4 4's puzzle

[tricky77puzzle]

Alright, so here's how the game is played:

You're trying to make as many numbers as possible using 4 4's or less, in an attempt to try to prove an old theory: "You can make any whole number using only four instances of the number four and commonly used mathematical symbols."

So so far, the symbols allowed are:
1. All standard mathematical operators (+-*/)
2. Factorials (n! = n * (n-1) * (n-2) * ... * 2 * 1)
3. Square roots, arbitrary roots (the index is included in the 4 4's)
4. Repeat signs (denoted with an apostrophe ' or grave accent `, ex. 0.3` = 0.33333.....)
5. Exponentiation (denoted with a caret ^, 2^8 = 2*2*2*2*2*2*2*2, 8 2's)
6. Decimals do not require a leading zero. So you could put .4 instead of 0.4.

I'll start:
0 = 4-4
1 = 4/4
2 = sqrt(4)
3 = 4 - 4/4
4 = 4
5 = 4 + 4/4
6 = 4 + sqrt(4)
7 = 4 + 4 - 4/4
8 = 4 + 4
9 = 4 + 4 + 4/4
10 = 4 + 4 + sqrt(4)

And the next person would continue from 11, 12, and so on, and introduce new symbols and explaining them if necessary.

[megaxxx]

Re: 4 4's puzzle

Stuck on 11 eh? Fail.

[tricky77puzzle]

Re: 4 4's puzzle

No I'm not stuck on 11. I just want the next person to continue.

11 = 44/4
12 = (44 + 4) / 4
13 = 44/4 + sqrt(4)
14 = (4!) / sqrt(4) + sqrt(4)
15 = 44/4 + 4
16 = 4 * 4
17 = 4 * 4 + 4/4
18 = 4 * 4 + sqrt(4)
19 = 4! - 4 - 4/4
20 = 4! - 4

And so on...

[Fractal_Monkey]

Re: 4 4's puzzle

I remember doing this as a personal project in high school, except that I used exactly 4 4s for each number. As far as I can recall, 113 is the smallest number that can't be done using those rules. Anyway, it goes something like...

21 = 4! - 4 + 4/4
22 = 4! - sqrt(4) (*4/4)
23 = 4! + 4/4 - sqrt(4)
24 = 4! = (4 + 4 + 4) * sqrt(4)
25 = 4! + 4/4 = 4! + sqrt(4) - 4/4

etc.

[rushyrulz]

Re: 4 4's puzzle

28= 4!(4/4)-4

[tricky77puzzle]

Re: 4 4's puzzle

29 = 4! + 4 + 4/4
30 = 4! + 4 + sqrt(4)
31 = 4! + 4/.4' - sqrt(4)
32 = 4 * 4 + 4 * 4
33 = 4! - 4/.4'
34 = 4! + 4 + 4 + sqrt(4)

[Kotarouchan]

Re: 4 4's puzzle

35 = 4! + 44/4
36 = 4 + 4 + 4 + 4!
37 = 4! + (4! + sqrt(4))/ (sqrt(4))
38 = 4! + 4! + 4/.4
39 = 4! + 4!/(4*.4)
40 = 44 - sqrt(4) - sqrt(4)
41 = (4! + sqrt(4))/.4 - 4!
42 = 44-4+sqrt(4)
43 = 44 - 4/4
44 = 44 - 4 + 4
45 = 44 + 4/4
46 = 44 + 4 - sqrt(4)
47 = 4! + 4! - 4/4
48 = 4! + 4! + 4 - 4

[Kit-]

Re: 4 4's puzzle

I was going to write an exciting proof, but then I realized that (4!!!!!!!!!)! + 4!!!!!!!!! + 3 is impossible to get anyways.

[.rouge]

Re: 4 4's puzzle

49 = 4! + 4! + 4/4
50 = 4! + 4! + sqrt(4)

[gabgabo]

Re: 4 4's puzzle

.rouge don't bump old topics.

[Doug31]

Re: 4 4's puzzle

I was going to write an exciting proof, but then I realized that (4!!!!!!!!!)! + 4!!!!!!!!! + 3 is impossible to get anyways.

That is possible. You just have to use (4!!!!!!!!!)! + 4!!!!!!!!! + 4 - the infinitieth root of 4 lolz.

[archbishopjabber]

Re: 4 4's puzzle

51= (4! - 4 + .4)/.4
52= sqrt(4)(4!) + 4

[Doug31]

Re: 4 4's puzzle

53= (4! - sqrt(4))/.4 - sqrt(4)
54= 4!/.4 - 4 - sqrt(4)
55= (4! - sqrt(4))/.4
56= 4!/.4 - 4
57= (4! - sqrt(4))/.4 +sqrt(4)
58= 4!/.4 - sqrt(4)
59= 4!/.4 - 4/4
60= 4!/.4
61= 4!/.4 + 4/4
62= 4!/.4 + sqrt(4)
63= (4! + sqrt(4))/.4 - sqrt(4)
64= 4!/.4 + 4
65= (4! + sqrt(4))/.4
66= 4!/.4 + sqrt(4) + 4
67= (4! + sqrt(4))/.4 + sqrt(4)
68= 4!/.4 + 4 + 4
69= (4! + sqrt(4))/.4 + 4
70= (4! +4)/.4