# Thread: Sort of a game

1. ## Sort of a game

Someone asked me how to get 6 with 2 2 2 using any mathematical operations ( +,-,*,/, ^2, ^3..., sqaure root, cube root,...)

For 2 2 2, 2+2+2=6
For 3 3 3, 3*3 -3=6 ( or 3+3 * (3^0))
For 4 4 4, square root 4 + square root 4 + square root 4=6
For 5 5 5, 5+ (5/5)=6
For 7 7 7, 7- (7/7)=6
For 8 8 8, cube root 8 + cube root 8 + cube root 8=6
For 9 9 9, (9+9)/(square root 9) OR square root 9 *square root 9-square root 9
Now am stuck on the last part, that is 1 1 1 to get 6

2. Hello,
Originally Posted by Kai
Someone asked me how to get 6 with 2 2 2 using any mathematical operations ( +,-,*,/, ^2, ^3..., sqaure root, cube root,...)

For 2 2 2, 2+2+2=6
For 3 3 3, 3*3 -3=6 ( or 3+3 * (3^0))
For 4 4 4, square root 4 + square root 4 + square root 4=6
For 5 5 5, 5+ (5/5)=6
For 7 7 7, 7- (7/7)=6
For 8 8 8, cube root 8 + cube root 8 + cube root 8=6
For 9 9 9, (9+9)/(square root 9) OR square root 9 *square root 9-square root 9
Now am stuck on the last part, that is 1 1 1 to get 6
(1+1+1)! (factorial)
3!=3x2x1=6

3. Was just about to provide an answer with my first post, got beaten to it tho

Just to be awquard I'll do it with 3 zeros

$\displaystyle (cos(0))!$

4. Originally Posted by craig

Just to be awquard I'll do it with 3 zeros

$\displaystyle (cos(0))!$
Wouldn't it be $\displaystyle (cos(0)+cos(0)+cos(0))!$ ?