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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]
     
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