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Thread: 1-100 using four nines

  1. #1

    1-100 using four nines

    hey my goal in school is to get to 1-100 only using four nines and it's due TOMMOROW somebody help me get to there and do it fast PLEASE. thank you to all those people who contributed i am greatly in your debt.

  2. #2
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    Hello, david3456777!

    Since you've shown us nothing,
    . . I assume you haven't got any of the answers (?)


    My goal in school is to get to 1-100 only using four nines.

    More information is required.

    I assume we can use: [tex] +\;-\;\times\;\div[/tex] and numbers like "99".
    . . but can we use exponents, square roots, factorials, decimal points, etc. ?


    [tex]\begin{array}{ccc} 1 &=& \frac{99}{99} \\ 2 &=& \frac{9}{9} + \frac{9}{9} \\ 3 &=& \frac{9+9+9}{9} \\ \vdots && \vdots \\ 10 &=& \frac{99-9}{9} \\ \vdots && \vdots \\ 100 &=& 99 + \frac{9}{9} \end{array}[/tex]


  3. #3
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    9/.9 + 9/.9 = 20
    9/.9 + 9*9 = 91

    9 - 9^(9-9) = 8
    I'm just an imagination of your figment !

  4. #4
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    Quote Originally Posted by david3456777 View Post
    hey my goal in school is to get to 1-100 only using four nines

    and it's due TOMMOROW Your emergency is not ours.

    somebody help me get to there The "help" you would receive should be the difference of efforts.
    Or, put another way, it is what the answers are, minus your work shown.
    So far 100% of the answers - 0% of the work from you = 100 % left of the work from you
    still remaining.



    and do it fast Again, read my first comment above. PLEASE.

    thank you to
    all those people who contributed You haven't contributed anything here so far.

    i am greatly in your debt.Why should you be in any of our debt!? You can be thankful
    when you earned the help first.


    None of your laziness and mismanagement of time should be rewarded.



    david3456777,

    read what is at this link and attempt to follow it:


    http://www.freemathhelp.com/forum/th...Before-Posting!!
    Last edited by lookagain; 05-15-2012 at 01:39 PM.

  5. #5
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    Quote Originally Posted by david3456777 View Post
    hey my goal in school is to get to 1-100 only using four nines and it's due TOMMOROW somebody help me get to there and do it fast PLEASE. thank you to all those people who contributed i am greatly in your debt.
    0 = 99 - 99
    1 = (9 - 9) + 9/9
    2 = 9/9 + 9/9
    3 = (9 + 9 + 9)/3
    4 = 9/.9 - [(sqrt(9) + sqrt(9)]
    5 = sqrt(9) + sqrt(9) - 9/9
    6 = (9 - 9) + 9 - sqrt(9) = sqrt(9 + 9 + 9 + 9)
    7 = sqrt(9) + sqrt(9) + 9/9
    8 = 99/9 - sqrt(9) = 9 + 9 - 9/.9
    9 = (9 - 9) + sqrt(9x9) = 9sqrt(9) - (9 + 9)
    10 = (9 - 9) + 9/.9
    11 = 9/.9 + 9/9 = 9 + (9 + 9)/9
    12 = (9 - 9) + 9 + sqrt(9) = sqrt(9) + sqrt(9) + sqrt(9) + sqrt(9)
    13 = 9 + sqrt(9) + 9/9
    14 = 99/9 + sqrt(9)
    15 = 9 + 9 - 9/sqrt(9)
    16 = 9/9 + sqrt(9) + sqrt(9)
    17 = 9 + 9 - 9/9
    18 = 9 + 9 + (9 - 9)
    19 = 9 + 9 + 9/9
    20 = 9/.9 + 9/.9
    21 = 9 + 9 = 9/sqrt(9)
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  6. #6
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    Quote Originally Posted by TchrWill View Post
    0 = 99 - 99
    1 = (9 - 9) + 9/9
    2 = 9/9 + 9/9
    3 = (9 + 9 + 9)/3 . . . . . Look at this again. Missing a "9" and can't have a "3."
    4 = 9/.9 - [(sqrt(9) + sqrt(9)]
    5 = sqrt(9) + sqrt(9) - 9/9
    6 = (9 - 9) + 9 - sqrt(9) = sqrt(9 + 9 + 9 + 9)
    7 = sqrt(9) + sqrt(9) + 9/9
    8 = 99/9 - sqrt(9) = 9 + 9 - 9/.9
    9 = (9 - 9) + sqrt(9x9) = 9sqrt(9) - (9 + 9)
    10 = (9 - 9) + 9/.9
    11 = 9/.9 + 9/9 = 9 + (9 + 9)/9
    12 = (9 - 9) + 9 + sqrt(9) = sqrt(9) + sqrt(9) + sqrt(9) + sqrt(9)
    13 = 9 + sqrt(9) + 9/9
    14 = 99/9 + sqrt(9)
    15 = 9 + 9 - 9/sqrt(9)
    16 = 9/9 + sqrt(9) + sqrt(9) . . . . Look at this again. Missing a decimal point with the second "9."
    17 = 9 + 9 - 9/9
    18 = 9 + 9 + (9 - 9)
    19 = 9 + 9 + 9/9
    20 = 9/.9 + 9/.9
    21 = 9 + 9 = 9/sqrt(9) . . . . Look at this again. The second equals sign must be a plus sign.


    And still the OP has shown no work.

  7. #7

    well........

    Quote Originally Posted by lookagain View Post
    And still the OP has shown no work.
    hey in my class we can do anything now even using fractorials floor and ceiling functions also so anything can help

  8. #8

    here are the numbers

    hey so yesterday i got a ton but i still got a few here they are and thanks to all those people who answered.
    38 41 42 43 44 46 47 49 50 52 56 58 59 61 62 64 65 67 66 68 70 74 76 86 88 92
    here are the rest of the numbers i need.

  9. #9
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    Quote Originally Posted by lookagain View Post
    And still the OP has shown no work.
    Thanks ever so much for identifying my typos.

    0 = 99 - 99
    1 = (9 - 9) + 9/9
    2 = 9/9 + 9/9
    3 = (9 + 9 + 9)/9
    4 = 9/.9 - [(sqrt(9) + sqrt(9)]
    5 = sqrt(9) + sqrt(9) - 9/9
    6 = (9 - 9) + 9 - sqrt(9) = sqrt(9 + 9 + 9 + 9)
    7 = sqrt(9) + sqrt(9) + 9/9
    8 = 99/9 - sqrt(9) = 9 + 9 - 9/.9
    9 = (9 - 9) + sqrt(9x9) = 9sqrt(9) - (9 + 9)
    10 = (9 - 9) + 9/.9
    11 = 9/.9 + 9/9 = 9 + (9 + 9)/9
    12 = (9 - 9) + 9 + sqrt(9) = sqrt(9) + sqrt(9) + sqrt(9) + sqrt(9)
    13 = 9 + sqrt(9) + 9/9
    14 = 99/9 + sqrt(9)
    15 = 9 + 9 - 9/sqrt(9)
    16 = 9/.9 + sqrt(9) + sqrt(9) 17 = 9 + 9 - 9/9
    18 = 9 + 9 + (9 - 9)
    19 = 9 + 9 + 9/9
    20 = 9/.9 + 9/.9
    21 = 9 + 9 + 9/sqrt(9)
    TchrWill

    No matter how insignificant it might appear, learn something new every day.

  10. #10
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    A few more . . .


    [tex]\begin{array}{c}(\sqrt{9})!(\sqrt{9})! + \sqrt{9} + \sqrt{9} \;=\;42 \\ \dfrac{9!}{((\sqrt{9})!)!(9+9)} \;=\;42 \\ (\sqrt{9})!(\sqrt{9})! + \dfrac{9}{.9} \;=\;46 \\ \dfrac{((\sqrt{9})!)!}{9} + 9 + \sqrt{9} \;=\;92 \end{array}[/tex]


    Can we use: /[tex]\log_{_{\sqrt{9}}}9 \,=\,2\,?[/tex]

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