4x4

arccos(4-4)-4-4=82

Originally posted by kbaumen
arccos(4-4)-4-4=82

Originally posted by mtthw
An alternative:

4!/.4 + 4! - sqrt(4) = 82

There's something about these inverse trig formulations that isn't quite as satisfying.

Originally posted by kbaumen
I dunno, I'm quite satisfied with 'em.

Originally posted by mtthw
Oh, it's just a personal thing. I'm not saying they shouldn't be allowed.

But even with them, it's kind a hard to get 83. I'm stuck on it for some 10 min.

(!4)^(sqrt(4)) + sqrt(4) = 83

Originally posted by Agerg
(!4)^(sqrt(4)) + sqrt(4) = 83

Originally posted by Agerg
(!4)^(sqrt(4)) + sqrt(4) = 83

subfactorial...the total permutations of n elements excluding those that contain elements in their natural place

ie: !3 = 2

123, 132, 213, 321, all contain at least one element where it should be leaving only 132, and 312

!1 = 0, !2 = 1, !3 = 2, !4 = 9...

(though you got me on the 3 fours though! 😵 )

Originally posted by Agerg
subfactorial...the total permutations of n elements excluding those that contain elements in their natural place

ie: !3 = 2

123, 132, 213, 321, all contain at least one element where it should be leaving only 132, and 312

(though you got me on the 3 fours though! 😵 )

(!4)^(4 - sqrt(4)) + sqrt(4) = 83

44*sqrt(4) - 4 = 84

Nevermind, it was wrong.

(!4)^(4 - sqrt(4)) + 4 = 85