Great job guys, 3 different way to solve 8? Nice
The one I had was Morillo's but they are all good, congrats guys.
not really 8! = 1*2*3*4*5*6*7*8Originally Posted by MaTiSpOrT Go to original post
(cos0+cos0+cos0)!=6
8-sqrt(sqrt(8+8))=6
sqrt(10-10/10)=6
Easy. I just added few +'s and a parenthesis in 45° angle.
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That's how I would do.
(0! + 0! + 0!)! = 6
(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
sqrt3(8) + sqrt3(8) + sqrt3(8) = 6
√9 * √9 - √9 = 6
I don't know how to write sqrt3 better, but I think you understand it anyway:
2^3 = 8 so sqrt3(8) = 2
Don't think you are allowed to use the extra '3' in the cube root though guunnersOriginally Posted by Guunners Go to original post
great problem though.
Also, if you are feeling cheap you can use the floor function [greatest integer less than or equal to] for just about any combination...![]()
If this is the case then we shouldn't be allowed to use the √ either because it's a sqrt2 but it's common so we just say sqrt. It's the same model though.
4^(1/2) = 2
8^(1/3) = 2
If I write curt(8) + curt(8) + curt(8) = 2 I didn't use any extra '3'.
I guess it depends on who sets the task and how they word it guunnersyou have a good point...
edit: just found this: http://testtube.com/scamschool/six-the-hard-way
suggests one way to read the problem is just numbers on a page and adding symbols without re-writing.
I rather like their approach, although it does preclude something like 2^2+2 = 6 which seems harsh, unless you write it 2x2+2=6.