 The comment is the justification, although i admit it was intended for people who have taken an undergrad complex analysis course. Listen, I've read your main comment and it seems like you aren't a total crank, so I'll add this: I am assuming that every moment in time belongs to exactly one day. If you want to assume otherwise, my proof isn't for you. But in that case, the modern world might not be for you, since so many technological and financial systems depend on this assumption!
@@OLJeffo Calculus would look at the distance between -1 and 1 and see zero differently than would a complex analyst, both probably have their uses. I wouldn't make assumptions about people so quickly.
I once tried doing this seemingly interesting topic as early as 5th grade when I tried it manually for times when hour and minute hand overlap. Anyways, recently in college (3rd year), we are being taught Aptitude where this clock topic also came. And during 5th grade, I did it this way: The hands will overlap between (starting at 12 PM): 1-2 PM 2-3 PM 3-4 PM 4-5 PM 5-6 PM 6-7 PM 7-8 PM 8-9 PM 9-10 PM 10-11 PM at 12 AM 11 times in 12 hours, so 22 times in 24 hours. 12/11=1.0909 Multiply decimal part by 60 for minutes and decimal of resultant by 60 for seconds. So about 1 hour, 5 min, 27 sec.
This is the simplest, most intuitive answer of all the comments. I feel like a lot of people assume the answer to be something and that throws them off, including me. But also before watching the video I didn't know if we count 12 twice or not
If you did that in fifth grade that's impressive. If I were your fifth grade teacher I'd have praised your methodical approach but I would have asked you to reconsider the decision to analyze half of the question and double that conclusion- and think about how that might not lead to the same conclusion as methodically following the problem to it's end. If after that you still seemed engaged I'd ask you to calculate for one quarter of the day and multiply that by four. (that would likely lead to an answer of 27 or 26.)
@@peterbaruxis2511the difference is that the two halves of a day are the exact same route on the clockface (one revolution) and thus are identical. Also, one quarter of the day would lead to the answer of 6x4=24
Something is missing here. 1:05:27 + 1:05:27 should be 2:10:54 as 7+7 is 4. So i assume that you actually calculated each step with the fraction and did not just add 1:05:27 each time
He should have explained that the rounding to the nearest second influences the result more than 27 seconds every hour because it accumulates every hour.
Per the calculation at 7:06 you would need to add in the additional 3/11 seconds to get an actual value. Note he says "precise" and inserts an approximation symbol. Bad form.
@@jrkorman "Note he says "precise" and inserts an approximation symbol." No, you're misrepresenting what he did. You're making it seem as if at 7:06 he first says "precise" and then inserts an approximation symbol. What he did, was _first_ give an approximation of the interval, in order to give a recognizable representation of how much "1 + 1/11 hours" is, and _then_ he proceeds to calculate and present all 22 "precise times" by repeatedly adding "this interval". Although it's sloppy that he doesn't explicitly mention which value he uses for repeatedly adding, it stands to reason that he would use the exact value of "1 + 1/11 hours" (or "1 hour, 5 minutes, 27 + 3/11 seconds") and not the approximation in order to determine the " _precise_ times" (which he then rounded to the nearest second, of course). Is he wrong to presume that the viewer already knows how to add a fractional value repeatedly (and then round the results to the nearest second)?
@@yurenchu assuming people know rounding, significant figures, and error analysis is bad for the level of instruction he is aiming for. That and he gets the instructions incorrect at 6:55 where he states that “we now want to round this to the nearest second” before doing the additive steps.
No, the _base_ of one hand always overlaps the _base_ of the other hand. But just as the base of a statue is not the statue, the base of a hand is not the hand.
I tackled this in a pretty similar way to solution 3. Let's say x is how many ticks the minute hand covers. The hour hand has some head start 5h, where h is the hour (ex. 1pm = 5 ticks, 3pm = 15 ticks) + x/12 which is how many ticks the hour hand covers past its head start, since it's 1/12 as fast. We can use x = 5h + x/12 to find where the minute hand will be when it covers as many ticks as the hour hand- in other words, when they meet. Doing a bit of algebra gets us 11x/12 = 5h, or x = 60h/11, where h is the hour. Just plug the hours into the equation and you have your answer (ex. if the hour is 2, x = 120/11 ticks or 10 + 10/11 minutes)
22 times. It is enough to stop the hour hand and allow the clock to rotate. Over the course of 12 hours, the face completes one rotation in the opposite direction to the movement of the clock hands. 12-1=11, 11*2=22
@@Nikioko 23 times per day would imply 46 times per two days, and 7*23 = 161 times per week, which are clearly not true (there are only 154 overlaps of hour hand and minute hand per week). If the time of 12:00:00 A.M. (= midnight) belongs to the present day, then the time of one second after 11:59:59 P.M. belongs to the _next_ day.
The second overlap happens around 1:05, pure logic suggests the only way both will be pointing at exactly 12 by the time noon arrives means the min hand fails to catch the hour hand once. And then simply that process is repeated for the pm so 22. Not sure if the interviewer would like my “no math needed” reasoning.
I got 1 Hour, 5 minutes, 27 seconds for the overlapping hands interval (assuming no second hand), and the hands overlap 11 times over 12 hours, and 22 times over the course of a full day.
If I was asked a riddle in a job interview, I'd be like "Am I here to have a serious interview or to play games? How 'bout some "Call of Duty" instead?"
@@ivaerz4977Nah you can convert a variety of problems into any given branch of what is mother science. For eg you can convert the clock problem from a seemingly pure logical one to a physics one. Just think about it ,lets say that the hour hand is a slow old ant moving at 1/12 th the speed of his younger child ant. They are moving in a circle(or rather the circle of death),how many times would they meet each other if the younger person travles at a speed of 12 units per hour if the circle's length is 12 units. The above is an easy question of relative speed/circular velocity.
There's another way to think about the first question which is faster than algebraically, radially, or graphically: The big hand completes 12 full rotations every 12 hour period. So, if the small hand stood still, it would overlap exactly 12 times. But since the small hand makes 1 full rotation *in the same direction* as the big hand, you can subtract 1 to get 11 overlaps. Multiply by 2 for a 24 hour period and you get 22.
This was actually something I spent a lot of time thinking about in high school, so I can say off the top of my head that the answer is 11. Or, 22 times in a 24 hour period.
If you imagine viewing the clock from a rotating frame of reference such that the hour hand is stationary, you are effectively cancelling 2 revolutions per day. Hence the minute hand will pass the hour hand 22 times in a day.
The problem with the question is purely semantic. It is ambiguous whether both the start and the endpoint count. I mean, can you prove 24;00 is Midnight but 0;00 is not?
I remember this problem from high school. I solved it this way, which impressed Miss Sullivan. We know that the hour hand will take one revolution to go from noon to midnight, and at those times the hands align. We also know that the time between each alignment is exactly the same. (To see this rotate clock so that the alignment that happens between 1:05 and 1:10 is now at noon. The next one will land at exactly the same.) So the amount of time between consecutive alignments will be standard. Call that time x. Also let n be the number of alignments in those 12 hours. Since 12:00 is an alignment at noon and midnight, we must have x*n=12, with n an integer. At 1:00 a complete alignment has not been made, so x>1 and hence n1 and a|12, then 12/a is an integer, call it m. Also b>1. x*n = x*a*b = 12 which implies that x*b = 12/a = m. So the b-th alignment happens exactly m hours have passed since 12, or at m o'clock. That means it happens on the hour, but the minute hand is on 12 an not at m. So with n relatively prime to 12, n = 1, 5, 7, or 11. At 1:00 the minute hand had not an alignment. At 1:30, the minute hand has swept over the 1:05 to 1:10 region where the hour hand slowly moves. So 90 minutes or 1.5 hours is too long for x. So 1.5 > x = 12/n. Solving this inequality, we get n>8. The only integer relatively prime to 12 between 8 and 12 is 11. n=11
Is your answer 11? are you asking others to substitute 11 for n in your equations? You do the math as your proof. Was Miss sullivan impressed with ( I don't know if your answer is 11 & she was impressed with that or if she was impressed at your formation of an equation which you did not calculate out to a conclusion.) So far to me Miss Sullivan seems easily impressed.
@@peterbaruxis2511 "Is your answer 11?" My very last thing I wrote was n=11. So the time will align 12/11 hours after noon. And Miss Sullivan was impressed with an 11 year old's logic. Most high school math teachers would be impressed with creative mathematical prowess. (And before you bring it up, I was 11 when I started high school.)
@@peterbaruxis2511he clearly showed the result of his calculation in his post and it was good. Don't know what your issue is. A point in time cannot belong to two days. It is never simultaneously two different days at the same time. No one ever says it is Saturday and Sunday right now. That's why the date and day on your phone changes at midnight.
I stand firmly by 23, since I think it’s actually important and worthwhile to double count midnight. 11 overlaps per 12 hours, plus one once you reach the 12 again. For starters, it’s like a solar eclipse. There isn’t only one instant of overlap, but a range of time for the totally. The minute hand is traditionally a little thinner than the hour hand, and there is a span of time when the entirety of the width of the minute hand is bounded by the wider hour hand. For a few seconds before and after midnight, the hands overlap, even though the angles of either hand are not identical. Recognizing that there is non-exact overlap is a useful physical observation. It seems related to fence post problems and off by one errors. I tend to think it’s better not to just assume you include or exclude the last fence post, but should consider why you want to include it or exclude it. When you just guess before hand, it seems more likely to lead to a mistaken, than considering each potential problem separately, and making a case for or against double counting. If you’re looking at only the exact angle matching with a very calculus-focused analysis, I guess, but maybe I’ve got a more horological view on the puzzle. I do have a background in math, but I’m also a watch collector.
By that argument, you could say the answer is one, as the hour and minute hands are always overlapping. As they meet in the middle of the clock face and spin about the same axis, one has to be overlapping the other. Clearly that's not what the question is asking for, though, so that's why I feel it better to treat it as a thought experiment and assume the hands to have 0 width. Either that or define "overlap" as "exact overlap of the center lines of the hands".
@@quigonkenny Even with zero-size hands, there's still a question about how we handle the interval. We can treat it as open on both ends, closed on both ends, or open on one and and closed on the other. There's a non-absurd argument for twenty-one overlaps in a day, if you have open intervals on both ends. There are no doubt times when it makes sense to exclude the endpoints. The case for including both midnights is basically the trig functions. We tend to say that Sine of an angle is zero at 0, pi, and 2*pi radians. It would seem strange to me to include only one endpoint. Understanding how the cycle works is part of the point, and I think that applies to clocks as well as triangles.
I totally agree. I thought it was 23 as well for that very reason. I knew that I was counting midnight twice because you can't assign it to one day or another. It happens simultaneously so it occurs on both days. One day isn't over until it goes past midnight, and before it goes past they must overlap. It must be counted at the start and at the end.
Both of these are a stretch, though, and you can safely rule them out logically without further clarity in the question. It is not ambiguous that a day does not end with midnight. As soon as we go from PM to AM, we've entered a new day. So 23:59:59.999... is the highest time in any day (excluding leap seconds) by international convention, which you can look up. The fact that physical hands have non-infinitesimal width is fun to consider but you can be sure that's not what the question means because a) the definition of overlap would be arbitrary (what portion must overlap?) b) different clocks have different hands thickness and, crucially, c) every infinitesimal moment during overlap could be considered a "time" they overlapped. Intuitively, removing those absurd or impractical possibilities means interpreting an entire single overlapping session as one "time" of overlap, and since the answer must be true for each day, we can't steal the midnight overlap from the next day.
@@parodoxis You are trying to apply Zeno's paradox. 0.999... is mathematically proven to be equal to one. It's an infinite series that cannot stop until it reaches the actual end of the measurement period. A 24 hour period does not end until the end of the 12th hour. You cannot stop counting until you actually reach midnight, at which time the hands overlap.
If you use Euler's formula to model the hour-hand as e^(i*X), we can model the minute-hand as e^(i*12X) since it's frequency is 12 times faster than the hour-hand. There's an obvious solution at X = 0. To get the rest, we can add the period to the hour-hand as e^(i*X + 2πn) where n is an integer. Set equal to each other and natural log both sides. The resulting solution is that the two hands should meet when X = 2πn/11. Look at the special case where n = 11. This is simply X = 2π, which is the trivial case of both hands pointing at 12 on the clock. This means that for every full rotation of the hour hand, there are 11 crossings between the two hands before they repeat the pattern. 2 full rotations is 1 day, so in total 22 crossings. To get the times, notice that hour 1 on the clock is π/6 radians past the 12 (hour 2 is 2π/6, hour 3 is 3π/6 = π/2, and so on ...). Dividing 2πn/11 (the crossings) by π/6 (1 hour) yields 12n/11 where n is in the range [1, 22]. To convert to seconds, just multiply by 3600 seconds per hour. E.g. the first crossing happens at (12 * 1 / 11) * 3600 seconds = 3927.27 seconds = 3600 seconds + 327.27 seconds = 1hr + 5 min + 27.27 seconds. Repeat for each n in [1, 22] (or [1, 11] and then add 12 hours)
If you think the answer is 22 and you think Euler's formula applies to this you either don't understand the formula or you do understand the formula but you don't know how to apply it.
It doesn't really matter when we agree the day starts. The clock hands overlap once every approx 1hr and 5 mins (little off, but it works). So in the span of 12 hrs, they overlap 11 times (counting the overlap at the begining of each 12 hr period). 11×2 is 22.
I remember we did this in school. If I remember correctly, everyone got it wrong. Our homework assignment was to add the seconds hand and figure out how many times any two or more overlap. But there would be no homework if someone could answer how many times all three overlapped at the same time right away. We had to do the homework. I don’t remember, but I can’t imagine I was too happy when I got home and figured out the answer.
It depends on your definition of "overlap". Mine is "when one hand covers the other by any amount." So, only once at 12:00:00 and continues until someone pulls the top hand off or the day ends. The real puzzle is how anyone thinks riddle solving skills are a good proxy for engineering skills. It's an example of short-cut thinking that says a lot about the organization and very little about the interviewee, IMHO.
how is understanding how a piece of engineering works not a proxy for engineering skills? "So, only once at 12:00:00" - huh? the minutes hand moves faster and passes by the hours hand every 65 minutes (and some seconds); and every time it happens there is some amount of overlap for a while; so what are you trying to say?
I disagree that is is a riddle type question: “Why are manhole covers round?” Is a riddle question: you either have already heard it, have an aha moment in the interview, or you don’t, and it doesn’t really tell you anything. If you started your response with “what do you meant by overlap, because they are always overlapping at the hub?”, no reasonable interviewer would think that was a bad question. It reminds me of a company I was an interviewer at, one of our standard questions was “If we gave you a basketball and a tape measure, how would you estimate how many basketballs you could fit in this room?”, and a candidate who was otherwise acceptable would always give a reasonable answer, and answer the follow up about whether they though their answer would give an under- or over- estimate. Plus it was directly analogous to the kinds of simulations we wrote, without us having to dump a bunch of our context on them.
@@dstrctd Short-cut thinking does sometimes work, I assume. But I wonder how often it was confirmed with real data? In my experience, some of the best engineers failed their interviews but were hired by someone with previous experience of their skills and work ethic. I.e. the interviews failed. Since you never hired anyone who was not "otherwise acceptable" you have no experience with counter examples. In other circles, this would be called "confirmation bias."
12:45 onwards - think of it simply like this. Imagine that the hour hand doesn't move but the minute hand moves (minute hand degrees per minute - hour hand degrees per minute) = 6° - 0.5° = 5.5°. Then divide that by the number of degrees to move as if the hour hand doesn't move - 90° / 3 = 30°. With that you can divide 30° with 5.5° to get (5.45 repeating) minutes. You can then multiply the remainder (0.4545454545) by 60 seconds to get 27.27 repeating seconds.
Even if you thought the hands met once an hour, your original rationale for “23” makes no sense. You started at 24, but then subtracted one because 12 midnight overlapped with the next day. But the only way there would be such overlap is if you counted both the midnight at the start of the day, and at the end of the day-and if you did that, you would have started with 25 occurrences, not 24. Subtracting the one for the overlap then brings you to 24 occurrences, not 23. So your wrong answer was, essentially, twice as wrong as what you thought it was.
The hands of a clock (hour and minute hands) overlap approximately every 65 minutes. In a 12-hour period, they overlap 11 times. Since a day consists of two 12-hour periods, the hands of the clock overlap 22 times in a full 24-hour day. REASONING: The minute hand completes one full revolution (360 degrees) every 60 minutes. The hour hand completes one full revolution every 12 hours, which means it moves 30 degrees every hour (or 0.5 degrees per minute). To find the time it takes for the hands to overlap, consider the following: Relative Speed: The minute hand moves faster than the hour hand. The minute hand moves 6 degrees per minute (360 degrees / 60 minutes), and the hour hand moves 0.5 degrees per minute. So, the relative speed between the two hands is 5.5 degrees per minute (6 - 0.5 degrees). Initial Position: At 12:00, both hands are at the same position. After some time, the minute hand will catch up to the hour hand again. Time to Overlap: For the minute hand to catch up and overlap with the hour hand, it needs to make up the difference in position, which increases as the hour hand moves forward. Every hour, the hour hand moves 30 degrees (since 12 hours = 360 degrees, so 360/12 = 30 degrees per hour). The minute hand must cover this 30-degree difference at the relative speed of 5.5 degrees per minute. So, the time 𝑡 t it takes for them to overlap is: 𝑡 = (30 degrees) / (5.5 degrees per minute) ≈ 5.4545 minutes Interval Between Overlaps: Since each overlap happens after the minute hand has gained an additional 30 degrees on the hour hand, this process takes approximately 65.45 minutes. Therefore, the hands of the clock overlap every approximately 65 minutes, resulting in 11 overlaps every 12 hours, or 22 overlaps in a full 24-hour day. - CHATGPT 4o
@@dante0817 You stand by what chatgpt say's. Ask chatgpt if twelve AM Tuesday ( the very beginning of the day ) is the same moment in time as the very end of the day on Monday.
Another line of reasoning can be similar to the lapping problem, where the minute hand has lapped the hour hand by (24-2)=22 days. The minute hand makes 24 revolutions per day while the hour hand makes 2 revolutions per day.
he already calculated it right at the beginning of the video, getting the wrong 24-1 = 23 as answer, because he made one small logical error : the clock has 2x12 hours and not 1x24, and thus "half the solution" is 12-1 = 11, for a final result of 22 (instead of 24-1 = 23) ps: i once saw a big wall clock on a house (i believe it was in venice/italy) that really had 24 hours for a single turn, and thus would result in that first answer of 24-1 = 23 :-) on the other hand, i am not entirely sure whether that clock really had multiple hands or only just one single hour hand, for an answer of ZERO overlaps :-)
If you look at the list of overlap times, it's clear by inspection that the seconds hand will not be at the same angle as the hour and minute hands at any of their overlaps, except when both are pointing to 12.
@MindYourDecisions what about the times any 2 of the 3 hands overlap on a smooth motion clock... and what times do they overlap on a stop action clock (all 3 hands)
@@chichi90504 On a smooth motion clock, the hour hand and minute hand overlap *22* times per day, the minute hand and seconds' hand overlap *1416* times per day, the hour hand and seconds' hand overlap *1438* times per day. Since gcd(22, 1416, 1438) = 2 , they overlap (all three simultaneously) only 2 times per day (namely at midnight and noon). So the number of times per day that _any two_ hands overlap, is 2 + (22-2) + (1416-2) + (1438-2) = *2872* times. On a stop action clock (each hand moves discretely, at 360°/60 = 6° increments), every pair of two hands still overlaps at the same frequency (= times per day) as with the smooth motion clock; but all three hands simultaneously overlap *22 times* a day, each time for a full second; namely at 12:00:00 , 1:05:05 , 2:10:10 , 3:16:16 , 4:21:21 , 5:27:27 , 6:32:32 , 7:38:38 , 8:43:43 , 9:49:49 , 10:54:54 (A.M. and P.M.).
@@chichi90504 For some reason my reply in this thread is not visible unless the comments are sorted in order of "Newest" (instead of in order of "Popularity").
23 is a legitimate answer. I knew that I was counting midnight twice because you can't assign it to one day or another. It happens simultaneously so it occurs on both days, and it's not the same midnight since they are a day apart from each other. One day isn't over until it goes past midnight, and before it goes past they must overlap. It must be counted at the start and at the end because it occurs at the start, and at the end.
By international standards, 0:00 is the start, and only the start, of any given day. Nothing is counted twice nor needs to be (disregarding leap seconds)
@@parodoxis What if you measured from midday on one day to midday on the next day? Would you still ignore the second time they overlap at the 12 o'clock position? The clock is doing the same thing. This isn't about international standards, and it isn't even about measuring time. It's about how many times the hands of this mechanical device overlap, if you start it with both hands overlapping at the top, and run it until the small hands revolves around the dial twice. You count the first overlap when it starts, and it finishes in the same position with the hands overlapping again. If you count the overlap at the start of the measuring period, then you have to count the overlap at the end of the period. The hands are in the same position both times.
@@user-bf8tv8xv4w If you start the clock at any time of the day where the hands are overlapped and count the overlaps over the course of 24 hours as measured by that clock you would get the same answer every time. 23, yes you need to count the beginning once and the end once.
@@user-bf8tv8xv4w "if you start... and run twice..." - if that were stated in the question, then yes, I'd agree, since you are focusing on a mere duration. But the question didn't say that, it said in "a day". That's a well known interval, part of a series of fixed periods which don't overlap. That difference matters, as I explain in my other reply to you.
@@parodoxis The question asked how many times "in A day". "A" day is denoting a singular stand alone period of time. The word "A" denotes a single unit. If the question had asked how many times "PER day", or how many times "DAILY", then the day is treated as a part of a series. Like the fence post dilemma. If they are spaced at one metre, and a fence is two metres long, then the question "How many posts in a two metre long fence", It must have three posts. One in the middle and one at each end. A panel will fall down if it doesn't have a post at each end. If the question is "How many posts per two metres" the the answer is two. One post for every metre. You see the difference between "A", and "PER"?
the answer is 23. that's because the answer most viewers learned at school is to a different question: how many times do the hands cross. in this question it is "overlap". the hands start overlapped and they finish overlapped.
@@parodoxis No it doesn't. A full day ends exactly at midnight. If it ends a moment before midnight, then it's a moment short of a full day. What if you counted for half a day, from midnight to midday? Do you count the overlap at the start of your measuring period, but ignore the overlap at the end? The hands are in the same position so you have to count both.
@@parodoxis One day ends the very same moment that the next day begins. Count the seconds at the beginning of this day but don't start counting until the last second of yesterday has fully elapsed.
I had a little chuckle during method 3 when Presh spoke about angles of 0.5 degrees when I remembered that degrees themselves are sometimes subdivided into 1/60th units called minutes and 1/3600th units called seconds, and we could talk about the rate of rotation of the hands as angular minutes per time minute and angular seconds per time second
11/12*24=22 times.I basically used the concept of relative speed. Basically the hour hand reduces the relative speed of the minute hand,its relative speed is reduced to 11/12 rotations per hour. so time taken for the first meet is 12/11 hours which is basically 1-05-27 for the second meet its 1-05 +12/11~2-11...etc and so on so its basically an arithmetic series.
My initial logic was to figure out when does the hand first overlap after midnight and I found out that it was around 1:05 AM which is 65 minutes after midnight. Then i just proceeded to calculated it like below (24*60)/65 = 22.15 (rounded down to 22 times) 24 hours in a day 60 minutes per hour 65 minutes per hand overlap
I think you've fallen victim to Zeno's paradox. The distance between markings on the clock lies in the interval, not the end points. There is no "little bit of time" between the end of one day and the start of the next. For the clock to transverse 24 hours it must start and end with the hands in the same position. If they start overlapped they must end overlapped. Hence the answer is 23.
@@jamessmith2522 The victim of the paradox is you. Clock works in cycles, not intervals. Cycles reset, so you cannot double count their ends. The only answer is 22.
Technically the end of one day and start of another are simultaneous, but we have to declare the time at exactly 12:00 and 0 seconds midnight to belong to exactly one day or the other by convention (i guess you could just say it's part of both days or neither or some other weird exception to every other time on the clock but the alternative is a lot simpler). the convention is that it belongs to the starting day rather than the ending one.
before watching the video my brain says once per hour so in a day 24 times - but knowing Presh's videos I know 40% of the time i am wrong if i go with gut assumptions ... so went to pen and paper and calculator to work out what I typed below === now to find out how much I am wrong by. if they start overlapped at 00:00 then it would be 01:05:27:273 then 02:10:54:546 then 03:16:21:82 and so on the two hands overlap every 65 minutes 27 seconds 273 milliseconds, don't think we need to go smaller than milliseconds - so not once every hour but once every hour, five minutes and almost a half minute. if they start overlapped at 00:00 and you do not count that then 1st =then it would be 01:05:27:273 2nd = then 02:10:54:546 3rd = then 03:16:21:82 4th = then 04:21:48:11 5th = then 05:27:15:14 6th = then 06:32:42:16 7th = 07:38:09:19 8th = 08:43:36:22 9th = 09:49:03:24 10th = 10:54:30:27 11th = 11:59:57:30 12th = 13:05:27: etc etc 13th = 14:10:54:5 14th = 15:16:21:8 as you can see we are now in the teens and the ordinal numbers for passing do not match the time numbers 15th = 16:21:48:11 16th = 17:27:15 17th = 18:32:42 18th = 19:38:09 19th = 20:43:36 20th = 21:49:03 21st = 22:54:30 22nd = 11:59:57 and the next time they pass will be the next day. as they start and end the day overlapped from the eleven/twenty-three crossing. -- (please note these are the start times for the hands to start to over lap, but the minute hand takes more than 3 seconds to completely pass the hour hand) sorry had to work it out with degrees arc minutes and arcseconds for my brain to understand it. it seems logical to be 24 but maths says no.. it is i think this term is right a logical fallacy for my brain at first. another way is there are 1440 minutes in a day.... divide that by a little over 65 and you will get 22 then reading the comments the Le Mans 24 hour race came to mind bobby and fred start the race at the same time, in the time it takes bobby to do 24 laps fred has done 2 and they pass the checkered line at the same time... b24 - f2 = 22 lap difference
Your list of overlap times is NOT the correct answer to the question. The question included "round the times to the nearest second". In the first interval, you correctly omitted the "3/11 seconds" fraction, because is was less than 0.5, however, in the second interval this fraction will sum up to "6/11 seconds" which is greater than 0,5 and the nearest second will be the following second, not the one before.
You are correct, but to my knowledge, all of his lists have 2:10:55, not 2:10:54, as the third overlap, so although he didn't mention the rounding issue, he did do the rounding correctly himself.
there’s a very simple solution - when the clock is at 11 the minute hand only touches the hour hand again at 12. since it’s travelling faster than the hour hand, if the minute hand intersects the hour hand strictly between 11 and 12 it will reach 12 faster than the hour hand, contradiction. at any other time it’s easy to see there must be an intersection of the two hands between time t and t+1. so the answer is 11 intersections in 12 hours or 22 in a day.
23 is illogical. There are only 12 hours on this clock! Not 24! So in 24 hours you use the clock twice (2 * 12 = 24). In 12 hours the hands meet 11 times. 2 * 11 = 22. It is as simple as that.
23 is perfectly logical, as he explained his reasoning and it made perfect sense. Something can be logical yet incorrect. My fridge stopped making ice or water flowing. A logical assumption is that it may have been too cold and the pipes froze. This particular case was not caused by frozen pipes. That's an example of logical but incorrect.
I approached this as a drt catch-up problem: the hour hand moves at the rate of 1 unit per hour; the minute hand at 12 units per hour. The question is how long it will take the minute hand to “lap,” or catch up with the hour hand given the difference in their rates (12 - 1 = 11) and the 12 unit distance between them. That is 12units/11, or 1 hour, 27 minutes and 3/11 sec. So that is the first time the hands overlap. Just keep adding this amount successively to get the exact times the hands will overlap in 24 hours. In that 24 hours or 24 units they will overlap 24// 12/11 times, or 22 times.
Def. of "overlap": to extend over so as to cover partly. The hour and minute hand are continuously overlapping at the pivot point, so they overlap an infinite number of times. Poorly worded question.
I have a slightly different way of calculating the exact times (or time intervals) at which the hands of the clock overlap (or in other words, the time interval between two successive overlaps). The value of the time interval between two successive overlaps can then be used to calculate the number of times the minute hand overlaps with (or crosses) the hour hand in 24 hours. We can use the analogy of relative speed (or velocity). Suppose a police car, moving at a constant speed of ‘v1’, is trying to chase down another car going in the same direction at a constant speed of ‘v2’ (v1 > v2). If the distance between the two cars is ‘s’, then how much time will the police car take to chase down the other car? In this example, since the two cars are moving in the same direction, the relative speed of the police car with respect to the other car is (v1 - v2), and therefore, the time required will be t = s/(v1 - v2). If the two cars happen to go around in the same direction (say, clockwise) in a loop (like formula 1 cars completing multiple laps on a race circuit!) of total loop distance/length ‘s’, then the two cars will cross each other every t = s/(v1 - v2) time interval. We can use the above analogy to calculate the time intervals at which the minute hand will overlap with (or cross) the hour hand. Since the minute hand covers 360 deg every 60 min, its (angular) speed v1 = 360/60 = 6 deg/min. Likewise, since the hour hand covers 360 deg every 12 hours (= 12x60 min), the (angular) speed of the hour hand will be v2 = 360/(12x60) = 0.5 deg/min. Note that the minute hand is analogous to the police car, the hour hand to the car being chased, and both hands are moving (rotating) in the same direction (i.e., clockwise). Hence, the relative (angular) speed of the minute hand with respect to the hour hand will be (v1 - v2) = 6 - 0.5 = 5.5 deg/min. The (angular) distance ‘s’ here is 360 deg. Therefore, the time interval between two successive overlaps (or crossings) of the two hands of the clock will be t = s/(v1 - v2) = 360/5.5 = 65.4545…. min = 65 min & 27.2727… sec. Now, from this, we can easily calculate the number of times the minute hand overlaps with (or crosses) the hour hand in 24 hours. We simply need to divide the total time (i.e., 24 hr = 24x60 min) by the time interval between two successive overlaps (i.e., 360/5.5 = 65.4545… min). Thus, the number of overlaps in 24 hours = 24x60/(360/5.5) = 24x60x5.5/360 = 22.
I did the same mistake as you (getting 23) but in a slightly different way. I knew there was no overlap in the 11 o'clock hour and completely glossed over the fact that the 11 o'clock hour occurs twice on a 12 hour clock! I can appreciate the 1st solution, but it relies too much on going through the hand movements process. I ended up liking the 2nd solution best and to find the times, you just have to set one of the intervals of the minute hand equations equal to the hour hand equation (y = 30x). Because of the constant rates, I used the 1 o'clock one (y = 360x - 360) and got the intersection at x = 360/330 which after translating into a time, comes out to 1 hour 5 minutes and 27.272727... seconds.
minute hand makes 24 rotations, hour hand makes 2 rotations in the same direction, meaning it overlaps 24 - 2 = 22 times. This is a valid way as you know that you are only dealing in full rotations, as both end up at the start after 24 hours.
When the minute hand rotates 60 notches, the hour hand won’t be there because it moved 5 notches in that time. So overlaps have to be over an hour apart, so the number of overlaps has to be less than 24 in a day.
Similar solution but slightly different view: on its way to each hour, the minute pointer will catch up once with the hour pointer (also on its way to 11h and also 12h). Only on its way from 0h to 1h the minute pointer already catched up with the hour pointer at the start. I find this easier to visualize
@7:40, as a programmer I immediately raised a red flag here. If you round off to 27 seconds and then keep adding that to the starting time, you accumulate an error which gets bigger each time. For example, it should be 2:10:56 seconds, because rounding the seconds (6/11) there would go to the next highest second, not the previous lowest second. So, you should use the full precision available and round off to whole seconds at each time point of interest. This will give you a slightly different and more accurate result. This may not be significant here, but it definitely could be if you were dealing with money, nuclear weapons, navigation, and a million other things that could be done in software. Anyway, a simple way to get the answer to the original question is to say that VERY roughly, the hands overlap every 1 hr 5 minutes, so 1440 minutes (1 day) / 65 minutes = 22.15... which rounds to 22 times. If I were under time pressure in an interview I'd just come up with the answer that way. Calculating the exact times and not falling into the accumulated rounding error trap would be something that could be discussed - demonstrating that you are aware of the pitfall is likely to be far more important than solving the exact times!
He did round to the next second, not to the previous second. 2:10:55 is the correct rounding of 2:10:54.545454.... ; your suggestion of 2:10:56 is actually wrong.
@@yurenchu Yes, I realised that later. Nevertheless, what he described, about rounding, then repeatedly adding, would lead to accumulated rounding errors. In practice perhaps that's not actually what he did, but rounded each time, as suggested. I didn't check each time individually. But it goes to show that describing an algorithm requires as much precision as the algorithm itself!
@@GRAHAMAUS At 7:11 , he says that he's going to calculate the _precise_ times. It then stands to reason that he of course would be using the exact value of "this interval" (1 + 1/11 hours), not the approximation (1:05:27) which he merely displayed to get a concrete feel of the size of the interval. Yes, his wording may be a bit sloppy (as he failed to mention explicitly which value he is repeatedly adding, and that he's rounding the outcomes to the nearest second), but you not checking if the displayed times were indeed off (and failing to see that 2:10:55 was _not_ calculated by merely adding 1:05:27 to 1:05:27 ; note that 27+27 = 54, not 55) is also sloppy. By the way, you don't have to check every listed time individually to determine if an accumulation of round-off errors after repeated addition has occurred. Just look at the last time in the list, 10:54:33 ; adding 1:05:27 to it results indeed in 12:00:00 , so apparently there is no accumulation of error, as apparently the exact value of the interval had been repeatedly added.
So you round-off, you come-up with approximately 22 & you conclude 22? & you claim to be a programmer and you apply this line of thinking to nuclear weapons and navigation? I hope you are not under time pressure the next time you have a nuclear or navigational decision to make. It's 23.
I knew it was 22 because it doesn't overlap at 1:00, it overlaps slightly after, and even more after 2:00, such that they won't overlap between 11:00 and 12:00, so the answer for one full rotation of the hour hand is 11, and twice in a day makes 22
While the answer is 11, the excact time each overlap occures depends on the clock design A couple of extra wheels could have the hour arm move just twice in every hour (so a small amount after the hour and then nothing until 3 seconds before the next hour) Plent of funky watch and clock movements have been built.
This is a slightly more complex version of the around the world in 80 days story, where the one traveling thought he was a day late as he forgot that by traveling around the world he saw the sun coming up one more time than the ones who stayed at the start/finish. This one is more complex as it goes round twice, this would be equal to the hypothetical twice around the world in 160 days novel.
Both hands start aligned at 12:00. After exactly 1 hour, the hour hand has moved on 1/12 of a turn = 30 degrees; but now the hour hand is pointing to the 1 while the minute hand is still pointing to the 12, and by the time the minute hand reaches the 1, the hour hand will have moved on a little further still. The two hands will have to come into alignment again sometime between 13:00 and 13:10; again sometime between 14:10 and 14:15; sometime between 15:15 and 15:20; between 16:20 and 16:25; between 17:25 and 17:30; between 18:30 and 18:35; between 19:35 and 19:40; between 20:40 and 20:45; between 21:45 and 21:50; and between 22:50 and 22:55. This is just before the hour hand reaches the 11, and the next time they line up again will be midnight; so there are 11 occasions when the hands align in 12 hours, meaning 22 occasions in a full day. Now, we can save the effort of summing up a bunch of infinite series, even if we know an identity for that, by noting that the speed of the hands is constant; so these alignments must be occurring at regular intervals of 86400/22 = 3927.2727..... seconds, i.e. at the following times of day rounded to the nearest second: 00:00:00, 01:05:27, 02:10:54, 03:16:21, 04:21:49, 05:27:16, 06:32:43, 07:38:10, 08:43:38, 09:49:05, 10:54:32; 12:00:00, 13:05:27, 14:10:54, 15:16:21, 16:21:49, 17:27:16, 18:32:43, 19:38:10, 20:43:38, 21:49:05 and 22:54:32.
Here’s my way of getting there. Suppose we start the day at 5 minutes past midnight. In every hour, the minute hand is going to “catch up” at some point where the hour hand is between that hour and the next. So the first on will be between 1:05 and 1:10. The second between 2:10 and 2:15. So when we get to the 11th crossing, it’s sometime between 11:55 and 12:00, but this one’s weird, because it really happens at 12, and the crossing that goes with 12 happens between 12:00 and 12:05, but it’s really right at 12:00 and is the same as the crossing that goes with 11. So there are 12 numbers, but 11 and 12 share the same crossing, so only 11 crossings. Whole thing happens again before we get back around to 5 minutes past midnight, so 22 crossings.
According to Zeno's paradox, the minute hand will never reach the hour hand if it has a head start 🙂 LOL ! BTW, I liked method 2 the most. For method 1 you started with a constant of 1:05:27 that came from nowhere, even though it made sense.
I think the correct answer should be 23 because per the analysis you used; starting from 12 midnight, the hour hand will overlap the minute hand at 12:00 pm and continue to 1:05pm and so on before it gets to 12 o'clock AM.
My solution was a lot like Solution 3. I figure, okay, at 1am, the hour hand is 1/12th of the way around the clock, while the minute hand is 0/12th of the way around the clock. The minute hand moves 12 times as fast. Therefore, I just need to find when: hour_starting_position + hour_speed = minute_starting_position + minute_speed 1/12 + x = 0 + 12x 1/12 = 11x 1/132 = x So the hour hand will travel 1/132th of a full rotation before the minute hand catches up. That's 1/11th of a single hour, which is 5.45 minutes, which is 5 minutes and 27 seconds. Same as solution 3, but thinking in terms of "percent of distance traveled" instead of "degrees".
Mathematically you're absolutely right. But, intuitively, which I also missed in the first place, it's still absolutely logical for the answer to be 22 times. The reason is: the clock is a 12 hour clock, so, you have two separate 12 hour periods in which the hours hand makes a full revolution, making it skip one overlap every 12 hours. If it was a 24 hour clock, your (and my) initial answer of 23 hours will be true. Only the last hour, the 23rd, would be skipped. Now, there are VERY unusual clocks in which the hours hand skips a full hour every time the minutes hand hits 12 o'clock. In this case, the answer would be: full 24 times.
Yes, but no. Noon count's once (it does not mark the beginning or the end of any day.) The midnight at the beginning of the day and the midnight at the end of the day each count once.
The answer is 2, since at every interval where the minute isn’t 0, the second hand is wildly displaced from the minute hand in the few seconds it could be considered overlapping (the nearest being in hours 3/8, where the seconds respectively are 22/38). This can be proven simply by noting that each hour has an overlapping minute and hour hand EXCEPT the 11th hour, which for the minute hand to catch up it would be the 12th hour. Therefore we can prove the relation between the number of hours to amount of overlapping can have a maximum of 11/12ths the hour count. However, doing this same technique with the minute and second hands results in 59 overlaps per minute, and since 11 and 59 do not have any common factors, their LCM can only be 59*11 which as inputs result in a full revolution of the slowest hand. Noting that a clock only tells time for half of a day, we can safely assume that all 3 hands will only overlap twice per day, both accounting for a fluid second hand. A stiff second hand would need more in-depth calculations, but could also be proven based on the results I initially calculated for. H(x)=floor(12x/11) // hour hand offset in relation to input hour x M(x)=floor(60x/11) // minute hand offset in relation to input hour x S(x)=round((3600x/11)%60) ± 1 // second hand offset in relation to input hour x, ±1 for possible inaccuracies
I like problems solved more simply. I got it wrong, but I wished I would have solved it like this: The hands obviously overlap on the 12 two times per day. (The third time doesn’t count because it would be the next day.) Then each hour they overlap a little more past the start of the hour, two times in one day, so that is 20 more times. Finally, the 11 hour overlap can’t be counted because it would be so close to 12 by the time the minute hand caught up. This would be casual not formal math. Of course, Presh would show the math behind what I said, that’s his channel. Embarrassingly, I forgot that the hour hand moves from hour to hour as much as the hour is finished. Elementary students forget this or think that the hour hand just sits on the hour until the next hour. So, I made an elementary mistake.
There is a very simple way (mathematical) to calculate how much that fraction (1/11) equals to in seconds. So, known that the hands overlap 11 times every 12 hours. That equals once every hour and 1/11 of an hour. Since there are 60 minutes in an hour and 60 seconds in a minute, there are 3600 seconds in an hour, and if you divide 3600 by 11 you get 327.272727¯. 300 seconds equal 5 minutes, so 327.27¯ = 5 minutes, 27.27¯ seconds. Therefore, the hands overlap ~ every one hour, 5 minutes and 27 seconds.
Many people want to say 24! BUT the actual answer is 22. Clock hands overlap at: 12:00, 1:05, 2:10, 3:15, 4:20, 5:25, 6:30, 7:35, 8:40, 9:45, and 10:50 TWICE A DAY (AM and PM). There's no overlap at 11:55 because the hour hand is moving closer toward 12 when the minute hand is at 11.
This is great but as a former interviewer at AWS, i didn't ask this question of any of the candidates i interviewed. We just have too much ground to cover amd not a whole lot of time to get the data points we need. In fact, like many of you, i guessed 24.
I solved the problem of how many times the hands of a clock overlap in a day by focusing on the observation that the minute hand moves at 60 minutes per hour, while the hour hand moves at just 5 minutes per hour (here, minutes refer to the markings on the clock dial, treated as units of distance). So, the minute hand is 55 minutes per hour faster than the hour hand. This reduces the question to: how long will it take the minute hand to complete a lap and overtake the hour hand? The problem becomes equivalent to asking how long it would take the minute hand to complete a lap if it moved at 55 minutes per hour. Viewing everything from the minute hand’s perspective, we know the speed is 55 minutes/hour, and the distance to cover is 60 minutes (clock markings). Therefore, it would take the minute hand 60/55 hours to make one full lap around the clock dial. From there, I created a function that takes the hour (h) as an argument and outputs the exact minute in that hour when the hands overlap: *roundDown(h mod 12 × (3600/55) mod 60 / sgn(11 - h mod 12))* The division is simply there to trigger an error if the argument is 11 (or 23), as the clock hands never overlap at 11 a.m. or 11 p.m. This problem also got me thinking about the saying, "even a broken watch is right twice a day." Well, not necessarily. As shown, you can arrange the hands of a clock in such a way that they never align during normal clock operation-like overlapping the hands somewhere between 11 and 12 o’clock!
Great explanation. Reminds me of the quarters thing where one quarter stays still, and the other quarter rolls around its perimeter, and how many times does it rotate.
The correct answer is 23 times. Midnight is the dividing line between two days, thus each midnight belongs to TWO days. There are two midnights, plus 21 other overlaps. This is like the fence post problem, each midnight belongs to two days, just as each fence post belongs to two fence panels. Also 12:00 am and 12:00 pm are both midnight! "am" means "ante meridiem" or "before midday". The only 12:00 that is before midday is midnight. Similarly, "pm" means "post meridiem" or "after midday". The only 12:00 that is after midday is midnight. You should distinguish the 12:00 times by using "noon" and "midnight" as suffixes, not "am" or "pm".
@@carultch Even if I accepted your point, which I don't, 23:59:59.9 belongs to the same day, and the hands are already overlapping. Clock hands aren't infinitely thin.
Another algebraic approach. The minute hand is moving at a rate 6°/minute so at time t minutes has move 6°t. The hour hand moves at 0.5°/minute, so at time t minutes has moved 0.5° t. The hands overlap whenever the difference between how far each has moved is a multiple of 360°. I.e., they overlap whenever (6°-0.5°) t = 360° k, where k is a non-negative integer. Solving for t, the hands overlap whenever t = 720/11 k. We want to know how many overlaps occur in the first day, so we want to find all non-negative integers k that give t with 0 ≤ t < 1440 (the number of minutes in a day). Plugging t = 720/11 k into that inequality we get 0 ≤ 720/11 k < 1440. Rearrange to give 0 ≤ k < 1440x11/720 = 22. There are 22 non-negative integers that satisfy that: 0, 1, 2, ..., 21, giving us 22 times where the hands overlap.
The simple way to think of this is that the question is asking how many full revolutions the minute hand makes _relative_ to the hour in 24 hours. In 24 hours, the hour hand makes 2 revolutions relative to the face, thus the face makes -2 revolutions relative to the hour hand. The minute hand makes 24 revolutions relative to the face, thus the minute hand makes 24 - 2 = 22 revolutions relative to the hour hand in 24 hours. There will 23 overlaps if we count both end points, or 22 if we don't. We can then simply divide 24 hours by 22 revolutions to calculate all the overlap times.
The problem with this question is, it doesn't say if the starting point of midnight on one day is counted or not, so if it is it will be 23 in a day, if not then 22. Like most questions they are quite ambiguous and sets it up for multiple answers that are correct in how you interpret the question.
In 12 hours, the hour hand makes a full turn. In the same time, the minute hand makes 11 turns more than the hour turn. So if we consider the hour hand to be fixed (and the scale moving backwards instead) and only consider the distance between hour hand and minute hand - then we get 11 revolutions by the minute hand in a period of 12 hours. Since both hands move at constant pace, it follows that the two hands must meet each other exactly every (12/11) hours. Or 22 in total in 24 hours.
22 times. It happens at 1/11th of a hour further each hour. In 12 hours, the little hand goes around 12 times, but the big hand also goes around once. If the big hand didn't move, the little hand would catch it 12 times and pass it. But since the big hand does move, it gets a little farther than 1 hour away each time and by the 12th time, it doesn't quite get there. This is easier to see if you start at any time OTHER than noon. 1) 1:05:27 3/11 seconds 2) 2:10:54 6/11 seconds 3) 3:16:21 9/11 seconds 4) 4:21:49 1/11 seconds 5) 5:27:16 4/11 seconds 6) 6:32:43 7/11 seconds 6) 7:38:10 10/11 seconds 8) 8:43:38 2/11 seconds 9) 9:49:05 5/11 seconds 10) 10:54:32 8/11 seconds 11) 12:00:00
An easy way to think of it is to start by thinking of if the Hour hand ticked, when they overlap, 12:00, 1:05 etc. then notice that you would have 11:55 and 12:00 as both potential times they overlap, despite being 5 minutes apart. When the hour hand is constantly slowly rotating, by the time the 11:55 comes around, the hour hand is basically on the 12. (This is due to the interactions of 12 revolutions for the minute hand, and 1 Revolution of the hour hand) Then double for 24hrs.
Every single hour, the hour hand moves one hour marker (one twelfth) and the minute hand moves 12 hour markers (12 twelfths) so each revolution around the clock (which is every 12 hours) will require an extra one twelfth distance to overlap both hands. Over the course of those 12 hours, we lose that one twelfth as a whole hour, which means 11 overlaps every 12 hours
The trick is to notice that between 11 and 1 they overlap only once. So, if you count the part from 11 to 1 as an "hour" then it is true, they overlap once in every hour, but now there are only 22 hours in a day. To simplify this further you can notice that our 12 is redundant regarding this question, so it is even easier to see why we count with 22 hours per day :)
I would agree with your intuition of 23. I think the last overlap counts, since the question asks about a real world clock with finite hands. Your excellent presentation details how to consider modeling an ideal clock. As such the original question is open to interpretation.
How can you agree with his intuition of 23? His intuition of 23 is that the answer is possibly 24 , but since we _mustn't_ count the "last" overlap, we must substract 1 , hence giving the answer 24 - 1 = 23 . Your motive for arriving at 23 is exactly the other way around, and counter to his intuition!
To me it was pretty obvious from the start that the hour hand was going to be able to "escape" from the minute hand twice a day giving you the answer of 22; and using symmetry you could get all the times easily. HOWEVER I suppose the interviewers would be more interested in some analytical explanation like the ones you showed us here. My favourite was method 4 because this puzzle has an element of the Zeno's paradox in it, so that infinite sum was very pleasant to my eyes; even that 60 at the beginning could be interpreted as 5/(12^-1). Awesome video yet again.
While trying to work the problem for a clock with a hand for the second, I figured out that 24 is a possible valid answer for this problem, depending on your definition of the clock. Since the size of the hands (in seconds) are not given, it can be expected to be irrelevant, so two close overlap are considered distinct, then if you define the clock as having a finite number of instantaneous step (as with a mechanical step by step mechanism), using the second method of the video but with now stair curve there is now another overlap : the last step before the end of the cycle. But these overlaps overlaps. There are two if you cont in number of step an only one in number of events (but the last+first event will be the only one with a 2 steps duration).
The hands of a watch/clock sit on a pinion. They are not 0 units of measure wide, so the minute hand, which is sitting on top of the hour hand in the hand stack (if the seconds hand were present it would be sitting on top of the minute hand), is constantly overlapping the hour hand. Pedantic? Yes. Missing the point? Yes. Correct? Also yes.
I think twenty two. If the hour hand was stationary, it would 24 times, but because the hour hand is moving away, it will take longer to interact, losing one interaction per rotation. Since it rotates twice per day, it should be 24 (hours per day) - 1 (overlap per rotation of hour hand) x 2 (rotations per day.) Or 24-2=22
24-((360+30)*24-360*24)/360 = 22 24 - (390*24 -360*24)/360 = 22 24 - 720/360 = 22 24 - 2 = 22 Full cycle around the clock = 360° One hour by hour hand = 360°/12=30° One hour by minute hand = 360° Each hour the minute hand need that additional 30° to catch hour hand. 360°+30°=390° So full distance that minute hand need to catch hour hand 24 times -> 390°*24=9360° Minute hand will travel 360° each hour, so 360°*24=8640° is the distance it will travel for exactly 24 hours, and 24 circles. 9360°-8640°=720° That means 720°/360°=2 is the number of circles it would need above the 24 full hours in order to meet the hour hand 24 times, and so 24 times it would then meet with that additional 30° - 2 so the number of times above 24 full circles of minute hand (so 24 hours) = 22 number of times it would meet the other hand without that 30° additional degrees each hour.
Here's another algebraic solution: 1. If X = # of overlaps, S = seconds in a day = 86,400, and P = the period in seconds between overlaps of the two hands, then: X = S/P = 86,400/P 2. Now calculate P (the period in seconds between overlaps): a. Get the rotation velocities (in degrees per second, or dps) of each hand, Vm and Vh: Vm = 6°/60seconds = 1/10 dps Vh = 30°/3,600seconds = 1/120 dps b. If Velocity = Degrees/Time, then Degrees = Time * Velocity: Dm° = T * Vm = T/10° Dh° = T * Vh = T/120° c. The hands overlap when the distances are equal, not counting full circles, so: Dm° mod 360° = Dh° mod 360° T/10° mod 360 = T/120° mod 360 d. Because we can't calculate a result with the mod operator, we take the first overlap after midnight to get P: the minute hand has traveled once around, while the hour hand has not, so we know that P = T when T/10 - 360 = T/120, solving for T we get T = 3927 + 3/11 3. From step 1: X = 86,400/P = 86,400/(3927 + 3/11) = 22
Intuitively: It will unlikely be 24 due to it being too simple. It should not be 23 either - since a 24 hour period involves 2 complete 12-hour cycles; meaning that the answer has to be an even number (as AM and PM are exactly the same case). So this leaves us with the answer probably being 22 (unless intuition fails us and the answer is actually just simply 24).
I think a better solution is - Mark a point on the dial of the clock... say at 3 or 4... Hour hand cuts this point 2 times in 24 hours and Minute hand cuts it 24 times in 24 hours.... Since both the hands are moving in same direction, their relative speed should be subtracted (24-2) = 22 times. (Note that this will work when the cycles are complete... e.g. for 12 hrs it will be (12-1) 11 times... that too is simple... but for 6 hours, we need to plot a point at appropriate location otherwise we will either get 5 or 6. Best will be to take the next available number on the dial. e.g. take 1 as the point if it is between 12:01AM to 12:01PM) If it was a defective clock and one hand was in clockwise direction and the other hand was in anti-clockwise direction... then the relative speed is in opposite direction and we have to add those, so in that case it will be 24+2 = 26 times!
I more or less used you method 3 but turned it into a straight line race with one thing at 60x/hr and the other at 5x/hr with a 5x headstart. Basically distance rather than degrees.
I remember that some old analog clocks don't have continuous movement on the minute hand, instead it skips forward in one minute increments. In fact the ratchet mechanism involved causes the hand to sweep back a few degrees for a moment before it snaps forward to the next minute. In this case you could have the hands overlap upto 3 times in an hour, depending on where the hour hand is compared to the distance of the back sweep.
The snarky solution to puzzle is that the hands are always overlapped when you consider the center point.
Arguably, at the center point it's not the hands that overlap, but the _bases_ of the hands that overlap. Just as the base of a statue is not the statue, the base of a hand is not the hand.
There's also 24h analog clocks, just to make things more annoying. And clocks with a second hand in addition to minute and hour. There's probably one that has fully seperate displays for each hand, as well.
Finally, a question that doesn’t fall back on semantics or ambiguous wording.
@@StephenByersJ Yeah but it still had specific definitions like what overlaps at 12 count and which don't which is why most probably got it wrong by just 1 or 2.
Not so fast. An electronic sensor system would get a different count than a mathematician because on a physical clock the two hands will overlap even when they are not pointing at the same point simply because the physical hands are too thick.
The overlaps would be longer, but the amount of them would remain the same except for the very last time, where the overlap would begin right before the deadline is hit.
I'd argue once might be a valid answer, as they're always overlapping at the center of the clock.
Jesus, divided by 1/2 is not ambiguous lol
The fact that you can form an infinite series, is an amazing illustration of how mathematicians defined real numbers, it's a limit of rational numbers, even though the minute hand eventually catches up, but that time stamp is never a precise one
The (limit of the) infinite sum is still rational though. The time stamp is "never precise" merely because we haven't divided the hour in a multiple of 11 minutes.
So I'd say that this particular infinite series has nothing to do with the definition of real numbers.
..hum the precice time is for n[1 to 22] = n(720)/11 minuits wich is about 65.45 minuits, or excatly 65 minuits, 27 seconds and (9090909090898153381683397656577 / 33333333333293229066172458074116) hundreds of a second (aprox 27.3) times n, where n is number of hours
..can also be written as n60(12/11) or n60 * (1 + 1/11)
...n is 11 for each rotation of the minuit hand n*2 rotations = 22
@@Patrik6920 The interval between two consecutive overlaps is exactly 65 minutes and (27 + 3/11) seconds.
(note: 27 + 3/11 = 27.27272727... )
And not "minuit", because _minuit_ is French and means "midnight".
@@yurenchu ment the english word...
minute-hand runs 24 laps, hour-hand runs 2 laps, so minute-hand run by hour-hand 22 times (24-2).
I don't see how that means you can just subtraction them. The minute hand passes over the hour hand between each number so isn't it 23 or 24?
@@maxhagenauer24it doesn’t - watch the video
@@maxhagenauer24 It never passes in the 11 or 1 o'clock hours because it is overlapping exactuly at 12.
@@eytanz Think through the statement carefully :)
@@maxhagenauer24if hour-hand doesnt move minute-hand overlaps it 24 times, however hour-hand does 2 laps so that's 2 less overlaps bcs it moved away
I used to think of this problem as a child. It is like the Achilles-tortoise parable. Think of 1 o'clock. the hour hand is the tortoise and Achilles is the minute hand. The minute hand will pass the hour hand even though it subdivides into an infinite number of increments. As in your last solution.
same.
Exactly same.
ah yeah, the old Achilles-tortoise parable that we all thought of as a child. Old news, everyone knows it as a way of telling time. I often say It's about Achilles past Tortoise, and then leave it to the stranger on the street to subdivide it into an infinite number of increments.
Likewise. 🙃
Yes! Method 4 was also the way I solved it long time ago.
The linear graph way you used to solve this is really neat!
It is but the answer is still 23.
@@peterbaruxis2511 15 min video explains how it isn't and yet you insist it is without any explanation. Watch the video again.
This is analogous to a general fact in astronomy, where there is 1 more sidereal days or months than synodic days or months per year (given the rotation of the planet around its axis is the same as around the sun, which is not true for Venus). Here a "synodic hour" is an overlap and a "sidereal hour" is a normal hour. Thus 11 "synodic hours" per one "clock year" or 11 overlaps for half a day.
Exactly what I was thinking. Applies to number of rotations of the Earth about its axis per year and number of rotations of the Moon about the Earth per year. If rotations are considered relative to the fixed stars, you get one more rotation in each case than the way we usually count, based on the position of the sun.
1:50 If you're initially assuming once for every hour of the day then that would be 24, one each for the 24 hours from 0 to 23 using 24 hour clock, so one of the midnights is already not included, otherwise you'd start with 25 hours in a day (0-23 and another 0) and subtract 1 to get 24.
Exactly. Even if the answer is wrong either way, Presh should've got 24 as their initial answer.
The hands don't meet during the 11 o'clock hour. They really do meet only 11 times in a 12 hour interval.
@@9adam4 Yes, we know. The point here is that Presh's intuitive reasoning to arrive at 23 ("Once every hour makes 24, but then I subtract 1 in order to not double count the end points") doesn't make sense.
@@yurenchu Where does he say the answer is 23? Missed that part.
@@9adam4 Look at the time stamp referenced in this thread's opening comment.
Following the idea of one of the comments. In 24 hours, the minute hand runs 24 circles, the hour hand runs 2 circles, so the minute hand overpasses the hour hand 24-2 = 22 times, or in other words, they overlap 22 times. The interval between two overlaps is apparently a constant, so the interval is 24/22 hours (h), or 12/11 (h), which is 1 hour 5 minutes 27 and 3/11 seconds, and the kth overlap simply corresponds to the time (12/11)k (h), k = 1, 2, 3, ..., 22. When k = 11, it is the noon time; when k = 22, it is the end of the day, or 24 (h). k can take the value 23 or more, but they represent overlaps of the next day.
it's 23.
Got it right the first time. I went empirically through the meeting points and no later than near the bottom of the clock I realized we can't have the minutes hand at the half hour mark and the hour hand at the full 6th hour mark. Then I sketched it up on a sheet of paper and saw that the meeting point happens at 12:00 as well as intermediately between every two hours except the first and the last.
This puzzle reminds me of the fact that even though we colloquially know that Earth rotates 365 days per year, making 365 days, it actually rotates 366 times per year if you look at it from a fixed reference frame outside of the solar system. Our reference frame normally is fixated on the sun, so it rotates one turn per year, adding one turn to the 365 turns we observe relative to the sun.
More precisely, 1 Orbit contains 365.242374 days, which would mean the earth rotates 366.242374 times.
We compensate with leap years. To approximate 365¼ we have a leap year every fourth year (so 365.25), to approximate the slight error in that (≈0.01) we skip it if the year is also divisible by 100 (so 365.24). To further compensate for the ≈0.002 error in that, we intentionally fail to skip it on years divisible by 400... fun stuff
Btw, a reference frame fixated on the sun is not "a fixed reference frame outside the solar system"... the sun is very much in the solar system (where you got your ≈366 rotations).
It would be the same outside the solar system too, until you get far enough away to consider the rotation of the galaxy and such.
@@parodoxis I referred to the fact that normally in our earthly life, our reference frame for counting days is based on the vector from Earth to Sun, and that vector rotates once per year around the Sun, as does Earth.
@@MrSaemichlaus right, and so did I, that's rotation # 366. No confusion there, did you instead mean to clarify your reference to using a frame "outside our solar system"? That's the part I didn't get. Either way, no worries.
This problem has a special place in my heart and still irks me to this day. Not sure where I first met it in HS, but the same as presented here. It didn't specify how many hands (I thought Hour / Minute / Second) as the question also specified find to the nearest second. So I went about solving it as though all 3 hands had to overlap =____=, got it wrong even though i thought it was kind of straight forward.
per min:
hour hand (0.5 degrees) | min hand (6 degrees) | sec hand (360 degrees)
relative speed (hour v min) 5.5 / min >> 360/5.5 = 720/11 to overlap (720 mins is 12 hrs, so every 12 hours there are 11 overlap)
relative speed (hour v sec) 0.5 / min >> 360/0.5 = 720 to overlap ... so once everyone 12 hours...
so all three overlap at exactly noon and midnight, twice.
aside: Saw this question again at at some years later (prob interview, don't remember), and remembering that it was poorly worded, i calculated the total overlaps (second v min, second v hour, min v hour) and got it wrong again =_____=
You did describe a more interesting problem though.
And the result is actually surprising (at least to me).
How do you know you got it wrong? How do I know you got it wrong. You don't state your answer to the question and you don't state what you believe to be the correct answer.
@@peterbaruxis2511 “so all three overlap at exactly noon and midnight, twice.” …?!
Writing this before I watch the video.
If you just want the number of times, you can quickly recognize that you only need to consider the first 12 hours because the next 12 hours are duplicated positions. Then, your logic of "subtract 1 for the minute hand passing the hour hand" does indeed give you the correct number of 11 times in a half day. Thus, there are 22 overlaps in a day.
If you want the actual times, use basic modular arithmetic.
Using the hour ticks as the unit of circular distance and denoting time passed T in units of hours, the hour hand travels at a rate of T and the minute hand travels at a rate of 12T (so example, when T = 1.5, then 1.5 hours has passed, the hour hand traveled 1.5 units and the minute hand traveled 18 units). The hour and minute hand are lined up when 12T = T mod 12, or in other words, when 11T = 0 mod 12. So just start enumerating the instances:
11T = 0 -> T = 0. So at 00:00, they line up.
11T = 12 -> T = 12/11. So when 12/11 hours passed, they line up again (the time would be 01:05.27)
At this point, you can keep going or just recognize that you can get the rest of the answers by adding an additional time delta of 01:05.27.
Keep on finding solutions for as long as T < 24 and you'll see that there are only 22 values of T which can satisfy the modular equation.
Another way, once you discover they overlap 22 times in a day, is to calculate the total seconds in the day, 24*60*60 = 86,400 seconds, divide by the 22 overlaps, and you get ~3927.2 seconds, which equates to 1 hour, 5 minutes, 27 seconds. Just keep adding them to get the times, to the nearest second, that the hands overlap.
I answered 23 based on the clock hands having non-zero width. That would account for the last (albeit not total) overlap starting before midnight. It should have said "total" or"full" overlap to make it clear the endpoint was not included. When doing a lap around a circle you could use a half-open interval, but most would think that both the start and the stop points are included, hence a closed interval.
You're right. If you start the clock at midnight then the hands are overlapping. If you stop it the following midnight then they are also overlapping. You can't just count the first time and ignore the second. It's an arbitrary decision to assign midnight as the start of a day, and not the end of the previous day. That's a human construct, not a mathematical calculation of how many times the hands overlap.
Either ignore both ends and only count the overlaps occurring in the duration of the day, or count both ends. You can't have half an overlap, and you can't arbitrarily assign it to a certain day.
I can argue 23 with zero width hands- look for my comment.
You can argue anything you want, doesn't mean it's valid@@peterbaruxis2511
So many people are doing this problem wrong. Step one: create a pre-planning meeting to establish a strategy for acquiring a random sample of different clocks. … step fourteen: determine which of the squired clocks comply with the ethical sourcing standards we were supposed to be using, discarding the rest …. step one hundred-four: apply for an out-of-budget request for an additional $1.2 million for the overtime required to prepare for the pre-meeting for the project ….step fifteen-thousand, six-hundred, forty-one: submit the final version of the 4,231-page environmental impact report to the internal review committee, now in the correct font.
Think of clock hand positions as complex numbers on the unit circle. If z is the position of the hour hand, the minute hand is at z^12. (Clock is on its side.). They meet when z = z^12. This resolves to z(z^11-1) = 0. Since z=0 is not the kind of solution we want, there are 11 distinct solutions on the unit circle. Thus 22 for each 24 hour period.
if your calculations are correct then your application is flawed- the number is 23.
@@peterbaruxis2511 You’ve reached a conclusion without presenting an argument. I’m sorry, that’s just not how it works in math.
@@OLJeffo Justify doing what you propose in your original comment. The argument is implied in the fact that the correct answer is 23 so either your math is wrong or your approach is wrong. That's how things work.
@@peterbaruxis2511 The comment is the justification, although i admit it was intended for people who have taken an undergrad complex analysis course. Listen, I've read your main comment and it seems like you aren't a total crank, so I'll add this: I am assuming that every moment in time belongs to exactly one day. If you want to assume otherwise, my proof isn't for you. But in that case, the modern world might not be for you, since so many technological and financial systems depend on this assumption!
@@OLJeffo Calculus would look at the distance between -1 and 1 and see zero differently than would a complex analyst, both probably have their uses. I wouldn't make assumptions about people so quickly.
I once tried doing this seemingly interesting topic as early as 5th grade when I tried it manually for times when hour and minute hand overlap.
Anyways, recently in college (3rd year), we are being taught Aptitude where this clock topic also came.
And during 5th grade, I did it this way:
The hands will overlap between (starting at 12 PM):
1-2 PM
2-3 PM
3-4 PM
4-5 PM
5-6 PM
6-7 PM
7-8 PM
8-9 PM
9-10 PM
10-11 PM
at 12 AM
11 times in 12 hours, so 22 times in 24 hours.
12/11=1.0909
Multiply decimal part by 60 for minutes and decimal of resultant by 60 for seconds.
So about 1 hour, 5 min, 27 sec.
This is what I came up with too.
This is the simplest, most intuitive answer of all the comments. I feel like a lot of people assume the answer to be something and that throws them off, including me. But also before watching the video I didn't know if we count 12 twice or not
If you did that in fifth grade that's impressive. If I were your fifth grade teacher I'd have praised your methodical approach but I would have asked you to reconsider the decision to analyze half of the question and double that conclusion- and think about how that might not lead to the same conclusion as methodically following the problem to it's end. If after that you still seemed engaged I'd ask you to calculate for one quarter of the day and multiply that by four. (that would likely lead to an answer of 27 or 26.)
@@peterbaruxis2511the difference is that the two halves of a day are the exact same route on the clockface (one revolution) and thus are identical. Also, one quarter of the day would lead to the answer of 6x4=24
As a programmer, I think about 3600 seconds per hour all the time.
12*3600/11 is your answer.
It's so easy.
That's also how I got there.
I think about 3.6e15 picoseconds per hours
@@thechessplayer8328 sure ya do
@@Gruuvin1 it’s what all HFT hardware is moving to these days. Rule of thumb is light travels 1 foot per 1000 picos
@@thechessplayer8328 🙄
Something is missing here. 1:05:27 + 1:05:27 should be 2:10:54 as 7+7 is 4. So i assume that you actually calculated each step with the fraction and did not just add 1:05:27 each time
He should have explained that the rounding to the nearest second influences the result more than 27 seconds every hour because it accumulates every hour.
Per the calculation at 7:06 you would need to add in the additional 3/11 seconds to get an actual value. Note he says "precise" and inserts an approximation symbol. Bad form.
@@jrkorman "Note he says "precise" and inserts an approximation symbol."
No, you're misrepresenting what he did. You're making it seem as if at 7:06 he first says "precise" and then inserts an approximation symbol.
What he did, was _first_ give an approximation of the interval, in order to give a recognizable representation of how much "1 + 1/11 hours" is, and _then_ he proceeds to calculate and present all 22 "precise times" by repeatedly adding "this interval". Although it's sloppy that he doesn't explicitly mention which value he uses for repeatedly adding, it stands to reason that he would use the exact value of "1 + 1/11 hours" (or "1 hour, 5 minutes, 27 + 3/11 seconds") and not the approximation in order to determine the " _precise_ times" (which he then rounded to the nearest second, of course).
Is he wrong to presume that the viewer already knows how to add a fractional value repeatedly (and then round the results to the nearest second)?
@@yurenchu assuming people know rounding, significant figures, and error analysis is bad for the level of instruction he is aiming for.
That and he gets the instructions incorrect at 6:55 where he states that “we now want to round this to the nearest second” before doing the additive steps.
The hands ALWAYS overlap. One hand is mounted above the other. That's the way analog clocks work.
No, the _base_ of one hand always overlaps the _base_ of the other hand. But just as the base of a statue is not the statue, the base of a hand is not the hand.
@@yurenchu Tell that to a clockmaker.
@@Hidyman A clockmaker would not say that the hands always overlap.
@@yurenchu but a statue with no base is broken debris
@@viksox13 Yeah, like a fish out of water.
I tackled this in a pretty similar way to solution 3. Let's say x is how many ticks the minute hand covers. The hour hand has some head start 5h, where h is the hour (ex. 1pm = 5 ticks, 3pm = 15 ticks) + x/12 which is how many ticks the hour hand covers past its head start, since it's 1/12 as fast. We can use x = 5h + x/12 to find where the minute hand will be when it covers as many ticks as the hour hand- in other words, when they meet. Doing a bit of algebra gets us 11x/12 = 5h, or x = 60h/11, where h is the hour. Just plug the hours into the equation and you have your answer (ex. if the hour is 2, x = 120/11 ticks or 10 + 10/11 minutes)
22 times. It is enough to stop the hour hand and allow the clock to rotate. Over the course of 12 hours, the face completes one rotation in the opposite direction to the movement of the clock hands. 12-1=11, 11*2=22
Nope. It's 23 times. You made the same mistake as with the fence problem: A 100 m fence has a pole every 1 m. How many poles are there?
@@Nikioko 100 if the fence is circular or anyway closed. Different problem though 🙂
@@Nikioko 23 times per day would imply 46 times per two days, and 7*23 = 161 times per week, which are clearly not true (there are only 154 overlaps of hour hand and minute hand per week).
If the time of 12:00:00 A.M. (= midnight) belongs to the present day, then the time of one second after 11:59:59 P.M. belongs to the _next_ day.
@@yurenchu Nope. 23 times in one day implies 45 times in two days and 155 in a week. You have to calculate correctly.
What do you mean by a rotation in the opposite direction?
The second overlap happens around 1:05, pure logic suggests the only way both will be pointing at exactly 12 by the time noon arrives means the min hand fails to catch the hour hand once. And then simply that process is repeated for the pm so 22. Not sure if the interviewer would like my “no math needed” reasoning.
The day wasn't over yet when your 22nd overlap occurred. Account for the rest of the day.
I got 1 Hour, 5 minutes, 27 seconds for the overlapping hands interval (assuming no second hand), and the hands overlap 11 times over 12 hours, and 22 times over the course of a full day.
If I was asked a riddle in a job interview, I'd be like "Am I here to have a serious interview or to play games? How 'bout some "Call of Duty" instead?"
Funny how I can solve this in seconds but still fail in my mathematics exam
Lol good pun or joke
Cuz this question is not Maths but logic
And? Did you use math to solve it? No? So whats your point
@@ivaerz4977Nah you can convert a variety of problems into any given branch of what is mother science.
For eg you can convert the clock problem from a seemingly pure logical one to a physics one.
Just think about it ,lets say that the hour hand is a slow old ant moving at 1/12 th the speed of his younger child ant.
They are moving in a circle(or rather the circle of death),how many times would they meet each other if the younger person travles at a speed of 12 units per hour if the circle's length is 12 units.
The above is an easy question of relative speed/circular velocity.
@@boltez6507still a logic puzzle. If you want to math it, go for it. But it is still logic
There's another way to think about the first question which is faster than algebraically, radially, or graphically:
The big hand completes 12 full rotations every 12 hour period. So, if the small hand stood still, it would overlap exactly 12 times. But since the small hand makes 1 full rotation *in the same direction* as the big hand, you can subtract 1 to get 11 overlaps. Multiply by 2 for a 24 hour period and you get 22.
This was actually something I spent a lot of time thinking about in high school, so I can say off the top of my head that the answer is 11. Or, 22 times in a 24 hour period.
A 22 hour period, also known as a day.
If you imagine viewing the clock from a rotating frame of reference such that the hour hand is stationary, you are effectively cancelling 2 revolutions per day. Hence the minute hand will pass the hour hand 22 times in a day.
The problem with the question is purely semantic. It is ambiguous whether both the start and the endpoint count.
I mean, can you prove 24;00 is Midnight but 0;00 is not?
This is neat, to solve this puzzle with so many approaches
I remember this problem from high school. I solved it this way, which impressed Miss Sullivan. We know that the hour hand will take one revolution to go from noon to midnight, and at those times the hands align. We also know that the time between each alignment is exactly the same. (To see this rotate clock so that the alignment that happens between 1:05 and 1:10 is now at noon. The next one will land at exactly the same.) So the amount of time between consecutive alignments will be standard. Call that time x. Also let n be the number of alignments in those 12 hours. Since 12:00 is an alignment at noon and midnight, we must have x*n=12, with n an integer.
At 1:00 a complete alignment has not been made, so x>1 and hence n1 and a|12, then 12/a is an integer, call it m. Also b>1. x*n = x*a*b = 12 which implies that x*b = 12/a = m. So the b-th alignment happens exactly m hours have passed since 12, or at m o'clock. That means it happens on the hour, but the minute hand is on 12 an not at m. So with n relatively prime to 12, n = 1, 5, 7, or 11.
At 1:00 the minute hand had not an alignment. At 1:30, the minute hand has swept over the 1:05 to 1:10 region where the hour hand slowly moves. So 90 minutes or 1.5 hours is too long for x. So 1.5 > x = 12/n. Solving this inequality, we get n>8. The only integer relatively prime to 12 between 8 and 12 is 11.
n=11
Is your answer 11? are you asking others to substitute 11 for n in your equations? You do the math as your proof. Was Miss sullivan impressed with ( I don't know if your answer is 11 & she was impressed with that or if she was impressed at your formation of an equation which you did not calculate out to a conclusion.) So far to me Miss Sullivan seems easily impressed.
@@peterbaruxis2511 "Is your answer 11?" My very last thing I wrote was n=11. So the time will align 12/11 hours after noon.
And Miss Sullivan was impressed with an 11 year old's logic. Most high school math teachers would be impressed with creative mathematical prowess. (And before you bring it up, I was 11 when I started high school.)
@@DrR0BERT my answer is 23.
@@peterbaruxis2511 OK. So you use a 24 hour clock. Awesome. That makes sense.
@@peterbaruxis2511he clearly showed the result of his calculation in his post and it was good. Don't know what your issue is. A point in time cannot belong to two days. It is never simultaneously two different days at the same time. No one ever says it is Saturday and Sunday right now. That's why the date and day on your phone changes at midnight.
I stand firmly by 23, since I think it’s actually important and worthwhile to double count midnight. 11 overlaps per 12 hours, plus one once you reach the 12 again.
For starters, it’s like a solar eclipse. There isn’t only one instant of overlap, but a range of time for the totally. The minute hand is traditionally a little thinner than the hour hand, and there is a span of time when the entirety of the width of the minute hand is bounded by the wider hour hand. For a few seconds before and after midnight, the hands overlap, even though the angles of either hand are not identical. Recognizing that there is non-exact overlap is a useful physical observation.
It seems related to fence post problems and off by one errors. I tend to think it’s better not to just assume you include or exclude the last fence post, but should consider why you want to include it or exclude it. When you just guess before hand, it seems more likely to lead to a mistaken, than considering each potential problem separately, and making a case for or against double counting. If you’re looking at only the exact angle matching with a very calculus-focused analysis, I guess, but maybe I’ve got a more horological view on the puzzle. I do have a background in math, but I’m also a watch collector.
By that argument, you could say the answer is one, as the hour and minute hands are always overlapping. As they meet in the middle of the clock face and spin about the same axis, one has to be overlapping the other.
Clearly that's not what the question is asking for, though, so that's why I feel it better to treat it as a thought experiment and assume the hands to have 0 width. Either that or define "overlap" as "exact overlap of the center lines of the hands".
@@quigonkenny Even with zero-size hands, there's still a question about how we handle the interval. We can treat it as open on both ends, closed on both ends, or open on one and and closed on the other.
There's a non-absurd argument for twenty-one overlaps in a day, if you have open intervals on both ends. There are no doubt times when it makes sense to exclude the endpoints.
The case for including both midnights is basically the trig functions. We tend to say that Sine of an angle is zero at 0, pi, and 2*pi radians. It would seem strange to me to include only one endpoint. Understanding how the cycle works is part of the point, and I think that applies to clocks as well as triangles.
I totally agree. I thought it was 23 as well for that very reason. I knew that I was counting midnight twice because you can't assign it to one day or another. It happens simultaneously so it occurs on both days.
One day isn't over until it goes past midnight, and before it goes past they must overlap. It must be counted at the start and at the end.
Both of these are a stretch, though, and you can safely rule them out logically without further clarity in the question.
It is not ambiguous that a day does not end with midnight. As soon as we go from PM to AM, we've entered a new day. So 23:59:59.999... is the highest time in any day (excluding leap seconds) by international convention, which you can look up.
The fact that physical hands have non-infinitesimal width is fun to consider but you can be sure that's not what the question means because a) the definition of overlap would be arbitrary (what portion must overlap?) b) different clocks have different hands thickness and, crucially, c) every infinitesimal moment during overlap could be considered a "time" they overlapped. Intuitively, removing those absurd or impractical possibilities means interpreting an entire single overlapping session as one "time" of overlap, and since the answer must be true for each day, we can't steal the midnight overlap from the next day.
@@parodoxis You are trying to apply Zeno's paradox. 0.999... is mathematically proven to be equal to one. It's an infinite series that cannot stop until it reaches the actual end of the measurement period.
A 24 hour period does not end until the end of the 12th hour. You cannot stop counting until you actually reach midnight, at which time the hands overlap.
If you use Euler's formula to model the hour-hand as e^(i*X), we can model the minute-hand as e^(i*12X) since it's frequency is 12 times faster than the hour-hand. There's an obvious solution at X = 0. To get the rest, we can add the period to the hour-hand as e^(i*X + 2πn) where n is an integer. Set equal to each other and natural log both sides. The resulting solution is that the two hands should meet when X = 2πn/11. Look at the special case where n = 11. This is simply X = 2π, which is the trivial case of both hands pointing at 12 on the clock. This means that for every full rotation of the hour hand, there are 11 crossings between the two hands before they repeat the pattern. 2 full rotations is 1 day, so in total 22 crossings.
To get the times, notice that hour 1 on the clock is π/6 radians past the 12 (hour 2 is 2π/6, hour 3 is 3π/6 = π/2, and so on ...). Dividing 2πn/11 (the crossings) by π/6 (1 hour) yields 12n/11 where n is in the range [1, 22]. To convert to seconds, just multiply by 3600 seconds per hour.
E.g. the first crossing happens at (12 * 1 / 11) * 3600 seconds = 3927.27 seconds = 3600 seconds + 327.27 seconds = 1hr + 5 min + 27.27 seconds. Repeat for each n in [1, 22] (or [1, 11] and then add 12 hours)
If you think the answer is 22 and you think Euler's formula applies to this you either don't understand the formula or you do understand the formula but you don't know how to apply it.
It doesn't really matter when we agree the day starts. The clock hands overlap once every approx 1hr and 5 mins (little off, but it works).
So in the span of 12 hrs, they overlap 11 times (counting the overlap at the begining of each 12 hr period).
11×2 is 22.
I remember we did this in school. If I remember correctly, everyone got it wrong. Our homework assignment was to add the seconds hand and figure out how many times any two or more overlap. But there would be no homework if someone could answer how many times all three overlapped at the same time right away. We had to do the homework. I don’t remember, but I can’t imagine I was too happy when I got home and figured out the answer.
It depends on your definition of "overlap". Mine is "when one hand covers the other by any amount." So, only once at 12:00:00 and continues until someone pulls the top hand off or the day ends. The real puzzle is how anyone thinks riddle solving skills are a good proxy for engineering skills. It's an example of short-cut thinking that says a lot about the organization and very little about the interviewee, IMHO.
how is understanding how a piece of engineering works not a proxy for engineering skills?
"So, only once at 12:00:00" - huh? the minutes hand moves faster and passes by the hours hand every 65 minutes (and some seconds); and every time it happens there is some amount of overlap for a while;
so what are you trying to say?
I agreer with you and the definition of overlap is "extend over so as to cover partly".
I disagree that is is a riddle type question: “Why are manhole covers round?” Is a riddle question: you either have already heard it, have an aha moment in the interview, or you don’t, and it doesn’t really tell you anything.
If you started your response with “what do you meant by overlap, because they are always overlapping at the hub?”, no reasonable interviewer would think that was a bad question.
It reminds me of a company I was an interviewer at, one of our standard questions was “If we gave you a basketball and a tape measure, how would you estimate how many basketballs you could fit in this room?”, and a candidate who was otherwise acceptable would always give a reasonable answer, and answer the follow up about whether they though their answer would give an under- or over- estimate. Plus it was directly analogous to the kinds of simulations we wrote, without us having to dump a bunch of our context on them.
@@dstrctd Short-cut thinking does sometimes work, I assume. But I wonder how often it was confirmed with real data? In my experience, some of the best engineers failed their interviews but were hired by someone with previous experience of their skills and work ethic. I.e. the interviews failed. Since you never hired anyone who was not "otherwise acceptable" you have no experience with counter examples. In other circles, this would be called "confirmation bias."
12:45 onwards - think of it simply like this. Imagine that the hour hand doesn't move but the minute hand moves (minute hand degrees per minute - hour hand degrees per minute) = 6° - 0.5° = 5.5°. Then divide that by the number of degrees to move as if the hour hand doesn't move - 90° / 3 = 30°. With that you can divide 30° with 5.5° to get (5.45 repeating) minutes. You can then multiply the remainder (0.4545454545) by 60 seconds to get 27.27 repeating seconds.
Even if you thought the hands met once an hour, your original rationale for “23” makes no sense. You started at 24, but then subtracted one because 12 midnight overlapped with the next day. But the only way there would be such overlap is if you counted both the midnight at the start of the day, and at the end of the day-and if you did that, you would have started with 25 occurrences, not 24. Subtracting the one for the overlap then brings you to 24 occurrences, not 23.
So your wrong answer was, essentially, twice as wrong as what you thought it was.
The hands of a clock (hour and minute hands) overlap approximately every 65 minutes. In a 12-hour period, they overlap 11 times. Since a day consists of two 12-hour periods, the hands of the clock overlap 22 times in a full 24-hour day.
REASONING:
The minute hand completes one full revolution (360 degrees) every 60 minutes.
The hour hand completes one full revolution every 12 hours, which means it moves 30 degrees every hour (or 0.5 degrees per minute).
To find the time it takes for the hands to overlap, consider the following:
Relative Speed: The minute hand moves faster than the hour hand. The minute hand moves 6 degrees per minute (360 degrees / 60 minutes), and the hour hand moves 0.5 degrees per minute. So, the relative speed between the two hands is 5.5 degrees per minute (6 - 0.5 degrees).
Initial Position: At 12:00, both hands are at the same position. After some time, the minute hand will catch up to the hour hand again.
Time to Overlap: For the minute hand to catch up and overlap with the hour hand, it needs to make up the difference in position, which increases as the hour hand moves forward. Every hour, the hour hand moves 30 degrees (since 12 hours = 360 degrees, so 360/12 = 30 degrees per hour).
The minute hand must cover this 30-degree difference at the relative speed of 5.5 degrees per minute. So, the time 𝑡
t it takes for them to overlap is:
𝑡 = (30 degrees) / (5.5 degrees per minute) ≈ 5.4545 minutes
Interval Between Overlaps: Since each overlap happens after the minute hand has gained an additional 30 degrees on the hour hand, this process takes approximately 65.45 minutes.
Therefore, the hands of the clock overlap every approximately 65 minutes, resulting in 11 overlaps every 12 hours, or 22 overlaps in a full 24-hour day.
- CHATGPT 4o
@@dante0817 You stand by what chatgpt say's. Ask chatgpt if twelve AM Tuesday ( the very beginning of the day ) is the same moment in time as the very end of the day on Monday.
Another line of reasoning can be similar to the lapping problem, where the minute hand has lapped the hour hand by (24-2)=22 days. The minute hand makes 24 revolutions per day while the hour hand makes 2 revolutions per day.
My brain went "ELEVEN!" until I read "in a DAY" and I went "oh it's 22 then"
he already calculated it right at the beginning of the video, getting the wrong 24-1 = 23 as answer, because he made one small logical error :
the clock has 2x12 hours and not 1x24, and thus "half the solution" is 12-1 = 11, for a final result of 22 (instead of 24-1 = 23)
ps: i once saw a big wall clock on a house (i believe it was in venice/italy) that really had 24 hours for a single turn, and thus would result in that first answer of 24-1 = 23 :-)
on the other hand, i am not entirely sure whether that clock really had multiple hands or only just one single hour hand, for an answer of ZERO overlaps :-)
The best part of the solution is your presentation of four methods. Method 2 is simple and surprises me.
Can you make one, in which you also account for the second’s hand( all three hands of a clock). Also great content!
Thanks! From the list of times I think only 12:00:00 am and pm would be overlapping.
If you look at the list of overlap times, it's clear by inspection that the seconds hand will not be at the same angle as the hour and minute hands at any of their overlaps, except when both are pointing to 12.
@MindYourDecisions what about the times any 2 of the 3 hands overlap on a smooth motion clock... and what times do they overlap on a stop action clock (all 3 hands)
@@chichi90504 On a smooth motion clock,
the hour hand and minute hand overlap *22* times per day,
the minute hand and seconds' hand overlap *1416* times per day,
the hour hand and seconds' hand overlap *1438* times per day.
Since gcd(22, 1416, 1438) = 2 , they overlap (all three simultaneously) only 2 times per day (namely at midnight and noon).
So the number of times per day that _any two_ hands overlap, is 2 + (22-2) + (1416-2) + (1438-2) = *2872* times.
On a stop action clock (each hand moves discretely, at 360°/60 = 6° increments), every pair of two hands still overlaps at the same frequency (= times per day) as with the smooth motion clock; but all three hands simultaneously overlap *22 times* a day, each time for a full second; namely at 12:00:00 , 1:05:05 , 2:10:10 , 3:16:16 , 4:21:21 , 5:27:27 , 6:32:32 , 7:38:38 , 8:43:43 , 9:49:49 , 10:54:54 (A.M. and P.M.).
@@chichi90504 For some reason my reply in this thread is not visible unless the comments are sorted in order of "Newest" (instead of in order of "Popularity").
23 is a legitimate answer. I knew that I was counting midnight twice because you can't assign it to one day or another. It happens simultaneously so it occurs on both days, and it's not the same midnight since they are a day apart from each other.
One day isn't over until it goes past midnight, and before it goes past they must overlap. It must be counted at the start and at the end because it occurs at the start, and at the end.
By international standards, 0:00 is the start, and only the start, of any given day. Nothing is counted twice nor needs to be (disregarding leap seconds)
@@parodoxis What if you measured from midday on one day to midday on the next day? Would you still ignore the second time they overlap at the 12 o'clock position? The clock is doing the same thing.
This isn't about international standards, and it isn't even about measuring time. It's about how many times the hands of this mechanical device overlap, if you start it with both hands overlapping at the top, and run it until the small hands revolves around the dial twice.
You count the first overlap when it starts, and it finishes in the same position with the hands overlapping again.
If you count the overlap at the start of the measuring period, then you have to count the overlap at the end of the period. The hands are in the same position both times.
@@user-bf8tv8xv4w If you start the clock at any time of the day where the hands are overlapped and count the overlaps over the course of 24 hours as measured by that clock you would get the same answer every time. 23, yes you need to count the beginning once and the end once.
@@user-bf8tv8xv4w "if you start... and run twice..." - if that were stated in the question, then yes, I'd agree, since you are focusing on a mere duration.
But the question didn't say that, it said in "a day". That's a well known interval, part of a series of fixed periods which don't overlap.
That difference matters, as I explain in my other reply to you.
@@parodoxis The question asked how many times "in A day". "A" day is denoting a singular stand alone period of time. The word "A" denotes a single unit.
If the question had asked how many times "PER day", or how many times "DAILY", then the day is treated as a part of a series.
Like the fence post dilemma. If they are spaced at one metre, and a fence is two metres long, then the question "How many posts in a two metre long fence", It must have three posts. One in the middle and one at each end. A panel will fall down if it doesn't have a post at each end. If the question is "How many posts per two metres" the the answer is two. One post for every metre.
You see the difference between "A", and "PER"?
the answer is 23.
that's because the answer most viewers learned at school is to a different question: how many times do the hands cross.
in this question it is "overlap".
the hands start overlapped and they finish overlapped.
Exactly.
The only way to get 22 is if you stop the clock before midnight. before it can run for a full day.
A full day ends the moment *before* midnight.
@@parodoxis No it doesn't. A full day ends exactly at midnight. If it ends a moment before midnight, then it's a moment short of a full day.
What if you counted for half a day, from midnight to midday? Do you count the overlap at the start of your measuring period, but ignore the overlap at the end? The hands are in the same position so you have to count both.
@@parodoxis One day ends the very same moment that the next day begins. Count the seconds at the beginning of this day but don't start counting until the last second of yesterday has fully elapsed.
@@parodoxis A day end's at the exact moment the next day begins.
I had a little chuckle during method 3 when Presh spoke about angles of 0.5 degrees when I remembered that degrees themselves are sometimes subdivided into 1/60th units called minutes and 1/3600th units called seconds, and we could talk about the rate of rotation of the hands as angular minutes per time minute and angular seconds per time second
Most people will catch "let's not count midnight twice", but the catch is to also not count noon twice.
The catch is that noon count's once and we are talking about two different midnights and each one count's once.
11/12*24=22 times.I basically used the concept of relative speed.
Basically the hour hand reduces the relative speed of the minute hand,its relative speed is reduced to 11/12 rotations per hour.
so time taken for the first meet is 12/11 hours which is basically 1-05-27 for the second meet its 1-05 +12/11~2-11...etc and so on so its basically an arithmetic series.
"I am Clockwork, master of time."
My initial logic was to figure out when does the hand first overlap after midnight and I found out that it was around 1:05 AM which is 65 minutes after midnight. Then i just proceeded to calculated it like below
(24*60)/65 = 22.15 (rounded down to 22 times)
24 hours in a day
60 minutes per hour
65 minutes per hand overlap
I think you've fallen victim to Zeno's paradox. The distance between markings on the clock lies in the interval, not the end points. There is no "little bit of time" between the end of one day and the start of the next. For the clock to transverse 24 hours it must start and end with the hands in the same position. If they start overlapped they must end overlapped. Hence the answer is 23.
@@jamessmith2522 The victim of the paradox is you. Clock works in cycles, not intervals. Cycles reset, so you cannot double count their ends. The only answer is 22.
Think logically. Per day you only get 12:00 AM once. With your logic, you would get it twice.
Technically the end of one day and start of another are simultaneous, but we have to declare the time at exactly 12:00 and 0 seconds midnight to belong to exactly one day or the other by convention (i guess you could just say it's part of both days or neither or some other weird exception to every other time on the clock but the alternative is a lot simpler). the convention is that it belongs to the starting day rather than the ending one.
Start counting at 1am of 1st jan and end to count at 1am of 2nd jan to not have problems at midnight
@@Fasalytch Same story - 1 am happens exactly once on 1st Jan, and exactly once on 2nd Jan, no double counting.
before watching the video my brain says once per hour so in a day 24 times - but knowing Presh's videos I know 40% of the time i am wrong if i go with gut assumptions ... so went to pen and paper and calculator to work out what I typed below === now to find out how much I am wrong by.
if they start overlapped at 00:00 then it would be 01:05:27:273 then 02:10:54:546 then 03:16:21:82 and so on
the two hands overlap every 65 minutes 27 seconds 273 milliseconds, don't think we need to go smaller than milliseconds - so not once every hour but once every hour, five minutes and almost a half minute.
if they start overlapped at 00:00 and you do not count that then
1st =then it would be 01:05:27:273
2nd = then 02:10:54:546
3rd = then 03:16:21:82
4th = then 04:21:48:11
5th = then 05:27:15:14
6th = then 06:32:42:16
7th = 07:38:09:19
8th = 08:43:36:22
9th = 09:49:03:24
10th = 10:54:30:27
11th = 11:59:57:30
12th = 13:05:27: etc etc
13th = 14:10:54:5
14th = 15:16:21:8
as you can see we are now in the teens and the ordinal numbers for passing do not match the time numbers
15th = 16:21:48:11
16th = 17:27:15
17th = 18:32:42
18th = 19:38:09
19th = 20:43:36
20th = 21:49:03
21st = 22:54:30
22nd = 11:59:57
and the next time they pass will be the next day. as they start and end the day overlapped from the eleven/twenty-three crossing. --
(please note these are the start times for the hands to start to over lap, but the minute hand takes more than 3 seconds to completely pass the hour hand)
sorry had to work it out with degrees arc minutes and arcseconds for my brain to understand it. it seems logical to be 24 but maths says no.. it is i think this term is right a logical fallacy for my brain at first.
another way is there are 1440 minutes in a day.... divide that by a little over 65 and you will get 22
then reading the comments the Le Mans 24 hour race came to mind bobby and fred start the race at the same time, in the time it takes bobby to do 24 laps fred has done 2 and they pass the checkered line at the same time... b24 - f2 = 22 lap difference
Your list of overlap times is NOT the correct answer to the question. The question included "round the times to the nearest second". In the first interval, you correctly omitted the "3/11 seconds" fraction, because is was less than 0.5, however, in the second interval this fraction will sum up to "6/11 seconds" which is greater than 0,5 and the nearest second will be the following second, not the one before.
You are correct, but to my knowledge, all of his lists have 2:10:55, not 2:10:54, as the third overlap, so although he didn't mention the rounding issue, he did do the rounding correctly himself.
there’s a very simple solution - when the clock is at 11 the minute hand only touches the hour hand again at 12. since it’s travelling faster than the hour hand, if the minute hand intersects the hour hand strictly between 11 and 12 it will reach 12 faster than the hour hand, contradiction.
at any other time it’s easy to see there must be an intersection of the two hands between time t and t+1. so the answer is 11 intersections in 12 hours or 22 in a day.
23 is illogical. There are only 12 hours on this clock! Not 24! So in 24 hours you use the clock twice (2 * 12 = 24). In 12 hours the hands meet 11 times. 2 * 11 = 22. It is as simple as that.
23 is perfectly logical, as he explained his reasoning and it made perfect sense.
Something can be logical yet incorrect.
My fridge stopped making ice or water flowing. A logical assumption is that it may have been too cold and the pipes froze.
This particular case was not caused by frozen pipes.
That's an example of logical but incorrect.
I approached this as a drt catch-up problem: the hour hand moves at the rate of 1 unit per hour; the minute hand at 12 units per hour. The question is how long it will take the minute hand to “lap,” or catch up with the hour hand given the difference in their rates (12 - 1 = 11) and the 12 unit distance between them. That is 12units/11, or 1 hour, 27 minutes and 3/11 sec. So that is the first time the hands overlap. Just keep adding this amount successively to get the exact times the hands will overlap in 24 hours. In that 24 hours or 24 units they will overlap 24// 12/11 times, or 22 times.
Def. of "overlap": to extend over so as to cover partly. The hour and minute hand are continuously overlapping at the pivot point, so they overlap an infinite number of times. Poorly worded question.
I have a slightly different way of calculating the exact times (or time intervals) at which the hands of the clock overlap (or in other words, the time interval between two successive overlaps). The value of the time interval between two successive overlaps can then be used to calculate the number of times the minute hand overlaps with (or crosses) the hour hand in 24 hours.
We can use the analogy of relative speed (or velocity). Suppose a police car, moving at a constant speed of ‘v1’, is trying to chase down another car going in the same direction at a constant speed of ‘v2’ (v1 > v2). If the distance between the two cars is ‘s’, then how much time will the police car take to chase down the other car? In this example, since the two cars are moving in the same direction, the relative speed of the police car with respect to the other car is (v1 - v2), and therefore, the time required will be t = s/(v1 - v2). If the two cars happen to go around in the same direction (say, clockwise) in a loop (like formula 1 cars completing multiple laps on a race circuit!) of total loop distance/length ‘s’, then the two cars will cross each other every t = s/(v1 - v2) time interval.
We can use the above analogy to calculate the time intervals at which the minute hand will overlap with (or cross) the hour hand. Since the minute hand covers 360 deg every 60 min, its (angular) speed v1 = 360/60 = 6 deg/min. Likewise, since the hour hand covers 360 deg every 12 hours (= 12x60 min), the (angular) speed of the hour hand will be v2 = 360/(12x60) = 0.5 deg/min. Note that the minute hand is analogous to the police car, the hour hand to the car being chased, and both hands are moving (rotating) in the same direction (i.e., clockwise). Hence, the relative (angular) speed of the minute hand with respect to the hour hand will be (v1 - v2) = 6 - 0.5 = 5.5 deg/min. The (angular) distance ‘s’ here is 360 deg. Therefore, the time interval between two successive overlaps (or crossings) of the two hands of the clock will be t = s/(v1 - v2) = 360/5.5 = 65.4545…. min = 65 min & 27.2727… sec.
Now, from this, we can easily calculate the number of times the minute hand overlaps with (or crosses) the hour hand in 24 hours. We simply need to divide the total time (i.e., 24 hr = 24x60 min) by the time interval between two successive overlaps (i.e., 360/5.5 = 65.4545… min). Thus, the number of overlaps in 24 hours = 24x60/(360/5.5) = 24x60x5.5/360 = 22.
I did the same mistake as you (getting 23) but in a slightly different way. I knew there was no overlap in the 11 o'clock hour and completely glossed over the fact that the 11 o'clock hour occurs twice on a 12 hour clock!
I can appreciate the 1st solution, but it relies too much on going through the hand movements process.
I ended up liking the 2nd solution best and to find the times, you just have to set one of the intervals of the minute hand equations equal to the hour hand equation (y = 30x). Because of the constant rates, I used the 1 o'clock one (y = 360x - 360) and got the intersection at x = 360/330 which after translating into a time, comes out to 1 hour 5 minutes and 27.272727... seconds.
minute hand makes 24 rotations, hour hand makes 2 rotations in the same direction, meaning it overlaps 24 - 2 = 22 times.
This is a valid way as you know that you are only dealing in full rotations, as both end up at the start after 24 hours.
When the minute hand rotates 60 notches, the hour hand won’t be there because it moved 5 notches in that time. So overlaps have to be over an hour apart, so the number of overlaps has to be less than 24 in a day.
Similar solution but slightly different view: on its way to each hour, the minute pointer will catch up once with the hour pointer (also on its way to 11h and also 12h).
Only on its way from 0h to 1h the minute pointer already catched up with the hour pointer at the start.
I find this easier to visualize
@7:40, as a programmer I immediately raised a red flag here. If you round off to 27 seconds and then keep adding that to the starting time, you accumulate an error which gets bigger each time. For example, it should be 2:10:56 seconds, because rounding the seconds (6/11) there would go to the next highest second, not the previous lowest second. So, you should use the full precision available and round off to whole seconds at each time point of interest. This will give you a slightly different and more accurate result. This may not be significant here, but it definitely could be if you were dealing with money, nuclear weapons, navigation, and a million other things that could be done in software. Anyway, a simple way to get the answer to the original question is to say that VERY roughly, the hands overlap every 1 hr 5 minutes, so 1440 minutes (1 day) / 65 minutes = 22.15... which rounds to 22 times. If I were under time pressure in an interview I'd just come up with the answer that way. Calculating the exact times and not falling into the accumulated rounding error trap would be something that could be discussed - demonstrating that you are aware of the pitfall is likely to be far more important than solving the exact times!
He did round to the next second, not to the previous second. 2:10:55 is the correct rounding of 2:10:54.545454.... ; your suggestion of 2:10:56 is actually wrong.
@@yurenchu Yes, I realised that later. Nevertheless, what he described, about rounding, then repeatedly adding, would lead to accumulated rounding errors. In practice perhaps that's not actually what he did, but rounded each time, as suggested. I didn't check each time individually. But it goes to show that describing an algorithm requires as much precision as the algorithm itself!
@@GRAHAMAUS At 7:11 , he says that he's going to calculate the _precise_ times. It then stands to reason that he of course would be using the exact value of "this interval" (1 + 1/11 hours), not the approximation (1:05:27) which he merely displayed to get a concrete feel of the size of the interval.
Yes, his wording may be a bit sloppy (as he failed to mention explicitly which value he is repeatedly adding, and that he's rounding the outcomes to the nearest second), but you not checking if the displayed times were indeed off (and failing to see that 2:10:55 was _not_ calculated by merely adding 1:05:27 to 1:05:27 ; note that 27+27 = 54, not 55) is also sloppy.
By the way, you don't have to check every listed time individually to determine if an accumulation of round-off errors after repeated addition has occurred. Just look at the last time in the list, 10:54:33 ; adding 1:05:27 to it results indeed in 12:00:00 , so apparently there is no accumulation of error, as apparently the exact value of the interval had been repeatedly added.
So you round-off, you come-up with approximately 22 & you conclude 22? & you claim to be a programmer and you apply this line of thinking to nuclear weapons and navigation? I hope you are not under time pressure the next time you have a nuclear or navigational decision to make. It's 23.
I knew it was 22 because it doesn't overlap at 1:00, it overlaps slightly after, and even more after 2:00, such that they won't overlap between 11:00 and 12:00, so the answer for one full rotation of the hour hand is 11, and twice in a day makes 22
"They never overlap ask Xeno"
*Extends hand to accept the job offer 🤝*
👏 Very clear and well explained video of a not-so-hard everyday problem. And the graphical method amaised me the most. 🙏
While the answer is 11, the excact time each overlap occures depends on the clock design
A couple of extra wheels could have the hour arm move just twice in every hour (so a small amount after the hour and then nothing until 3 seconds before the next hour)
Plent of funky watch and clock movements have been built.
This is a slightly more complex version of the around the world in 80 days story, where the one traveling thought he was a day late as he forgot that by traveling around the world he saw the sun coming up one more time than the ones who stayed at the start/finish. This one is more complex as it goes round twice, this would be equal to the hypothetical twice around the world in 160 days novel.
Both hands start aligned at 12:00. After exactly 1 hour, the hour hand has moved on 1/12 of a turn = 30 degrees; but now the hour hand is pointing to the 1 while the minute hand is still pointing to the 12, and by the time the minute hand reaches the 1, the hour hand will have moved on a little further still. The two hands will have to come into alignment again sometime between 13:00 and 13:10; again sometime between 14:10 and 14:15; sometime between 15:15 and 15:20; between 16:20 and 16:25; between 17:25 and 17:30; between 18:30 and 18:35; between 19:35 and 19:40; between 20:40 and 20:45; between 21:45 and 21:50; and between 22:50 and 22:55. This is just before the hour hand reaches the 11, and the next time they line up again will be midnight; so there are 11 occasions when the hands align in 12 hours, meaning 22 occasions in a full day. Now, we can save the effort of summing up a bunch of infinite series, even if we know an identity for that, by noting that the speed of the hands is constant; so these alignments must be occurring at regular intervals of 86400/22 = 3927.2727..... seconds, i.e. at the following times of day rounded to the nearest second: 00:00:00, 01:05:27, 02:10:54, 03:16:21, 04:21:49, 05:27:16, 06:32:43, 07:38:10, 08:43:38, 09:49:05, 10:54:32; 12:00:00, 13:05:27, 14:10:54, 15:16:21, 16:21:49, 17:27:16, 18:32:43, 19:38:10, 20:43:38, 21:49:05 and 22:54:32.
The times in your comment are not rounded to the nearest second, they are rounded down to the previous second.
Here’s my way of getting there. Suppose we start the day at 5 minutes past midnight. In every hour, the minute hand is going to “catch up” at some point where the hour hand is between that hour and the next. So the first on will be between 1:05 and 1:10. The second between 2:10 and 2:15. So when we get to the 11th crossing, it’s sometime between 11:55 and 12:00, but this one’s weird, because it really happens at 12, and the crossing that goes with 12 happens between 12:00 and 12:05, but it’s really right at 12:00 and is the same as the crossing that goes with 11. So there are 12 numbers, but 11 and 12 share the same crossing, so only 11 crossings. Whole thing happens again before we get back around to 5 minutes past midnight, so 22 crossings.
Ask yourself this; when did the 22nd crossing occur? Was the day over yet at that point? What happened at the exact moment that the day ended?
According to Zeno's paradox, the minute hand will never reach the hour hand if it has a head start 🙂 LOL ! BTW, I liked method 2 the most. For method 1 you started with a constant of 1:05:27 that came from nowhere, even though it made sense.
Zeno's paradox is not reality.
The first half of the question is super easy. The second half (the exact time) is more difficult.
I think the correct answer should be 23 because per the analysis you used; starting from 12 midnight, the hour hand will overlap the minute hand at 12:00 pm and continue to 1:05pm and so on before it gets to 12 o'clock AM.
My solution was a lot like Solution 3. I figure, okay, at 1am, the hour hand is 1/12th of the way around the clock, while the minute hand is 0/12th of the way around the clock. The minute hand moves 12 times as fast. Therefore, I just need to find when:
hour_starting_position + hour_speed = minute_starting_position + minute_speed
1/12 + x = 0 + 12x
1/12 = 11x
1/132 = x
So the hour hand will travel 1/132th of a full rotation before the minute hand catches up. That's 1/11th of a single hour, which is 5.45 minutes, which is 5 minutes and 27 seconds. Same as solution 3, but thinking in terms of "percent of distance traveled" instead of "degrees".
Mathematically you're absolutely right. But, intuitively, which I also missed in the first place, it's still absolutely logical for the answer to be 22 times. The reason is: the clock is a 12 hour clock, so, you have two separate 12 hour periods in which the hours hand makes a full revolution, making it skip one overlap every 12 hours. If it was a 24 hour clock, your (and my) initial answer of 23 hours will be true. Only the last hour, the 23rd, would be skipped. Now, there are VERY unusual clocks in which the hours hand skips a full hour every time the minutes hand hits 12 o'clock. In this case, the answer would be: full 24 times.
Yes, but no. Noon count's once (it does not mark the beginning or the end of any day.) The midnight at the beginning of the day and the midnight at the end of the day each count once.
The answer is 2, since at every interval where the minute isn’t 0, the second hand is wildly displaced from the minute hand in the few seconds it could be considered overlapping (the nearest being in hours 3/8, where the seconds respectively are 22/38). This can be proven simply by noting that each hour has an overlapping minute and hour hand EXCEPT the 11th hour, which for the minute hand to catch up it would be the 12th hour. Therefore we can prove the relation between the number of hours to amount of overlapping can have a maximum of 11/12ths the hour count. However, doing this same technique with the minute and second hands results in 59 overlaps per minute, and since 11 and 59 do not have any common factors, their LCM can only be 59*11 which as inputs result in a full revolution of the slowest hand. Noting that a clock only tells time for half of a day, we can safely assume that all 3 hands will only overlap twice per day, both accounting for a fluid second hand. A stiff second hand would need more in-depth calculations, but could also be proven based on the results I initially calculated for.
H(x)=floor(12x/11) // hour hand offset in relation to input hour x
M(x)=floor(60x/11) // minute hand offset in relation to input hour x
S(x)=round((3600x/11)%60) ± 1 // second hand offset in relation to input hour x, ±1 for possible inaccuracies
I like problems solved more simply. I got it wrong, but I wished I would have solved it like this:
The hands obviously overlap on the 12 two times per day. (The third time doesn’t count because it would be the next day.)
Then each hour they overlap a little more past the start of the hour, two times in one day, so that is 20 more times.
Finally, the 11 hour overlap can’t be counted because it would be so close to 12 by the time the minute hand caught up.
This would be casual not formal math. Of course, Presh would show the math behind what I said, that’s his channel.
Embarrassingly, I forgot that the hour hand moves from hour to hour as much as the hour is finished. Elementary students forget this or think that the hour hand just sits on the hour until the next hour. So, I made an elementary mistake.
There is a very simple way (mathematical) to calculate how much that fraction (1/11) equals to in seconds. So, known that the hands overlap 11 times every 12 hours. That equals once every hour and 1/11 of an hour. Since there are 60 minutes in an hour and 60 seconds in a minute, there are 3600 seconds in an hour, and if you divide 3600 by 11 you get 327.272727¯. 300 seconds equal 5 minutes, so 327.27¯ = 5 minutes, 27.27¯ seconds. Therefore, the hands overlap ~ every one hour, 5 minutes and 27 seconds.
Many people want to say 24! BUT the actual answer is 22. Clock hands overlap at: 12:00, 1:05, 2:10, 3:15, 4:20, 5:25, 6:30, 7:35, 8:40, 9:45, and 10:50 TWICE A DAY (AM and PM). There's no overlap at 11:55 because the hour hand is moving closer toward 12 when the minute hand is at 11.
it's 23.
This is great but as a former interviewer at AWS, i didn't ask this question of any of the candidates i interviewed. We just have too much ground to cover amd not a whole lot of time to get the data points we need. In fact, like many of you, i guessed 24.
I solved the problem of how many times the hands of a clock overlap in a day by focusing on the observation that the minute hand moves at 60 minutes per hour, while the hour hand moves at just 5 minutes per hour (here, minutes refer to the markings on the clock dial, treated as units of distance). So, the minute hand is 55 minutes per hour faster than the hour hand. This reduces the question to: how long will it take the minute hand to complete a lap and overtake the hour hand? The problem becomes equivalent to asking how long it would take the minute hand to complete a lap if it moved at 55 minutes per hour. Viewing everything from the minute hand’s perspective, we know the speed is 55 minutes/hour, and the distance to cover is 60 minutes (clock markings). Therefore, it would take the minute hand 60/55 hours to make one full lap around the clock dial.
From there, I created a function that takes the hour (h) as an argument and outputs the exact minute in that hour when the hands overlap:
*roundDown(h mod 12 × (3600/55) mod 60 / sgn(11 - h mod 12))*
The division is simply there to trigger an error if the argument is 11 (or 23), as the clock hands never overlap at 11 a.m. or 11 p.m.
This problem also got me thinking about the saying, "even a broken watch is right twice a day." Well, not necessarily. As shown, you can arrange the hands of a clock in such a way that they never align during normal clock operation-like overlapping the hands somewhere between 11 and 12 o’clock!
Great explanation. Reminds me of the quarters thing where one quarter stays still, and the other quarter rolls around its perimeter, and how many times does it rotate.
The correct answer is 23 times. Midnight is the dividing line between two days, thus each midnight belongs to TWO days. There are two midnights, plus 21 other overlaps. This is like the fence post problem, each midnight belongs to two days, just as each fence post belongs to two fence panels.
Also 12:00 am and 12:00 pm are both midnight! "am" means "ante meridiem" or "before midday". The only 12:00 that is before midday is midnight. Similarly, "pm" means "post meridiem" or "after midday". The only 12:00 that is after midday is midnight. You should distinguish the 12:00 times by using "noon" and "midnight" as suffixes, not "am" or "pm".
Midnight only belongs to the following day.
@@carultch Even if I accepted your point, which I don't, 23:59:59.9 belongs to the same day, and the hands are already overlapping. Clock hands aren't infinitely thin.
@@jerry2357 They are for purposes of this question.
Another algebraic approach. The minute hand is moving at a rate 6°/minute so at time t minutes has move 6°t. The hour hand moves at 0.5°/minute, so at time t minutes has moved 0.5° t. The hands overlap whenever the difference between how far each has moved is a multiple of 360°. I.e., they overlap whenever (6°-0.5°) t = 360° k, where k is a non-negative integer.
Solving for t, the hands overlap whenever t = 720/11 k.
We want to know how many overlaps occur in the first day, so we want to find all non-negative integers k that give t with 0 ≤ t < 1440 (the number of minutes in a day). Plugging t = 720/11 k into that inequality we get 0 ≤ 720/11 k < 1440. Rearrange to give 0 ≤ k < 1440x11/720 = 22. There are 22 non-negative integers that satisfy that: 0, 1, 2, ..., 21, giving us 22 times where the hands overlap.
The simple way to think of this is that the question is asking how many full revolutions the minute hand makes _relative_ to the hour in 24 hours.
In 24 hours, the hour hand makes 2 revolutions relative to the face, thus the face makes -2 revolutions relative to the hour hand. The minute hand makes 24 revolutions relative to the face, thus the minute hand makes 24 - 2 = 22 revolutions relative to the hour hand in 24 hours. There will 23 overlaps if we count both end points, or 22 if we don't. We can then simply divide 24 hours by 22 revolutions to calculate all the overlap times.
The problem with this question is, it doesn't say if the starting point of midnight on one day is counted or not, so if it is it will be 23 in a day, if not then 22. Like most questions they are quite ambiguous and sets it up for multiple answers that are correct in how you interpret the question.
In 12 hours, the hour hand makes a full turn.
In the same time, the minute hand makes 11 turns more than the hour turn.
So if we consider the hour hand to be fixed (and the scale moving backwards instead) and only consider the distance between hour hand and minute hand - then we get 11 revolutions by the minute hand in a period of 12 hours.
Since both hands move at constant pace, it follows that the two hands must meet each other exactly every (12/11) hours. Or 22 in total in 24 hours.
22 times. It happens at 1/11th of a hour further each hour. In 12 hours, the little hand goes around 12 times, but the big hand also goes around once. If the big hand didn't move, the little hand would catch it 12 times and pass it. But since the big hand does move, it gets a little farther than 1 hour away each time and by the 12th time, it doesn't quite get there. This is easier to see if you start at any time OTHER than noon.
1) 1:05:27 3/11 seconds
2) 2:10:54 6/11 seconds
3) 3:16:21 9/11 seconds
4) 4:21:49 1/11 seconds
5) 5:27:16 4/11 seconds
6) 6:32:43 7/11 seconds
6) 7:38:10 10/11 seconds
8) 8:43:38 2/11 seconds
9) 9:49:05 5/11 seconds
10) 10:54:32 8/11 seconds
11) 12:00:00
An easy way to think of it is to start by thinking of if the Hour hand ticked, when they overlap, 12:00, 1:05 etc. then notice that you would have 11:55 and 12:00 as both potential times they overlap, despite being 5 minutes apart. When the hour hand is constantly slowly rotating, by the time the 11:55 comes around, the hour hand is basically on the 12.
(This is due to the interactions of 12 revolutions for the minute hand, and 1 Revolution of the hour hand)
Then double for 24hrs.
Every single hour, the hour hand moves one hour marker (one twelfth) and the minute hand moves 12 hour markers (12 twelfths) so each revolution around the clock (which is every 12 hours) will require an extra one twelfth distance to overlap both hands. Over the course of those 12 hours, we lose that one twelfth as a whole hour, which means 11 overlaps every 12 hours
The trick is to notice that between 11 and 1 they overlap only once. So, if you count the part from 11 to 1 as an "hour" then it is true, they overlap once in every hour, but now there are only 22 hours in a day. To simplify this further you can notice that our 12 is redundant regarding this question, so it is even easier to see why we count with 22 hours per day :)
I would agree with your intuition of 23. I think the last overlap counts, since the question asks about a real world clock with finite hands. Your excellent presentation details how to consider modeling an ideal clock. As such the original question is open to interpretation.
How can you agree with his intuition of 23? His intuition of 23 is that the answer is possibly 24 , but since we _mustn't_ count the "last" overlap, we must substract 1 , hence giving the answer 24 - 1 = 23 . Your motive for arriving at 23 is exactly the other way around, and counter to his intuition!
To me it was pretty obvious from the start that the hour hand was going to be able to "escape" from the minute hand twice a day giving you the answer of 22; and using symmetry you could get all the times easily. HOWEVER I suppose the interviewers would be more interested in some analytical explanation like the ones you showed us here. My favourite was method 4 because this puzzle has an element of the Zeno's paradox in it, so that infinite sum was very pleasant to my eyes; even that 60 at the beginning could be interpreted as 5/(12^-1). Awesome video yet again.
But it isn't 22.
While trying to work the problem for a clock with a hand for the second, I figured out that 24 is a possible valid answer for this problem, depending on your definition of the clock. Since the size of the hands (in seconds) are not given, it can be expected to be irrelevant, so two close overlap are considered distinct, then if you define the clock as having a finite number of instantaneous step (as with a mechanical step by step mechanism), using the second method of the video but with now stair curve there is now another overlap : the last step before the end of the cycle.
But these overlaps overlaps. There are two if you cont in number of step an only one in number of events (but the last+first event will be the only one with a 2 steps duration).
The hands of a watch/clock sit on a pinion. They are not 0 units of measure wide, so the minute hand, which is sitting on top of the hour hand in the hand stack (if the seconds hand were present it would be sitting on top of the minute hand), is constantly overlapping the hour hand.
Pedantic? Yes. Missing the point? Yes. Correct? Also yes.
I think twenty two. If the hour hand was stationary, it would 24 times, but because the hour hand is moving away, it will take longer to interact, losing one interaction per rotation. Since it rotates twice per day, it should be 24 (hours per day) - 1 (overlap per rotation of hour hand) x 2 (rotations per day.) Or 24-2=22
My friend got hired by Google even though she failed this test and others. They just liked her.
24-((360+30)*24-360*24)/360 = 22
24 - (390*24 -360*24)/360 = 22
24 - 720/360 = 22
24 - 2 = 22
Full cycle around the clock = 360°
One hour by hour hand = 360°/12=30°
One hour by minute hand = 360°
Each hour the minute hand need that additional 30° to catch hour hand.
360°+30°=390°
So full distance that minute hand need to catch hour hand 24 times -> 390°*24=9360°
Minute hand will travel 360° each hour, so 360°*24=8640° is the distance it will travel for exactly 24 hours, and 24 circles.
9360°-8640°=720°
That means 720°/360°=2 is the number of circles it would need above the 24 full hours in order to meet the hour hand 24 times, and so 24 times it would then meet with that additional 30° - 2 so the number of times above 24 full circles of minute hand (so 24 hours) = 22 number of times it would meet the other hand without that 30° additional degrees each hour.
Here's another algebraic solution:
1. If X = # of overlaps, S = seconds in a day = 86,400, and P = the period in seconds between overlaps of the two hands, then:
X = S/P = 86,400/P
2. Now calculate P (the period in seconds between overlaps):
a. Get the rotation velocities (in degrees per second, or dps) of each hand, Vm and Vh:
Vm = 6°/60seconds = 1/10 dps
Vh = 30°/3,600seconds = 1/120 dps
b. If Velocity = Degrees/Time, then Degrees = Time * Velocity:
Dm° = T * Vm = T/10°
Dh° = T * Vh = T/120°
c. The hands overlap when the distances are equal, not counting full circles, so:
Dm° mod 360° = Dh° mod 360°
T/10° mod 360 = T/120° mod 360
d. Because we can't calculate a result with the mod operator, we take the first overlap after midnight to get P: the minute hand has traveled once around, while the hour hand has not, so we know that
P = T when T/10 - 360 = T/120, solving for T we get T = 3927 + 3/11
3. From step 1:
X = 86,400/P = 86,400/(3927 + 3/11) = 22
You forgot to count the last fencepost.
Intuitively: It will unlikely be 24 due to it being too simple. It should not be 23 either - since a 24 hour period involves 2 complete 12-hour cycles; meaning that the answer has to be an even number (as AM and PM are exactly the same case). So this leaves us with the answer probably being 22 (unless intuition fails us and the answer is actually just simply 24).
What about option 5: "None of the above" ?
I think a better solution is -
Mark a point on the dial of the clock... say at 3 or 4... Hour hand cuts this point 2 times in 24 hours and Minute hand cuts it 24 times in 24 hours....
Since both the hands are moving in same direction, their relative speed should be subtracted (24-2) = 22 times.
(Note that this will work when the cycles are complete... e.g. for 12 hrs it will be (12-1) 11 times... that too is simple... but for 6 hours, we need to plot a point at appropriate location otherwise we will either get 5 or 6. Best will be to take the next available number on the dial. e.g. take 1 as the point if it is between 12:01AM to 12:01PM)
If it was a defective clock and one hand was in clockwise direction and the other hand was in anti-clockwise direction... then the relative speed is in opposite direction and we have to add those, so in that case it will be 24+2 = 26 times!
I more or less used you method 3 but turned it into a straight line race with one thing at 60x/hr and the other at 5x/hr with a 5x headstart. Basically distance rather than degrees.