The Math Clock for Geeks

Try telling the time on this wall clock!

When I was young, I always had a hard time figuring out the time based on the positioning of the clock’s hands. I remember my Mom trying to teach me in so many different ways, but I just was not able to get a hold of it. With that, math was not my strongest subject either, so I am happy this clock was not around then, or I would have NEVER learned to tell the time.

This clock presents the hours by different mathematical equations, so it would have made my life simply horrible. The equations seem fairly easy for me now, but there is one that appears to be a little more complicated than the rest of them: the 5th hour, for it adds the “factorial” into the mess. Without knowing that 5 o’clock belonged there, I may have never figured it out…would you?

Via: CNET News

90 thoughts on “The Math Clock for Geeks

  1. Alex.

    MIT statisics class agrees .9999 repeting is 1 and the door problem Tim and simi posted are correct but if u have any question just contact MIT class of 2009 alum except Logan lyxwler

  2. Jen.

    I have one of the Unit Circle. I can barely remember a thing from high school, but I can draw one of those puppies in under a minute.

  3. World Time & Zone links.

    Sums it all up. I guess those who like maths are have a good time, (pun intended).

    Tring to look of 0.00001 errors but is the time correct? It is for some some-times but for others by the time the light travels form the clock to the eyes it will be incorrect LOL

    Click above link for World Time & Zone links

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  11. Zwaan.

    An infinitely reducing series always equals 1, otherwise, nothing would move, re the old problem of the second hand moving from 11 to 12. First it goes half-way, then a quarter of the remaing half, then 1/8, then 1/16, and so forth. It would never actually reach 12, except it sort of does, doesn’t it?

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  15. basicmath.

    Anyone who thinks that .9999… is a different number from 1 is simply unaware of the DEFINITION of a real number. Read up on any construction of the real numbers (via Cauchy sequences or Dedekind cuts) and you’ll understand why 1=.99999…. If you don’t know what either Cauchy sequences or Dedekind cuts are, you’re working under faulty assumptions.

  16. Geegee.

    As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality.(Albert Einstein)

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  18. printzen.

    All the talk about whether .9 recurring is or is not equal to 1 is unnecessary with regard to the all nines clock as 7 is better represented using 3 nines by 9/.9 – √9 = 7 exactly. Also the clock should more properly be referred to as the 3 nines clock as each number is represented by 3 nines, not any number of nines and not less than 3 nines (12 for example can be represented by 9 + √9 and only needs 9 + 9/√9 if the aim is to represent each number on the clock face by a combination of 3 nines exactly).

  19. Monty Hall.

    Larry and Bob, I’m afraid you’re both wrong. I won’t waste my energy explaining why, but I’ll refer you to which has LOTS of explanations why your analysis is flawed. Don’t worry… it took many, many great statisticians (and humble, little me) a long time to understand and accept this problem and it’s seemingly unintuitive solution.

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  21. Larry.

    Hi Simi:

    Actually Bob’s correct. There’s no reason for the probability ‘set free’ when the quizmaster eliminates one of the doors to preferentially go to the doors you didn’t pick. Instead it’s like the eliminated door never existed in the first place (all of the remaining doors including the one you picked go from 20% to 25%). Continuing your example, if the quizmaster then eliminated doors 3 & 4 (leaving only your door #1 and door #5) your theory is that door #5 is at 80% and door #1 is at 20%, which isn’t the case (it’d be like coming into the game when three of the doors were already open and being asked to pick a door – you’d pick one of the two unopened doors and have a 50-50 chance of being right). Another way to look at it: there are 5 doors, and you pick door #1 but I pick #5. Initially we each have a 20% chance of being right. If doors 2, 3, & 4 are then eliminated in succession, we would each have the same probability of being right at each and every stage (we’d each go from 20% to 25% to ~33% and finally to 50%). There’s no way for you to be stuck at 20% while I move up to 80% (and if there was, what if I picked door #1 and you picked door #5?).
    Hope this helps,

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  24. simi.

    hello bob

    solving problem sets in statistics is all about dealing with the new information that has been handed to us. the timing and the information content in most cases will effect the odds. for arguement sake lets say there were 5 doors with one prize in them, u pick one doo(door number 1) which holds a 20% chance of holding the prize, now the other 4 doors hold an accumulative 80% chance of holding the prize ( picture it in ur mind as two separate units). now the new information comes into place since we are told that a certain 1 of the 4 doors does not hold the prize ( door num 2) the accumulative odds dont change they are only split equaly amongst 3 remaining doors thus just under 27% while the first pick is still at 20%. just like in poker if u know for sure one of your outs has been mucked because a card flashed u can rely on that information to calculate ur new odds of winning the hand.

  25. Bob.

    When there are 3 possible doors the odds that you have chosen the right door are 1/3.
    Once the quizmaster has eliminated one of the doors the odds that you have chosen the right door increase to 1/2.
    These are the same odds for the other remaining door ….
    There is no “smart move” You can change your choice or stick to your original one and the odds are the same.

  26. jake3988.

    Oh god, not the incessant idiots that keep claiming that .9 repeating isn’t 1.

    Look, think about it. How many zeros will you need to put before the 1 in the decimal? infinite! You can’t have an infinite number and then have a 1, it’ll never actually occur.

    Thus, .9999 is 1.

  27. Tim.

    To switch to door number 3 is the smart choice.

    At the start, there is a 1/3 chance of picking the right door. When a door is picked, the is a 1/3 chance it will be the door you chose, and a 2/3 chance it will be one of the other doors.

    When the host eliminates one of the other doors, the 2/3 chance changes to the only remaining door. Therefore there is a 1/3 chance it will be the door you chose, and a 2/3 chance it will be the other door. So switch to double your chances of being right 😀

  28. simi.

    here a cute riddle for u guys.

    in a game show the are 3 doors that lead to 3 diff rooms. only one room holds a prize. the contender is given a choice by the host and picks door number 1. the host quickly adds that the prize is not in room number 2, and gives the contender a second choice between door number 1 and door number 3. what would be the smart choice?

  29. simi.

    1 5 6 7 how to get to 21
    1 – 5/7 = 2/7
    6 divided by 2/7 = 21
    the other 30$ room riddle is cute….
    it has nothing to do with maths but with semantics. the equation 27+2 is 29 is absurd
    since the 27 itself includes the 2 that the clerk pockets, if she hands them back 3 buck
    the total is no longer 30 it becomes 27 and that 27 includes the clerks 2 bucks obviously

  30. Jon.

    I’m so glad I stumbled across this page. I WILL have one of these clocks!

    Also, I can quite confidently say that 0.999…=1. I’ll go ahead and trust my maths degree for that one.

    For the sake of argument we can try to imagine what people mean by an infinitesimal real number: a positive number smaller than any other positive number given. Let’s define this number by x, so that given any number y, x is less than y. In fact, rather than ‘an’, let’s say ‘the’ infinitesimal, as there can be no number smaller. Then that is to say that 1-x=0.999… is the closest real number to the number one without itself being the number one. Then 0.999… is a unique decimal representation of a real number.

    Now let’s assume that 1-x (i.e. 0.999…) and 1 are different numbers. Then taking the average of the two numbers gives us a new number, say z, such that z=(0.999…+1)/2=(1-x+1)/2=1-x/2. This new number must be closer to the number one than 1-x is, as with x being a positive number it must be true that x is greater than x/2. Here we reach a contradiction. x was earlier defined to be the smallest possible difference between 0.999… and 1, but through that assumption we have contradicted ourselves in concluding that a smaller number must exist. Hence the assumption was wrong. So 0.999… and 1 cannot be different numbers, and so must be the same number. Hence 0.999… and 1 must be distinct decimal representations of the same real number. []

    It may be worth adding that (using the same definition of x) if I were to say that there is a number 1+x so that 1+x=1.000…1 then this would immediately ring alarm bells as it suggests that it is a finite number. The only way to express this as an infinitely recurring decimal would be 1.000… which is obviously equal to one, with comparably little confusion.

  31. The Smartest Man Possible.

    >> The difference between 1 and .9 repeating is
    >> Lim(x->0), i.e. the smallest positive number
    >> possible.

    First, you assume Lim(x)(x->0) != 0. To humor you, I’ll assume that as well.

    Next, you claim that Lim(x)(x->0) is the smallest positive number possible.

    But then, Lim(x)(x->0)/2 is half of the “smallest positive number possible.” Oops! A contradiction!

    I challenge you come up with the smallest positive number possible. I will then divide that number by 2 and laugh at your awkwardly huge number. And then divide it again just to be obnoxious about it! 🙂

    0.(9) is 1.

  32. phord.

    @peace: the clock is made of equations using three nines.

    A similar problem I encountered once: “Using nine nines and any combination of plus, minus, divide and multiply (and parentheses), one can reach many integer solutions. For example, 1 can be expressed as 1=9/9+(9-9)*9-9+9-9+9. What is the smallest non-negative integer which cannot be reached in this manner?”

    The challenge was to solve this problem programmatically. It was a fun diversion. Later when I went to an interview with Microsoft they asked me several similar questions on the spot. In a break, I asked them the 9 9’s question. One of them was really intrigued and emailed me the answer the next day. He had figured it out by hand. 😯


  33. The0retico.

    Hmm, i think it does not work like this…
    From the existence of supreme in real number, there is a number between racional number 1/3*3 and racional number 1. So there is a difference between 0.(9) and 1, it is exactly the supreme of

  34. Andy.

    love the clock! 😀 love the 0.(9) debate! i agree with the top mathematicians that:

    0.(9) = 1

    now there seems to be a lot of americans commenting on this post, …and i’m british, …but may i say, it’s MATHS!! not MATH!!



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  36. Dave.

    “Meh” – you may be right if you are stating that using addition on 1/3 or 1/9 fractions is flawed mathematics, but the real proof has already been posted. And this isn’t “making it look like there is validity in the statement”, nor is it flawed. Point nine recurring does equal one.

    x = 0.(9)
    10x = 9.(9)
    10x – x = 9
    9x = 9
    x = 1

  37. Meh.

    Its seems this debate about .9 = 1 is a bit ridiculous. Yes, when you do the math on paper you can make it look like there is validity in the statement. But you are actually rounding down the infinity and infinity is a pretty big number. And don’t use the paper to prove it. It is a flaw in the graphical representation of maths. Not some sort of weird loop hole in maths itself.

  38. Dano.

    all you people talking about 9s repeating, if you measure the clock you’ll see that the lines are .0000000… off

  39. ....

    for people who don’t “believe” that .(9) == 1
    and if you don’t feel like reading it all, here’s a different way of thinking about .(9) equalling one
    1/9 = (.1)
    2/9 = (.2)
    3/9 = (.3)
    4/9 = (.4)
    5/9 = (.5)
    6/9 = (.6)
    7/9 = (.7)
    8/9 = (.8)
    9/9 = (.9) and 9/9 obviously equals 1

  40. allan.

    ok.. i completely understand the clock.. yet.. f*ck that clock!
    it looks good as an ornament, but as a clock.. never mind..
    i don’t understand the 7th hour
    9-sqrt(9)+0.999 = 5.00000..1, not 7 (0.o)

  41. peace.

    Seems like some are overly complicated…not that I mind math, it’s fun, but for example,with the 3 o’clock hour, why include the ‘+9-9’ as opposed to just leaving it out? Again, its a very simple thing either way, but it just adds one more thing to it.

    Either way, very cool and tons of fun

  42. Bret.

    To #14 Anonymous: The mathematic world is “perfect” unlike computers. Computers fudge numbers because you can’t possibly store the true value of 0.(3) in a computer. You just can’t, it would require an infinite number of bits to store that precision. So yes, 0.(3) exists… math came before computers.

  43. Dan.

    Assume 0.(9) does not equal 1.

    Then 1-0.(9) = epsilon where epsilon > 0. Now, I’ll let you choose any arbitrarily small epsilon that you want. For the sake of illustration, let’s say you choose 0.001

    Take 1 – epsilon/2. This number is greater than 0.(9) but less than 1.

    Let’s check this. 1 – epsilon/2 = 1-0.001/2 = 0.9995.

    Now 0.9995 9 or 1 – 0.(9) does not equal epsilon. As 5

  44. Bastian.

    The hotel story is fine, it’s just that last line that you wrote that is wrong “27+2=29”. The men spent 27, 2 of which was pocketed by the woman. The other 3 is in their wallets. 25 (cost of room) + 2 (amount woman pocketed) + 3 (amount returned) = 30. Nothing funny there, just bad storytelling.

  45. Jimmie James.

    Three men walk into a hotel.
    The room costs $30.
    Each man pays $10.
    The clerk realizes that she has overcharded them.
    The actual cost of the room is $25.
    She now has $5 in change.
    She decides that $5 cannot be split beteen three men.
    She pockets $2, and gives each man $1 in change.
    Each man has now spent $9 on the room.
    The men spent $27, the woman pocketed $2.
    Yet the three men intitially gave $30 total.

    $27 + $2 = $29.


  46. Mike.

    1/3 does not equal 0.(3) – but it’s close.

    Just because a computer can’t divide 1 by 3 equally doesn’t mean the human race should forget how to.

  47. Anonymous.

    Well, it’s a good thing that you don’t have to assume 1/3 = 0.(3).

    Do the calculation by hand and get back to us when you find the true answer.

  48. James.

    Divide 1 by 3, see what you get. I am programmer also, #10 is a perfect example of the validity of the statement

  49. Anonymous.

    See though, I don’t agree that 1/3 = 0.(3). There’s still that smallest value keeping it from reaching a true 1/3. You guys are basically saying “It’s close enough, we’ll just assume.” and, especially in computer programming (I use this reference as I’m a programmer) you can *not* assume any values, especially through mathematics.

    1/3 = 1/3
    0.(3) = 0.(3)

    but 1/3 != 0.(3)

  50. Luís R..

    And what is the smallest positive number possible? You might as well point out the end of infinity.

    Like others said before, 0.(9) = 1. If you can agree that 1/3 = 0.(3), then

    3 * 1/3 = 3 * 0.(3) = 0.(9) = 3/3 = 1

  51. Bradley.

    .9 repeating is exactly one.

    1 / 3 = .33333333 (repeating)
    This logically implies that
    .33333 (repeating) * 3 = 1
    .33333 (repeating) * 3 = .999 repeating.
    by substitution:

    1 = .999 (repeating).

  52. Lee.


    1/3 = .3333(repeating)
    2/3 = .6666(repeating)

    1/3 + 2/3 = .3333(repeating)+ .6666(repeating)
    3/3 = .9999(repeating)
    1 = .9999 (repeating)

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  54. Sir Lordington the nth.

    The difference between 1 and .9 repeating is Lim(x->0), i.e. the smallest positive number possible.

  55. Tal Siach.

    Hi Sir Jorge and fat, thank you for coming.
    @fat: You are right, factorials are not difficult, especially this one, but as a child…I was not very fond of any math equations.
    When I grew up, I actually learned to love Math, and you reminded me of a cool little riddle. Maybe you or others would like to give it a try.
    You need to arrive at a total of 21 by using only 1,5,6,7. You may use addition, subtraction, multiplication and division only, and must use all four numbers. Good Luck!


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