Game of 4 4's

[golfinha]

37 = (4! : 4)^(SQRT(4))+log0(4)

[golfinha]

38 = 4!+4x4-SQRT(2)

[Craft]

37 = (4! : 4)^(SQRT(4))+log0(4)

Hmm, what is log0(4) ? (I don't recognize the notation)

Is it 0-base logarithm? If so, is there a definition for that (I can't remember)? And if there is, you seem to need the number 0 to define the function here, so it involves using a number other than 4...

[golfinha]

37 = (4! : 4)^(SQRT(4))+log0(4)

Hmm, what is log0(4) ? (I don't recognize the notation)

Is it 0-base logarithm? If so, is there a definition for that (I can't remember)? And if there is, you seem to need the number 0 to define the function here, so it involves using a number other than 4...

log0(n)=1


but if we can't use the log formula, we shouldn't use the SQRT - it envolves a 2

[Craft]

log0(n)=1


but if we can't use the log formula, we shouldn't use the SQRT - it envolves a 2

Well, the difference is that using sqrt-notation doesn't need the number 2 because the notation itself bears the meaning. I believe using operator 'log' generally means the same as 'log10', so 'log' could be used but only meaning 'log10'. Notation 'ln' means the e-based logarithm. Other logarithms need the base to be written, like log5 or log0 etc. but then there is a number other than 4...

Another example is the factorial notation, n! . It involves all the numbers from 1 to n, but the notation doesn't need other numbers than n, and thus 4! is allowed.

[golfinha]

Ok, can anyone else try to find another result for 37?

[golfinha]

Well, I found it, without other numbers, only math formulas


37=4!+4x4-INT(LN(4!))

[Craft]

Yes, but I don't like the int-operator either, because you can fix almost anything with that (see my remark on the previous page)

(Am I being a bit too strict here... )

[golfinha]

Ok Ok

Without the INT will you accept it?, if not I can't think of anything else today


If so the answer is
37=4!+4x4-LN(4!)

If not I don't know

[Craft]

Without the int it becomes only approximate, a little less than 37, because ln(4!) is appr. 3.178.

[golfinha]

I guive up, for me 37 will be just like 31

But if ther is an acceptable answer I want to know it

[lazza]

35 = (((4-i^4)!)^sqrt4) - i^4 I think

Your calculation is correct, BUT:
the complex unit i is a number, not a mathematical operator or function, and so it can't be used (using it is the same as using any other number than 4).

Hmmm... I thought the same myself, when androl used it, but then decided I wouldn't mention it, as I could use it myself

[Elmo]

Hmm, what is log0(4) ? (I don't recognize the notation)

x = log0(4) is the solution to the equation 0^x = 4, which is impossible, of
course, because 0^x is always 0 (for every positive x).

39 doesn't cross my mind, but 4*(4+4+sqrt4) = 40

[golfinha]

Here goes a summary of the 4 4's game:

0 = 4-4+4-4
1 = 4/4+4-4
2 = 4/4+4/4
3 =( 4 x 4 - 4) : 4
4 = (4-4)x4+4
5 = (4x4+4):4
6=(4+4)/4+4
7=44/4-4
8=4+4+4-4
9=4+4+(4/4)
10=(44-4)/4
11=44/Sqrt(4*4)
12=(44+4)/4
13=?
14=4 + 4 + 4 + sqrt4
15=(4 x 4) - (4/4)
16=(4 x 4) + 4 - 4
17=( 4 x 4 ) + ( 4/4 )
18= 4 x 4 + 4 - sqr4
19=44/4 + 4 + 4
20=4*(4+(4/4))
21=(44 - sqrt4)/sqrt4
22=4 x 4 + 4 + sqrt4
23=(44 + sqrt4) / sqrt4
24 = 4! + (4-4)/4
25 = 4! + (sqrt4 + sqrt4)/4
26=4! + ((4 + 4)/4)
27=4! + (4- (4/4))
28=44 - (4 x 4)
29=4! + 4 + 4/4
30=4 x (4+4) - sqrt4
31 = ?
32 = 4*4+4*4
33 = 4!+4*sqrt(4)+i^4
34 = 4!+4+4+SQRT(4)
35 = ?
36 = 4! + 4 x 4 - 4
37 = ?
38 = 4!+4x4-SQRT(2)
39 = ?
40 = 4*(4+4+sqrt4)


There are already 5 missing

[lazza]

I think that will be 6 missing.... I can't do 41 either...

42 = 44 - 4 + sqrt4 or = 4! + 4! - 4 - sqrt4