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Funkwelle
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29 May 2013, 3:57 am

31 = (4! + 4)/4 + 4!



Krabo
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29 May 2013, 4:34 am

Congratulations!

32 = [√(4)]^(4/4 + 4)



GGPViper
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29 May 2013, 4:43 am

Would this be valid (digital root)?

33 = 4! + dr(4!+4!+4!)



Krabo
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29 May 2013, 5:20 am

Right! The digital root of any positive integer is defined with the floor function, thus

dr(n) = n – 9*|(n–1)/9| where the pipes represent the floor function.

It is a mathematical function and it meets the criteria of the OP, so why not!



Jono
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29 May 2013, 6:36 am

I see a lot of people using square roots. Does the square root function really meet those criteria? I'm asking this because the square root is really a number raised to the power of 1/2, so it has a 2 in it.



Krabo
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29 May 2013, 6:54 am

Well, in a context like this those criteria are not meant to be strict. I only wanted to rule out all sorts of ad hoc functions, as I indicated on page 1, bottom. Yes, the square root includes a 2 in it but because it is a long-time convention to camouflage it with the root symbol, it is all right. The above mentioned digital root function, too, is all right because it was not invented by any of us to serve just the purpose of this thread. The Dirac delta function would be welcomed but because δ(4,4) = 1 consumes two 4's it fares no better than 4/4 = 1.



Funkwelle
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29 May 2013, 7:02 am

And logarithms?

Quote:
Typically the "log" operators are not allowed, since there is a way to trivially create any number using them.

(From wikipedia, can't post links yet)



Krabo
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29 May 2013, 7:07 am

Incredible! There is a whole Wikipedia article on Four Fours! I swear I didn't know! This has been my private hobby who knows for how long. All I can say is that obviously I re-invented it for myself.

Now I need to breathe into a paper bag.



GGPViper
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29 May 2013, 8:39 am

Well, onwards, then! With digital roots, it becomes a lot easier... and a lot cheesier.

34: 4! + 4 + 4 + √4
35: 4! +4 + dr(4x4)
36: 44 - 4 - 4 - This one was tough as hell :evil: ... :D
37: dr((4!)!) x 4 + (4/4)
38: dr((4!)!) x 4 + 4 - √4
39: dr((4!)!) x 4 + dr(4!/√4)
40 = (√4+√4)/√(4%)*√4 - Or, less dramatically: 44 - √4 - √4

I'm pretty sure that several of those could be done more elegantly than by using dr((4!)!), which yields a digital root of 9, but I *really* wanted to use a number larger than 6.2 x 10^22.



Krabo
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29 May 2013, 9:35 am

GGPViper wrote:
36: 44 - 4 - 4 - This one was tough as hell :evil: ... :D


What about 36 = [√(4)]*(4*4) + 4, but yours is of course more elegant.



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29 May 2013, 9:40 am

Krabo wrote:
I think it ends here. Odd integers seem to require "4/4" and it's not possible to express either 30 or 32 with two 4's alone. Thank you all for your contribution.

We could, of course, define a function φ(x) = 7x and write
31 = φ(4) + 4 – 4/4
but this wouldn't be in the spirit of the problem.


Very fancy. You have invoked the Ghost of Euler.

ruveyn



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29 May 2013, 10:22 am

Now that you mentioned, Euler published several works when totally blind. His last words were "I die."



GGPViper
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29 May 2013, 1:57 pm

41: 4! + 4! - dr(4x4)
42: 44 - 4 +√4
43: 44 - (4/4)
44: 44 - 4 + 4
45: 44 + (4/4)
46: (4!+4!) - 4 + √4
47: (4!+4!) - (4/4)
48: (4!+4!) x (4/4)
49: (4!+4!) + (4/4)
50: (4!+4!) + 4 - √4



ruveyn
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29 May 2013, 5:09 pm

1/2 = 4/(4+4) so doing square roots is o.k.

ruveyn



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29 May 2013, 5:11 pm

Also any integer can be generated by 4's.

4 +....+4 - n times divided by 4 = n. So any integer n can be generated by 4's.

ruveyn



Funkwelle
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29 May 2013, 11:56 pm

ruveyn wrote:
1/2 = 4/(4+4)

ruveyn wrote:
Also any integer can be generated by 4's.

4 +....+4 - n times divided by 4 = n. So any integer n can be generated by 4's.

ruveyn


Krabo wrote:
Four 4's
You are given a set of four numbers, they are all 4's.


You're supposed to use four 4's, not more or less. Exactly four 4's.