4 4's puzzle

tricky77puzzle · 07-26-2007, 12:15 PM

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 · 07-26-2007, 02:40 PM

Stuck on 11 eh? Fail.

tricky77puzzle · 07-27-2007, 04:42 PM

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 · 07-27-2007, 04:55 PM

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** · 07-27-2007, 05:02 PM

28= 4!(4/4)-4

tricky77puzzle · 07-28-2007, 04:26 PM

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** · 07-29-2007, 04:03 AM

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- · 07-29-2007, 06:57 AM

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

.rouge · 09-26-2007, 12:44 AM

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

**gabgabo** · 09-27-2007, 04:51 PM

.rouge don't bump old topics.

**Doug31** · 09-28-2007, 12:57 AM

Quote:

Originally Posted by Kit-

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 · 10-3-2007, 06:45 PM

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

**Doug31** · 10-6-2007, 05:59 PM

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

07-26-2007, 12:15 PM	#1
tricky77puzzle FFR Player Join Date: Jul 2007 Posts: 8	4 4's puzzle 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 = 2222222*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.

07-26-2007, 02:40 PM	#2
megaxxx FFR Player Join Date: Jun 2006 Location: Louisiana Age: 36 Posts: 329	Re: 4 4's puzzle Stuck on 11 eh? Fail. __________________ GET A JOB LOSER.

07-27-2007, 04:42 PM	#3
tricky77puzzle FFR Player Join Date: Jul 2007 Posts: 8	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...

07-27-2007, 04:55 PM	#4
Fractal_Monkey FFR Player Join Date: Apr 2005 Location: Manchester, England Posts: 464	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.

07-27-2007, 05:02 PM	#5
rushyrulz Digital Dancing! Join Date: Feb 2006 Location: 80 billion club, NE Age: 31 Posts: 12,981	Re: 4 4's puzzle 28= 4!(4/4)-4 __________________

07-28-2007, 04:26 PM	#6
tricky77puzzle FFR Player Join Date: Jul 2007 Posts: 8	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)

07-29-2007, 04:03 AM	#7
Kotarouchan I only need 3 seconds :) Join Date: Feb 2005 Location: South California Age: 36 Posts: 2,030	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 __________________

07-29-2007, 06:57 AM	#8
Kit- Private College Join Date: Feb 2006 Location: Lol badger Posts: 536	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. __________________ <img src="Bent Lines" />

09-26-2007, 12:44 AM	#9
.rouge Banned Join Date: Sep 2007 Posts: 20	Re: 4 4's puzzle 49 = 4! + 4! + 4/4 50 = 4! + 4! + sqrt(4)

09-27-2007, 04:51 PM	#10
gabgabo FFR Veteran Join Date: Feb 2007 Location: Canada Age: 31 Posts: 1,811	Re: 4 4's puzzle .rouge don't bump old topics.

10-3-2007, 06:45 PM	#12
archbishopjabber FFR Player Join Date: Dec 2005 Age: 80 Posts: 268	Re: 4 4's puzzle 51= (4! - 4 + .4)/.4 52= sqrt(4)(4!) + 4 __________________ "Knowing information legitimately lessens genuine error. Ordinarily, research generates excellent benefit understanding social history." "Guide to Freedom." Vol. 9. Page 11 http://www.infinite-story.com/story/2726/

10-6-2007, 05:59 PM	#13
Doug31 Falcon Paaaauuuunch!!!!!! Join Date: Jun 2004 Location: Washington Age: 36 Posts: 6,811	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 __________________

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