# Sort of a game

• Jan 4th 2009, 10:00 PM
Kai
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 (Thinking)
• Jan 5th 2009, 12:11 AM
Moo
Hello,
Quote:

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 (Thinking)

(1+1+1)! (factorial)
3!=3x2x1=6 :D
• Jan 5th 2009, 02:29 AM
craig
Was just about to provide an answer with my first post, got beaten to it tho (Wink)

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

$(cos(0))!$
• Jan 25th 2009, 08:57 AM
tom ato
Quote:

Originally Posted by craig

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

$(cos(0))!$

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