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.
@@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 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.
The hand doesn't disappear and reappear in the new position, it still has to overlap the hand. This is the kind of complication I'd expect from a junior or intermediate engineer
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.
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.
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".
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.
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.
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.
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 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)
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.
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
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.
Personally I got it from a rule that also applies to planets the normal direction AKA the same direction as they orbit. They always have exactly one more rotation per year than they have solar days for the same reason any time you have two things that rotate in the same direction there is going to be one extra period of rotation of the faster one in each complete rotation of the slower one between each return to the same relative position. The faster one need to catch up with the fact that the slower one has now also completed one rotation, the distant stars appear to rotate around the poles precisely one more time every year than our own sun does for this reason because Earth rotation needs that extra rotation to catch up with the orbital motion over the period of a year.
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.
but by that logic, they overlap ALL the time. Look close to the center of the clock. Notice both hands are surrounding the centerpoint? They ALWAYS overlap, at least a little bit. BTW, that would result in only a single overlap, not 23.
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.
**the big hand does not complete 12 full rotations every 12 hour period; it completes ONE full rotation every 12 hour period. But yeah your idea is correct regardless of this typo. So 1:xx has overlap (somewhere), 2:xx has overlap (so 2 total), 3:xx has overlap (so 3 total)... thus hour # =total number of overlaps, from 1 to 10 equals 10 overlaps. Until 11:55, as the minute hand gets closer to the hour hand, the hour hand moves away and gets closer to 12, only overlapping at noon, so count+1. Repeat for 1-10pm, only don't add midnight tomorrow, add midnight this morning/starting point count+1. Thus 10+1+10+1=22. Yep we had the same idea, just reworded it.
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.
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 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.
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.
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.
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.
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.
This confused me too. I understand the 3/11ths will eventually add up and exceed the 27 seconds to become 28, but it doesn't happen at 2:10:55, it would be another 2 hours before it would be a 28 second addition.
Yes - 22 is the correct answer. I was in the hospital for a few days back in 2022, and a large clock was mounted on the wall opposite my bed. I had to stay awake for long periods of time every day, and this very topic intrigued me. The hospital's clock stacked the hour, minute, and second hands at 11:59:59, and the three clicked together to 12:00:00. This synchronized movement was something I had never heard anyone mention (and I am pretty old now!).
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.
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?
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
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.
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.
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.
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.
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."
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.
I did it with the first merhod. Reasoning was: As a starting point, I knew the max overlaps in 12h is 12 times, and then knowing that the hour hand is always moving and will be a difference of 1/12th at the last overlap, this means I need to subtract 1 every 12. Now I know it’s 11 overlaps, I can calculate the next overlap at any point is 60 + 60/11
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").
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
@@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.
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.
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.
Why not count midnight twice? Hands don't have zero width. So overlap starts just before midnight and noon, and ends just after. 0 to 24 = 25 overlaps.
Exactly. Think about what a single complete cycle of the clock represents, which is half a day. It removes complications and makes the problem easier to solve to 11 times per half day, which can simply be multiplied by 2 to get the answer for a full day. Though this only works if a half day ends after 11:59:59.999... AM/PM (a reasonable assumption) and the hands are infinitely narrow (which is not a given and should be clear in the requirements).
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.
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 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.
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.
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.
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 :-)
I’d fail the interview on account of how thoroughly I overcomplicated it for myself, but on the bright side I got the right answer. I ended up with an equivalent to solution 4, and it was only after I got the result that I realized infinite sums weren’t needed to figure out where the hour hand would be at each crossing.
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.
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.
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.
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
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.
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.
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?"
You’d be surprised how the skill to solve this is closely related to engineering. In other words, the precise relationship between different values in a set.
My route was similar to "Method 2" but a little more brute force using logical deduction first, then math. I deduced 11-overlaps-every-12-hours, and used this to divide 12hours into the interval to find the times overlaps occur. We know 12:00am and 12:00pm will both be at least 2 overlaps (and another overlap resetting at 12:00am the next day), so we can solve for only a 12-hour period, then double our answer for the full 24-hours count. If we assume constant hand speeds, we know the hour hand will be "slightly ahead" each time the minute hand reaches a given hour# (hour# x5 minutes ahead), so we can deduce that our interval should be approximately 1:05:00. This also tells us that we cannot reach 12 overlaps in a 12hour period as that would have to be exactly every 60minutes, so our answer MUST be
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.
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.
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.
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!
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
Well I'm pleased (and a bit surprised) that I actually got the first part right straight away, just thinking it through (and I suppose visualising it) in my head. But my thought process was that they would overlap 12 times as the minute hand moves round the clock, except the hour hand is also moving round once in that time, which effectively takes away one of the times. Then multiply that by 2 for a full 24-hour day.
Oooh and then I got the second part as soon as you said how far the clock needs to advance. I realised how that calculation could (easily) be done and how that would give the full answer. But before that I was thinking I'd have to write some complex code to figure it out lol.
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".
I feel this answer is really determined by if the clock in question has sweep hands (continuous smooth motion with infinite many positions) or step hands (distinct positions that the hands stop at with finite positions). On a clock with a sweeping hand, there are infinitely many positions, so in that time between 11:59 and 12, then yes the only location that they will overlap is at the 12. But on a clock with with step hands, at the position before moving to midnight, since there are only so many locations the hands will be in, they will actually over lap at the 11:59 position. And then again at the 12:00 position. So in a single day according to a clock with step hands, there will be a full 24 overlaps.
There's also another solution using trigonometric approach: Angular speeds of big and little for 1 turn : Tb = 12*Tl with Tl = 2pi/(12*3600) Time : t Phase difference between Big and Little is 0 if Cos(Tb*t) = Cos(Tl*t + 2k*pi) for any k belonging to natural integrals ==> t = 2k*pi/(11*Tl) gives the time spent by the big hand in seconds from 00:00:00 to the next intersection of big and little for k = 1, t = 3927,27s = 1h05min27,27s
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 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.
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.
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
22 and 23 can both be correct. if a day is 24h then mathematically 00:00 is both the start and end of a single day in order to perform 2x360 degrees circle.
The velocity of the minute hand is 12 times larger than the velocity of the hour hand. Say there are 3600 discrete points where hour hand increments between any two hour marks. Then the velocity of the hour hand is 1 pt/s. The minute hand velocity is 12pt/s. So at 1pm the distance is 3600, velocity diff is 11pt/s. 3600/11 = 327.273 i.e. 5min, 27sec . To compute the second intersection just multiply 327.273 by 2 since initial distance is 7200. That’s it.
1. The hour and minute hands overlap roughly once every hour, but not exactly on the hour. The only time they overlap exactly is at 12:00. 2. In a 12-hour cycle, the hands overlap 11 times because each overlap happens a little later than the previous one (e.g., around 1:05, 2:10, 3:15, etc.), with the last overlap occurring just before 12:00 again. 3. In a 24-hour day, the cycle repeats twice, giving you 22 overlaps in total (11 overlaps per 12-hour cycle). So, the clock’s hands overlap 22 times over the course of a full day.
Starting at midnight, the hands are overlapping so we're at 1. Between midnight and 1, the hands never overlap. Somewhere between 1 and 2, the hands must overlap again because the minute hand started behind the hour hand, and after 10 minutes is guaranteed to pass it. So we're at 2 overlaps going into 2am. This line of reasoning (between hours n and n+1, after 5(n+1) minutes they must overlap) will work all the way up until we go from 11am to 12pm. The hands never overlap between the hours but instead at 12pm. So in a 12hr cycle they overlap 11 times, meaning in a 24hr cycle they overlap 22 times.
Use relative velocity: Hour hand speed : 30 degrees/ hr Min hand speed: 360 degrees/ hr Start of the day: both are at 0 degrees In the frame of the hour hand: Velocity of minute hand: 360 - 30 degrees / hr = 330 degrees/ hr In order to surpass the hour hand , the minute hand needs 360 degrees to cover. So in 24 hours no of times it will cross hour hand is: (24 * 330 ) / 360= 22
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.
The hour hand fully rotates twice a day and the minute hand 24 times. Reducing the amount of overlaps by 2 from 24. Actually got it right first time by this reasoning.
In Fig. 10.141, ABC is a triangle in which ZB = 24C. Dis a point on side Bo AD bisects ∠BAC and AB = CD. BE is the bisector of ZB. The measure of BAC (a) 72° (b) 73° (C) 74 (D) 95
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.
It does depend on your clock to some extent though. In some clocks there is a smooth transition from 11:59:00 to 12:00:00 with an analog sweep, rather than discrete position steps. However, in some clocks, the transition of the hour and minute hands from the 11:59 to 12:00 is a discrete step; in this case, the hands will over lap at BOTH 11:59 and 12:00, adding 11:59 AM and 11:59 PM to the total.
They overlap once per hour. Day starts at 12 aka 0. 0 to 24 is 25 times, because you include 0. Also since hands don't have zero width over lap begins and ends at different times. So while 24hrs is a new day, the overlap started just slightly before midnight. It's kind of like an eclipse.
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 simplicity of the divide by 11 got me XD. Great answer. When I saw hands catching up to hands, my brain quickly went to the sum of an infinite geometric series, same as that presented in solution 4. I solved it in hours though. Minutes seems like a weird choice, but to each their own.
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.
Think about number of overlaps when hour hand is on or just after each digit, and tally. Start at 12pm, so: 1 > Hour hand >= 12 - I 2 > Hour hand >= 1 - I 3 > Hour hand >= 2 - I … 11 > Hour hand >= 10 - I 12 > Hour hand >= 11 - doesn’t happen (as it hits 12am midnight. Now repeat) 1 > Hour hand >= 12 - I 2 > Hour hand >= 1 - I 3 > Hour hand >= 2 - I … 11 > Hour hand >= 10 - I 12 > Hour hand >= 11 - doesn’t happen as it would hit noon again But we don’t include the next noon as we included it at the start ==> total is 11+11=22
The hands of the clock overlap 0 times, or an infinite number of times throughout the day (depending on whether it counts if it is already overlapped when you start counting). This is because the base of the two hands constantly overlap at the center of the clock. o·ver·lap verb (overlaps, overlapping, overlapped) | ˌōvərˈlap | [with object] extend over so as to cover partly: the canopy overlaps the house roof at one end | [no object] : the curtains overlap at the center when closed. • [no object] cover part of the same area of interest, responsibility, etc.: their duties sometimes overlapped. • [no object] partly coincide in time: two new series overlapped. noun | ˈōvərˌlap | a part or amount which overlaps: an overlap of about half an inch. • a common area of interest, responsibility, etc.: there are many overlaps between the approaches | there is some overlap in requirements. • a period of time in which two events or activities happen together: an overlap of shifts.
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.
This one’s quite easy tbh, although doing it in your head can be a bit hard. The hands overlap every time the hours and minutes hands are approximately on the same hour mark: 00:00, 01:05, 02:10, etc, so they overlap every 1 hour and 5 minutes or 65 minutes. A day has 1440 minutes, 1440/65=22.15 times, 22 because overlaps can’t be partial. The correct answer is 22.
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.
Had to laugh so hard, this perfectly describes my employer
I see you've worked on government projects.
You're one funny man, I'll give you that
Definitely describes our govt.
ROFLMAO!
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
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.
@@yurenchu Nah
The hand doesn't disappear and reappear in the new position, it still has to overlap the hand. This is the kind of complication I'd expect from a junior or intermediate engineer
That's not a hand, its a wrist! ;-)
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.
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.
@@peterbaruxis2511 ad ignorantum
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...
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.
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.
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.
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 knew it boiled down to modular arithmetic, but couldn't remember enough to write the equation correctly
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)
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.” …?!
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
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.
Exactly the way I came to the same conclusion. Why Peter thinks the answer is 23 is beyond me, he cant have seen the video :)
Personally I got it from a rule that also applies to planets the normal direction AKA the same direction as they orbit. They always have exactly one more rotation per year than they have solar days for the same reason any time you have two things that rotate in the same direction there is going to be one extra period of rotation of the faster one in each complete rotation of the slower one between each return to the same relative position. The faster one need to catch up with the fact that the slower one has now also completed one rotation, the distant stars appear to rotate around the poles precisely one more time every year than our own sun does for this reason because Earth rotation needs that extra rotation to catch up with the orbital motion over the period of a year.
This answer I think is the simplest to understand, curious why he didn't chose to show it.
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
but by that logic, they overlap ALL the time. Look close to the center of the clock. Notice both hands are surrounding the centerpoint? They ALWAYS overlap, at least a little bit. BTW, that would result in only a single overlap, not 23.
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.
No algebra, plotting, degree measuring or exact overlap times needed.
@@StomDoth Except, the specific times were requested as part of the solution, weren't they?
@@undercoveragent9889 Good point. I only answered the part of the question I was interested in.
Yeah this is how I came to the answer myself when I saw this video and was happily surprised when i got the answer correct
**the big hand does not complete 12 full rotations every 12 hour period; it completes ONE full rotation every 12 hour period.
But yeah your idea is correct regardless of this typo.
So 1:xx has overlap (somewhere), 2:xx has overlap (so 2 total), 3:xx has overlap (so 3 total)... thus hour # =total number of overlaps, from 1 to 10 equals 10 overlaps. Until 11:55, as the minute hand gets closer to the hour hand, the hour hand moves away and gets closer to 12, only overlapping at noon, so count+1.
Repeat for 1-10pm, only don't add midnight tomorrow, add midnight this morning/starting point count+1.
Thus 10+1+10+1=22.
Yep we had the same idea, just reworded it.
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.
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.
The best part of the solution is your presentation of four methods. Method 2 is simple and surprises me.
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 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.
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.
I expect only mathematicians would be able to solve this interview question.
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 🙄
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?
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.
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.
This confused me too. I understand the 3/11ths will eventually add up and exceed the 27 seconds to become 28, but it doesn't happen at 2:10:55, it would be another 2 hours before it would be a 28 second addition.
Yes - 22 is the correct answer. I was in the hospital for a few days back in 2022, and a large clock was mounted on the wall opposite my bed. I had to stay awake for long periods of time every day, and this very topic intrigued me. The hospital's clock stacked the hour, minute, and second hands at 11:59:59, and the three clicked together to 12:00:00. This synchronized movement was something I had never heard anyone mention (and I am pretty old now!).
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.
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?
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
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.
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.
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.
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.
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."
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.
I just figured they'd overlap every 65 minutes. 1440 minutes in a day, divide 1440/65 = 22.15 meaning it only manages 22 overlaps
Just watched the video. Wrong calculation right answer lol
I did it with the first merhod.
Reasoning was:
As a starting point, I knew the max overlaps in 12h is 12 times, and then knowing that the hour hand is always moving and will be a difference of 1/12th at the last overlap, this means I need to subtract 1 every 12.
Now I know it’s 11 overlaps, I can calculate the next overlap at any point is 60 + 60/11
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").
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
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
👏 Very clear and well explained video of a not-so-hard everyday problem. And the graphical method amaised me the most. 🙏
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.
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.
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.
Why not count midnight twice? Hands don't have zero width. So overlap starts just before midnight and noon, and ends just after. 0 to 24 = 25 overlaps.
@@phalcon23 Because at the second midnight it's the next day, at 23:59, it's still the same day.
But the hands will start to overlap slightly before midnight. And they will stop overlapping slightly after. @@Dexaan
Exactly. Think about what a single complete cycle of the clock represents, which is half a day. It removes complications and makes the problem easier to solve to 11 times per half day, which can simply be multiplied by 2 to get the answer for a full day. Though this only works if a half day ends after 11:59:59.999... AM/PM (a reasonable assumption) and the hands are infinitely narrow (which is not a given and should be clear in the requirements).
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.
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 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.
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.
"I am Clockwork, master of time."
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.
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 :-)
I’d fail the interview on account of how thoroughly I overcomplicated it for myself, but on the bright side I got the right answer. I ended up with an equivalent to solution 4, and it was only after I got the result that I realized infinite sums weren’t needed to figure out where the hour hand would be at each crossing.
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.
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.
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.
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
So 22% of your viewers wear mechanical wrist watches
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.
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.
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?"
You’d be surprised how the skill to solve this is closely related to engineering. In other words, the precise relationship between different values in a set.
My route was similar to "Method 2" but a little more brute force using logical deduction first, then math. I deduced 11-overlaps-every-12-hours, and used this to divide 12hours into the interval to find the times overlaps occur.
We know 12:00am and 12:00pm will both be at least 2 overlaps (and another overlap resetting at 12:00am the next day), so we can solve for only a 12-hour period, then double our answer for the full 24-hours count.
If we assume constant hand speeds, we know the hour hand will be "slightly ahead" each time the minute hand reaches a given hour# (hour# x5 minutes ahead), so we can deduce that our interval should be approximately 1:05:00. This also tells us that we cannot reach 12 overlaps in a 12hour period as that would have to be exactly every 60minutes, so our answer MUST be
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.
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.
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.
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!
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
Well I'm pleased (and a bit surprised) that I actually got the first part right straight away, just thinking it through (and I suppose visualising it) in my head. But my thought process was that they would overlap 12 times as the minute hand moves round the clock, except the hour hand is also moving round once in that time, which effectively takes away one of the times. Then multiply that by 2 for a full 24-hour day.
Oooh and then I got the second part as soon as you said how far the clock needs to advance. I realised how that calculation could (easily) be done and how that would give the full answer. But before that I was thinking I'd have to write some complex code to figure it out lol.
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".
I feel this answer is really determined by if the clock in question has sweep hands (continuous smooth motion with infinite many positions) or step hands (distinct positions that the hands stop at with finite positions). On a clock with a sweeping hand, there are infinitely many positions, so in that time between 11:59 and 12, then yes the only location that they will overlap is at the 12. But on a clock with with step hands, at the position before moving to midnight, since there are only so many locations the hands will be in, they will actually over lap at the 11:59 position. And then again at the 12:00 position. So in a single day according to a clock with step hands, there will be a full 24 overlaps.
There's also another solution using trigonometric approach:
Angular speeds of big and little for 1 turn :
Tb = 12*Tl with Tl = 2pi/(12*3600)
Time : t
Phase difference between Big and Little is 0 if Cos(Tb*t) = Cos(Tl*t + 2k*pi) for any k belonging to natural integrals
==> t = 2k*pi/(11*Tl) gives the time spent by the big hand in seconds from 00:00:00 to the next intersection of big and little
for k = 1, t = 3927,27s = 1h05min27,27s
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 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.
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.
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
Definitely a very clear and understandable series of explanations. Well done.
22 and 23 can both be correct.
if a day is 24h then mathematically 00:00 is both the start and end of a single day in order to perform 2x360 degrees circle.
The velocity of the minute hand is 12 times larger than the velocity of the hour hand. Say there are 3600 discrete points where hour hand increments between any two hour marks. Then the velocity of the hour hand is 1 pt/s. The minute hand velocity is 12pt/s.
So at 1pm the distance is 3600, velocity diff is 11pt/s. 3600/11 = 327.273 i.e. 5min, 27sec . To compute the second intersection just multiply 327.273 by 2 since initial distance is 7200. That’s it.
1. The hour and minute hands overlap roughly once every hour, but not exactly on the hour. The only time they overlap exactly is at 12:00.
2. In a 12-hour cycle, the hands overlap 11 times because each overlap happens a little later than the previous one (e.g., around 1:05, 2:10, 3:15, etc.), with the last overlap occurring just before 12:00 again.
3. In a 24-hour day, the cycle repeats twice, giving you 22 overlaps in total (11 overlaps per 12-hour cycle).
So, the clock’s hands overlap 22 times over the course of a full day.
The correct answer should be once. They are always overlapping as they are both on the same pin in the centre of the clock
Starting at midnight, the hands are overlapping so we're at 1. Between midnight and 1, the hands never overlap.
Somewhere between 1 and 2, the hands must overlap again because the minute hand started behind the hour hand, and after 10 minutes is guaranteed to pass it. So we're at 2 overlaps going into 2am.
This line of reasoning (between hours n and n+1, after 5(n+1) minutes they must overlap) will work all the way up until we go from 11am to 12pm. The hands never overlap between the hours but instead at 12pm. So in a 12hr cycle they overlap 11 times, meaning in a 24hr cycle they overlap 22 times.
Use relative velocity:
Hour hand speed : 30 degrees/ hr
Min hand speed: 360 degrees/ hr
Start of the day: both are at 0 degrees
In the frame of the hour hand:
Velocity of minute hand: 360 - 30 degrees / hr = 330 degrees/ hr
In order to surpass the hour hand , the minute hand needs 360 degrees to cover.
So in 24 hours no of times it will cross hour hand is:
(24 * 330 ) / 360= 22
OMG velocity, degrees- gimme a break. You're not counting the last fencepost!
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.
The hour hand fully rotates twice a day and the minute hand 24 times.
Reducing the amount of overlaps by 2 from 24.
Actually got it right first time by this reasoning.
In Fig. 10.141, ABC is a triangle in which ZB = 24C. Dis a point on side Bo AD bisects ∠BAC and AB = CD. BE is the bisector of ZB. The measure of BAC
(a) 72°
(b) 73°
(C) 74
(D) 95
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?
The first half of the question is super easy. The second half (the exact time) is more difficult.
It does depend on your clock to some extent though. In some clocks there is a smooth transition from 11:59:00 to 12:00:00 with an analog sweep, rather than discrete position steps. However, in some clocks, the transition of the hour and minute hands from the 11:59 to 12:00 is a discrete step; in this case, the hands will over lap at BOTH 11:59 and 12:00, adding 11:59 AM and 11:59 PM to the total.
They overlap once per hour. Day starts at 12 aka 0. 0 to 24 is 25 times, because you include 0.
Also since hands don't have zero width over lap begins and ends at different times. So while 24hrs is a new day, the overlap started just slightly before midnight. It's kind of like an eclipse.
22 is right (24 turns - 2 turns). But based on the definition 23 can also be right if you count the starting point as first overlap.
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 simplicity of the divide by 11 got me XD. Great answer. When I saw hands catching up to hands, my brain quickly went to the sum of an infinite geometric series, same as that presented in solution 4. I solved it in hours though. Minutes seems like a weird choice, but to each their own.
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.
Think about number of overlaps when hour hand is on or just after each digit, and tally.
Start at 12pm, so:
1 > Hour hand >= 12 - I
2 > Hour hand >= 1 - I
3 > Hour hand >= 2 - I
…
11 > Hour hand >= 10 - I
12 > Hour hand >= 11 - doesn’t happen (as it hits 12am midnight. Now repeat)
1 > Hour hand >= 12 - I
2 > Hour hand >= 1 - I
3 > Hour hand >= 2 - I
…
11 > Hour hand >= 10 - I
12 > Hour hand >= 11 - doesn’t happen as it would hit noon again
But we don’t include the next noon as we included it at the start
==> total is 11+11=22
The hands of the clock overlap 0 times, or an infinite number of times throughout the day (depending on whether it counts if it is already overlapped when you start counting). This is because the base of the two hands constantly overlap at the center of the clock.
o·ver·lap
verb (overlaps, overlapping, overlapped) | ˌōvərˈlap | [with object]
extend over so as to cover partly: the canopy overlaps the house roof at one end | [no object] : the curtains overlap at the center when closed.
• [no object] cover part of the same area of interest, responsibility, etc.: their duties sometimes overlapped.
• [no object] partly coincide in time: two new series overlapped.
noun | ˈōvərˌlap |
a part or amount which overlaps: an overlap of about half an inch.
• a common area of interest, responsibility, etc.: there are many overlaps between the approaches | there is some overlap in requirements.
• a period of time in which two events or activities happen together: an overlap of shifts.
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.
This one’s quite easy tbh, although doing it in your head can be a bit hard.
The hands overlap every time the hours and minutes hands are approximately on the same hour mark: 00:00, 01:05, 02:10, etc, so they overlap every 1 hour and 5 minutes or 65 minutes. A day has 1440 minutes, 1440/65=22.15 times, 22 because overlaps can’t be partial.
The correct answer is 22.