Puzzlefor you all!!!!

So here comes a new puzzle for everyone....
you have to use the mathematical operators in between the numbers to get the correct output....

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
8 8 8 =6
9 9 9 =6

its very simple ,you all can easily solve it....

Replies

  • Kaustubh Katdare
    Kaustubh Katdare
    (1+1+1)! = 6
    2+2+2 = 6
    3x3-3 = 6
    4+4-(sqrt 4) = 6 [positive square root]
    (5/5)+5 = 6
    6-6+6 = 6
    7-(7/7) = 6

    - bah -

    (squrt 9) x (sqrt 9) - (sqrt 9) =6
  • mahul
    mahul
    I hope to answer the 8,

    8-sqrt(8/(cube root(8))) = 6
  • Kaustubh Katdare
    Kaustubh Katdare
    mahul
    I hope to answer the 8,

    8-sqrt(8/(cube root(8))) = 6
    If I were not bored, I'd do -

    cuberoot (8) + cuberoot (8) + cuberoot (8) = 6 ๐Ÿ˜
  • mahul
    mahul
    Wow! that's a lot better.... ๐Ÿ˜€
  • devesh
    devesh
    wow...
    such quick replies..
    CE rocks!!!!
  • Kaustubh Katdare
    Kaustubh Katdare
    Ayan, take this as yet-another warning. Further spamming will lead to banning. If you want to make test posts, go here -

    #-Link-Snipped-#

    of course, instead of posting 'jdfsdlks...', try posting something that makes sense.
  • apple6
    apple6
    #-Link-Snipped-#,i think i am so late but..,
    (1!+1!+1!)! =6
    2+ 2+ 2 =6
    3 *3- 3 =6
    4 +4-sqrt(4) =6
    (5/5)+ 5 =6
    6 +6- 6 =6
    7-(7/7) =6
    cuberoot(8)+cuberoot(8)+cuberoot(8) =6
    sqrt(9)*sqrt(9) -sqrt(9)=6
  • apple6
    apple6
    #-Link-Snipped-#, thank you..

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