[Guest]

2+2+2=6

what mathematical operations to use to get the same result?

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

[Guest]

1+1+1 <= 6 (less than or equal to)

2+2+2 = 6

3+3+3 >= 6

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9+9+9 >= 6

[Guest]

i wish the life could be that easy

..assume left side has to equal to 6

Edited March 22, 2008 by smile4me

[Guest]

But here are some:

2+2+2=6 or a bunch of others

3*3-3=6

5/5+5=6

6-6+6=6 or 6*6/6=6

7-7/7=6

Edited March 22, 2008 by Noct

[EventHorizon]

(1+1+1)! = 3! = 6

2+2+2 = 6

3*3-3 = 9-3 = 6

4! / sqrt(4*4) = 24 / 4 = 6

5 + (5/5) = 5 + 1 = 6

6-6+6 = 0+6 = 6

7 - (7/7) = 7 - 1 = 6

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

sqrt(9*9)-sqrt(9) = 9-3 = 6

square roots are valid...right?

Edited March 22, 2008 by EventHorizon

[Guest]

2 * 2 + 2 = 6

sqrt(4) * sqrt(4) +sqrt(4) = 6

6 * 6 / 6 = 6

8 - 4th root(8 + 8) = 6

9 - 4th root(9 * 9) = 6

[Guest]

For the ones i didn't get:

(1+1+1)!=6

4!/(square root(4)*square root(4))=6

cube root of 8+cube root of 8+cube root of 8=6

or square root (8+8)+ cube root (8)=6

(square root of (9) * square root (9))-square root (9)=6

[itachi-san]

Can we only place mathematical symbols between the numbers or can we really mess with them?

[bonanova]

1 1 1 ?

2 + 2 + 2 = 6

3x3 - 3 = 6

4 + sqrt[sqrt[4x4]] = 6

5 + 5/5 = 6

6 + 6 - 6 = 6

7 - 7/7 = 6

8 - sqrt[sqrt[8 + 8]] = 6

sqrt[9]x sqrt[9] - sqrt[9] = 6

[Guest]

(1+1+1)!= 6

2+2+2 = 6

3x3-3 = 6

4+sqrt(sqrt(4x4))=6

5+5/5 = 6

6+6-6 = 6

7-7/7 = 6

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

sqrt(9)x sqrt(9)-sqrt(9)=6

Edited March 23, 2008 by storm