4x4

4x4

Posers and Puzzles

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k
Standard memberkbaumen

Sigulda, Latvia

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11 Jun 07

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

m
Standard membermtthw

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11 Jun 07

Originally posted by kbaumen
arccos(4-4)-4-4=82
An alternative:

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

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

k
Standard memberkbaumen

Sigulda, Latvia

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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.
I dunno, I'm quite satisfied with 'em.

m
Standard membermtthw

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Originally posted by kbaumen
I dunno, I'm quite satisfied with 'em.
Oh, it's just a personal thing. I'm not saying they shouldn't be allowed.

C
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1 edit

Originally posted by mtthw
Oh, it's just a personal thing. I'm not saying they shouldn't be allowed.
feel free not to use them ~ numbers up to 2178 are all possible without!

Edit

Apparently.

k
Standard memberkbaumen

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

A
Standard memberAgerg
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(!4)^(sqrt(4)) + sqrt(4) = 83

C
SubscriberCauselessOne

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Originally posted by Agerg
(!4)^(sqrt(4)) + sqrt(4) = 83
er.. what is (!4) ?

k
Standard memberkbaumen

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1 edit

Originally posted by Agerg
(!4)^(sqrt(4)) + sqrt(4) = 83
What is '!4'? is that meant to be '4!'? In that case your expression is flawed.

4!=24; sqrt(4)=2;
24^2+2 =/= 83

EDIT: And that's only three 4's.

A
Standard memberAgerg
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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! 😵 )

k
Standard memberkbaumen

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1 edit

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! 😵 )
Nice. I found an article in wikipedia about this. This can really be helpful. Only you need four, not three 4's in the expression.

Oops, didn't see your edit.

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

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Standard memberAgerg
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44*sqrt(4) - 4 = 84

k
Standard memberkbaumen

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1 edit

Nevermind, it was wrong.

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