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Four Nines Puzzle

  1. Aug 4, 2005 #1

    ACG

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    Hi! I have a puzzle here and was wondering if you guys could solve it.

    Using exactly four nines each time and no other numbers or letters, find expressions which evaluate to every integer between 0 and 132. You may put parentheses wherever you wish and uses whatever mathematical symbols you want other than ceiling and floor.

    If I remember correctly, the hardest number for me was (I think) 68.

    Example:

    0 = 99-99.

    Good luck!

    ACG

    P.S.If you want a hint, read on.



















    spoiler space.






















    _
    Hint: .9 = 1.
     
  2. jcsd
  3. Aug 4, 2005 #2

    James R

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    Err... no it doesn't.
     
  4. Aug 4, 2005 #3
    ACG is trying to say that .9 repeating = 1, since you can use the repeating symbol as a mathematical symbol.

    ~Lyuokdea
     
  5. Aug 5, 2005 #4

    Galileo

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    9/9+9-9=1

    9/9+9/9=2
     
  6. Aug 5, 2005 #5
    [itex]68 = 69 - \frac{9}{9}[/itex]
     
  7. Aug 6, 2005 #6

    ACG

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    The first few are fairly easy. However, the higher the numbers get, the harder they are. By the time you get to 50-60 you have to start to get VERY creative. Rest assured, it can be done.
     
  8. Aug 8, 2005 #7
    jimmysnyder, that's a smart one ^^
     
  9. Aug 8, 2005 #8
    but are we allow to rotate the 9? i thoght it woulda changed it into a 6 as it clearly reads six nine not nine nine.

    Sqrts allowed?
     
  10. Aug 8, 2005 #9
    hmmmm.... for 2 I thought using .9999 was cheap so i did:


    ((9*.9)/9)*9
     
  11. Aug 9, 2005 #10
    i dont undertand what does .9=1 or .99999 meant
     
  12. Aug 9, 2005 #11
    I'm not sure but I think it means:

    1 = .9...

    so

    4 = .9... + .9... + .9... + .9...
     
  13. Aug 9, 2005 #12
    I think they are refering to .9 (with a dot above the 9 to signify a recurring number)

    I don't really agree because it will never equal 1 it just will be very close.

    In general terms is it cheating to use the shortened form of a square root (e.g. remove the 2)?
     
  14. Aug 9, 2005 #13
    19 down 113 to go :(
     
  15. Aug 10, 2005 #14

    ACG

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    One may not turn the 9's upside down. Sorry about that! :frown:

    Square root is legal (and indeed, is crucial). Any mathematical symbol which does not involve letters (sin, cos, etc.) is permitted. Ceiling and floor are prohibited.

    The .9=1 thing is

    .9 with a bar over it (.9 bar) = .999999... = 1.

    ACG
     
  16. Aug 10, 2005 #15

    ACG

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    Here's another hint -- it took me maybe 3 minutes to figure out the solutions for all of the numbers except maybe five or six. Those six took an hour more. Clearly, there's a trick. But what is it?
     
  17. Aug 10, 2005 #16
    3minutes? wow i'm over and hour and i've gotten maybe 3/4s


    hour later and i'm down to these #
    44,46,65,67,68,74,76,116,
    and 116+(i do have a few of these numbers but not alot)


    5min later and i'm left with
    65,67,78,74,76,116,118,131 i give up for now
     
    Last edited: Aug 10, 2005
  18. Aug 11, 2005 #17
    Is the bar the same as having a dot above the number?

    p.s. as I said before, it will still never=1 it will just be really close.

    Why would you say it equals 1?
     
  19. Aug 11, 2005 #18
    Proof of 0.999999999... = 1

    Let 0.9999999999... = x

    Therefore, we have
    10x = 9.99999999999...
    10x = 9 + 0.9999999999...
    10x = 9 + x
    9x = 9
    x = 1

    x = 0.9999999999999... = 1
     
  20. Aug 11, 2005 #19
    :biggrin: Nice bit of trickery :biggrin:
    but 0.99999<1 just like 1.000001>1
     
  21. Aug 11, 2005 #20
    what is [itex]1 - .9999... ?[/itex]
     
  22. Aug 11, 2005 #21

    ACG

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    I'm willing to post the solutions on the board if everyone agrees.

    In the meantime, I found a second problem.

    It's like the Four Nines Problem, except that it's got two 4's and two 9's.

    You can get all the numbers from 0 to 232. This is quite a bit harder as there are more options.
     
  23. Aug 12, 2005 #22
    Mmmm, good point :frown:

    It's still 0.0001 (with a recuring dot above the zero) which is >0

    I agree it's close enough to zero that it makes no practical difference but I still can't agree that it's zero.
     
  24. Aug 12, 2005 #23
    There is no such mathematical notation. The dot over the zero literally means there is no 1, just 0s. Besides, just as a practical matter, I don't see how you can justify that final 1 in terms of the subtraction problem I set you.
     
  25. Aug 12, 2005 #24
    I used to use a recuring dot when I was at school 20 odd years ago.

    I never used it in this context but in a way you're dealing with infinities so if you want to have a number that has an infinate number of nines after the decimal then I can have an infinate number of zeros before the 1 :)

    Anytime you fix the number of nines then the sum can be calculated.
     
  26. Aug 12, 2005 #25
    When I said that there was no such mathematical notation, I was not talking about the dot, I was talking about the 1 at the right hand end after the dot.

    After good, before bad. I'm allowed to put as many 9s as I want after the decimal point because there is a decimal point, but you are not allowed to put any zeros before the last 1 because there is no last 1. That is what infinity means. in = no, finite = end, infinite = no end = no last digit.

    Q: What comes after the last zero?
    A: There is no last zero.
    Q: What comes before the last one?
    A: There is no last one.

    I still don't see where that 'last' 1 comes from anyway. Certainly not from the subtraction problem I asked you to perform.
     
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