20050928, 16:45  #1 
Jan 2005
Caught in a sieve
18B_{16} Posts 
Classic problem: Four 4's
I found a version of the four 4's problem in an old book awhile back. Four those who haven't seen it befour, the challenge is to create fourmulas, using no more than four 4's, that evaluate to other integers.
This version is more stringent on the operations allowed than most. The operations allowed are listed below. The book claims that all positive integers up to 119 can be created this way. I have been unable to create three of those integers, so I'd like to see what you can come up with. If the book is correct, though, all integers from 1 through at least 130 can be created. Allowed unary operations: (each example in parentheses uses one 4) Square (4^2)  this is the only place where 2's are allowed. Also note below that exponentiation other than this is not allowed. Factorial (4!) Decimal (.4) Allowed binary operations: (each example in parentheses uses two 4's) + (4+4)  (44) * (4*4) / (4/4) Concatenate (44); also before (4.4) or after (.44) a Decimal. Any number of parentheses are also allowed. Standard order of operations applies when parentheses are missing. Since it's usually easy to fill out an expression with more 4's (e.g. 8 = 4+4 or 4^2/4+4^2/4), I'll determine best solutions first by fewest 4's, and second by shortest total length. Wacky, would you mind setting up a scoreboard from 0 to 160 or so? I'll start populating it: Code:
N: 4's,len expression (Author, date) 8: 2, 3: 4+4 (Ken_g6, 9/28) 
20050928, 19:22  #2  
"Richard B. Woods"
Aug 2002
Wisconsin USA
17014_{8} Posts 
Quote:
Notation: I'll use 4^{sq} to denote the unary squaring, so as not to introduce any "2". Here's some easy or obvious ones, not necessarily of minimal length, just to initiate a few table entries: 4 = 4 6 = 4! / 4 10 = 4 / .4 11 = 44 / 4 16 = 4^{sq} 24 = 4! 32 = 4^{sq} + 4^{sq} 36 = 4!^{sq} / 4^{sq} 44 = 44 48 = 4! + 4! 64 = 4^{sq} * 4 72 = 4! * 4  4! 80 = 4^{sq} * 4 + 4^{sq} 96 = 4! * 4 100 = ( 4 / .4 )^{sq} 110 = 44 / .4 111 = 44.4 / .4 128 = 4^{sq} * 4 + 4^{sq} * 4 144 = 4!^{sq} / 4 256 = 4^{sq} * 4^{sq} Last fiddled with by cheesehead on 20050928 at 19:36 

20050928, 19:42  #3  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
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20050928, 19:51  #4 
Jun 2003
The Texas Hill Country
441_{16} Posts 
I'll set up a scoreboard soon.
Obviously: 0 = 44 1 = 4/4 2 = (4+4)/4 3= (4+4+4)/4 5 = 4 + 4/4 7 = 4 + 4  4/4 8 = 4 + 4 9 = 4 + 4 + 4/4 12 = 4 + 4 + 4 
20050928, 20:10  #5 
Jan 2005
Transdniestr
111110111_{2} Posts 
13 = 4sq 4 + 4/4
15 = 4sq  4/4 Last fiddled with by grandpascorpion on 20050928 at 20:10 
20050928, 20:51  #6  
Jun 2005
Near Beetlegeuse
2^{2}×97 Posts 
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20050928, 20:56  #7 
Jan 2005
Transdniestr
503_{10} Posts 
17 = 4sq + 4/4
18 = 4/.4 + 4 + 4 19 = 4!  4  4/4 20 = 4!  4 21 = 4! 4 + 4/4 23 = 4!  4/4 25 = 4! + 4/4 26 = 4sq + 4/.4 27 = 4! + 4  4/4 28 = 4sq + 4sq  4 29 = 4! + 4 + 4/4 30 = (4+4/4)!/4 31 = 4sq + 4sq  4/4 32 = 4sq + 4sq 33 = 4sq + 4sq + 4/4 34 = 4! + 4/.4 35 = 4! + 44/4 37 = 4!sq / 4sq + 4/4 39 = 4! + 4sq  4/4 40 = 4! + 4sq 41 = 4! + 4sq + 4/4 95 = (4! * 4sq4)/4 96 = 4! * 4sq/4 97 = (4! * 4sq+4)/4 109 = (44 .4)/.4 120 = (4+4/4)! 121 = (44/4)sq Last fiddled with by grandpascorpion on 20050928 at 21:01 
20050928, 21:30  #8 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
7 = (4!+4)/4
60 = 4!/.4 
20050928, 21:49  #9 
Jan 2005
Transdniestr
111110111_{2} Posts 
14 = 4sq  (4+4)/4
22 = 4!  (4+4)/4 43 = 44  4/4 45 = 44 + 4/4 47 = 4! + 4!  4/4 49 = 4! + 4! + 4/4 52 = 4! + 4! + 4 56 = 4*4sq44 60 = 4*4sq4 63 = 4*4sq4/4 65= 4*4sq+4/4 81 = ((44/4)sq)sq Last fiddled with by grandpascorpion on 20050928 at 21:54 
20050928, 22:00  #10 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
hmm. The way I remember this was that you had to use exactly 4 4's
I don't think squaring should be included, I can't logically fathom why that's allowed. Perhaps they mean square roots? Also, is (.4...) or .4 with an overbar allowed? (4/.4..) is a nice way to get 9 :) Has anyone ever looked at a systematic way of enumerating all possible answers? 
20050928, 22:31  #11  
Jan 2005
Caught in a sieve
18B_{16} Posts 
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P.S. There's a better solution for 12 (and thus 3). Edit: I guessed 160 as a rough upper limit on the contiguous integers from 1 that have four 4's solutions. We can go higher if anyone wants. Last fiddled with by Ken_g6 on 20050928 at 22:35 

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