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Math Help - Puzzle needs to be solved

  1. #1
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    Post Puzzle needs to be solved

    Put Any Mathematic sign:

    2 2 2 = 6
    3 3 3 = 6
    4 4 4 = 6
    5 5 5 = 6
    6 6 6 = 6
    7 7 7 = 6
    8 8 8 = 6
    9 9 9 = 6
    1 1 1 = 6

    Let me know the answer.. 2,3 and 6 are solved..
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  2. #2
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    Quote Originally Posted by tilake1 View Post
    Put Any Mathematic sign:

    2 2 2 = 6
    3 3 3 = 6
    4 4 4 = 6

    \sqrt{4}+\sqrt{4}+\sqrt{4}

    5 5 5 = 6

    5+\frac{5}{5}

    6 6 6 = 6
    7 7 7 = 6

    7-\frac{7}{7}

    8 8 8 = 6

    \sqrt[3]{8}+\sqrt[3]{8}+\sqrt[3]{8}

    9 9 9 = 6

    9-\frac{9}{\sqrt{9}}

    1 1 1 = 6

    (1+1+1)!=6

    Let me know the answer.. 2,3 and 6 are solved..
    .
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  3. #3
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    Thanks...
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  4. #4
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    Hello, tilake1!

    I couldn't find an elementary expression for the 8's . . .


    . . \begin{array}{ccc}<br />
(1 + 1 + 1)! &=& 6 \\ \\[-3mm]<br />
2 + 2 + 2 &=& 6 \\ \\[-3mm]<br />
3! \times 3 \div3 &=&  6\\ \\[-3mm]<br />
\sqrt{4}+\sqrt4+\sqrt4 &=& 6 \\ \\[-3mm]<br />
5 + (5 \div 5) &=& 6 \\ \\[-3mm]<br />
6 \times 6 \div 6 &=&  6 \\ \\[-3mm]<br />
7 - (7 \div 7) &=& 6 \\ \\[-3mm]<br />
\log_{\sqrt{\!\!\sqrt{8}}}(8\sqrt{8}) &=& 6\\ \\[-2mm]<br />
(9 + 9) \div \sqrt9 &=& 6 <br />
\end{array}
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  5. #5
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    2 other ways for the 4's: 4 + 4 - sqrt(4) = 6 and 4/.4 - 4 = 6

    On the 8's; since sqrt is allowed, then cubert should be allowed: cubert(8) + cubert(8) + cubert(8) = 6
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  6. #6
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    8^0\left(log_28+log_28\right)=6
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    8^0\left(log_28+log_28\right)=6
    That's cheating Archie; you're using extra numbers, 0 and 2 !!
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  8. #8
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    Quote Originally Posted by Wilmer View Post
    That's cheating Archie; you're using extra numbers, 0 and 2 !!
    That sinks the factorial then!
    cos that's a multiplication by 2, written with a footprint.
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  9. #9
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    \displaystyle\huge\left[\left(\frac{d}{dx}[8]\right)!+\left(\frac{d}{dx}[8]\right)!+\left(\frac{d}{dx}[8]\right)!\right]!=6

    which holds for all constants.
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  10. #10
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    Take this Archie:
    (8 + 8 + 8) / CEILING(pi) = 6
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  11. #11
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    Define S to be the "Sum of the digits of a number".

    S(8+8+8)=6
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  12. #12
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    Quote Originally Posted by Archie Meade View Post
    Define S to be the "Sum of the digits of a number".

    S(8+8+8)=6
    Define G\equiv 6. Then G(a+a+a) = 6 and this
    holds for all constant a's
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  13. #13
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    HaHa , and


     a^{ \log_a(2) } + a^{ \log_a(2) } + a^{ \log_a(2) } = 6
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  14. #14
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    Quote Originally Posted by Soroban View Post
    . . \begin{array}{ccc}<br />
3! \times 3 \div3 &=&  6\\ \\[-3mm]<br />
\end{array}
    3 * 3 - 3 = 6 ; keep it simple!
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  15. #15
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    Another for the three 1's; arrange 1st two as a V:
    V1 = 6
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