devesh
Branch Unspecified
07 Aug 2012

# 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....
10 years ago
(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

Branch Unspecified
10 years ago
I hope to answer the 8,

8-sqrt(8/(cube root(8))) = 6
10 years ago
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 😁

mahul

Branch Unspecified
10 years ago
Wow! that's a lot better.... 😀

devesh

Branch Unspecified
10 years ago
wow...
such quick replies..
CE rocks!!!!
10 years ago
Ayan, take this as yet-another warning. Further spamming will lead to banning. If you want to make test posts, go here -

CE - Newbie Training Center - CrazyEngineers Forum

of course, instead of posting 'jdfsdlks...', try posting something that makes sense.

apple6

Branch Unspecified
6 years ago
devesh,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

Branch Unspecified
6 years ago
Prashanth_p@cchi, thank you..