1. #11
    morrillo's Avatar Senior Member
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    [sqrt(8/8+8)]! = [sqrt(9)]! = 3! = 6
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  2. #12
    MaTiSpOrT's Avatar Senior Member
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    haha!! that way is easier...
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  3. #13
    QcChopper's Avatar Senior Member
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    Great job guys, 3 different way to solve 8? Nice

    The one I had was Morillo's but they are all good, congrats guys.
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  4. #14
    Originally Posted by MaTiSpOrT Go to original post
    That one is the hardest one...
    √ ( ( 8 ! + 8 ! ) + √ ( 8 ! ) ) = 6
    √ ( ( 16 + 16) + 4) = 6
    √ ( 32 + 4) = 6
    √ 36 = 6
    Hope that´s the result. lol
    not really 8! = 1*2*3*4*5*6*7*8
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  5. #15
    morrillo's Avatar Senior Member
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    (cos0+cos0+cos0)!=6

    8-sqrt(sqrt(8+8))=6

    sqrt(10-10/10)=6
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  6. #16
    mutetus's Avatar Trials Developer
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    Easy. I just added few +'s and a parenthesis in 45° angle.
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  7. #17
    Guunners's Avatar Senior Member
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    That's how I would do.

    (0! + 0! + 0!)! = 6
    (1 + 1 + 1)! = 6
    2 + 2 + 2 = 6
    3 * 3 - 3 = 6
    √4 + √4 + √4 = 6
    (5 / 5) + 5 = 6
    6 + 6 - 6 = 6
    7 - (7 / 7) = 6
    sqrt3(8) + sqrt3(8) + sqrt3(8) = 6
    √9 * √9 - √9 = 6

    I don't know how to write sqrt3 better, but I think you understand it anyway:
    2^3 = 8 so sqrt3(8) = 2
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  8. #18
    morrillo's Avatar Senior Member
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    Originally Posted by Guunners Go to original post
    That's how I would do.

    (0! + 0! + 0!)! = 6
    (1 + 1 + 1)! = 6
    2 + 2 + 2 = 6
    3 * 3 - 3 = 6
    √4 + √4 + √4 = 6
    (5 / 5) + 5 = 6
    6 + 6 - 6 = 6
    7 - (7 / 7) = 6
    sqrt3(8) + sqrt3(8) + sqrt3(8) = 6
    √9 * √9 - √9 = 6

    I don't know how to write sqrt3 better, but I think you understand it anyway:
    2^3 = 8 so sqrt3(8) = 2
    Don't think you are allowed to use the extra '3' in the cube root though guunners
    great problem though.

    Also, if you are feeling cheap you can use the floor function [greatest integer less than or equal to] for just about any combination...
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  9. #19
    Guunners's Avatar Senior Member
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    If this is the case then we shouldn't be allowed to use the √ either because it's a sqrt2 but it's common so we just say sqrt. It's the same model though.

    4^(1/2) = 2
    8^(1/3) = 2

    If I write curt(8) + curt(8) + curt(8) = 2 I didn't use any extra '3'.
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  10. #20
    morrillo's Avatar Senior Member
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    I guess it depends on who sets the task and how they word it guunners you have a good point...

    edit: just found this: http://testtube.com/scamschool/six-the-hard-way
    suggests one way to read the problem is just numbers on a page and adding symbols without re-writing.
    I rather like their approach, although it does preclude something like 2^2+2 = 6 which seems harsh, unless you write it 2x2+2=6.
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