20070921, 22:43  #1 
6809 > 6502
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Aug 2003
101×103 Posts
5·1,873 Posts 
Four 1's puzzle
Figured I would start this one.
Like the 4 4's. Using four 1's and standard mathematical operations formulate equations with the answer being each of the numbers from 0 to 100 and on. Every calculation MUST use exactly all four 1's and no other digits. The allowable operations are: Class 1
Class 2
Class 3
Solution scoring: 100 points per class 3 operator 10 points per class 2 operator 1 point per class 1 operator (parenthesis per pair), except the following: 1/2 point per decimal 0 points for concatenation The goal is to get as low as score solution as possible. First 3 to start: 0 = 11  11 1 = 11 / 11 2 = 1/1 + 1/1 Last fiddled with by Uncwilly on 20070921 at 22:47 
20070921, 23:35  #2 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 
I'm putting the score for it next to the number.
A few in order: 3, 3 = 1 * 1 + 1 + 1 3, 4 = 1 + 1 + 1 + 1 And some easy ones out of order: 2, 9 = 11  1  1 3, 10 = 11  (1 * 1) 2, 11 = 11 * 1 * 1 3, 12 = 11 + (1 * 1) 2, 13 = 11 + 1 + 1 1, 22 = 11 + 11 1, 110 = 111  1 1, 111 = 111 * 1 1, 112 = 111 + 1 1, 121 = 11 * 11 0, 1111 = 1111 10, 285311670611 = 11 ^ 11 I'm not so sure that very many numbers can be made with four 1's, just because of how it's so low and many operations don't, or barely, change the number you start with. 
20070922, 01:40  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{6}×5×19 Posts 
10 = 11  1 / 1 (2 points)
12 = 11 + 1 / 1 (2 points) 100 = 11 / .11 (1.5 points) 
20070922, 03:27  #4 
Aug 2002
Ann Arbor, MI
1B1_{16} Posts 
I agree that we need to add at least one degree of fudging to get everything as as combination of some fixed amount of 1's. I was thinking different number bases, but sadly for all b>1.

20070922, 04:12  #5 
Jun 2003
7·167 Posts 
5 = (1 + 1 * 1)/.1 (4.5 points)
6 = (1 + 1 + 1 * 1)! (14 points) 7 = (1 + 1 + 1)! + 1 (14 points) 8 = 1/.1  1  1 (3.5 points) 
20070922, 05:23  #6 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{6}·5·19 Posts 

20070922, 05:26  #7 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{6}·5·19 Posts 
5 = 1*1/(.1+.1) (5 points)
10 = 11/1.1 (1.5 points) 20 = 1/.1+1/.1 (4 points) 24 = (1+1+1+1)! (14 points) 30 = (1+1+1)/.1 (4.5 points) 55 = 11/(.1+.1) (4 points) 60 = (1+1+1)!/.1 (14.5 points) 109 = 11/.11 (2.5 points) 1110 = 111/.1 (1.5 points) = 11/(11) (3 points) Last fiddled with by retina on 20070922 at 06:13 
20070923, 04:55  #8 
Jun 2003
7×167 Posts 

20070923, 06:10  #9 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{6}·5·19 Posts 
1 = 1/.1  11 (2.5 points)
21 = 1/.1 + 11 (2.5 points) Uncwilly, this puzzle is really hard. I hope you plan on eventually posting the solutions, because I think everyone is stuck. There must be something fundamental that we all seem to have missed so far. So many gaps in the range 0100. Perhaps a small hint is in order. 
20070924, 04:02  #10 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
13700_{8} Posts 
99 = 1/.1/.11 (4 points)
101 = 1/.1/.1+1 (4 points) 200 = (1+1)/.1/.1 (5 points) 719 = (1+1+1)!!1 (24 points) 720 = (1+1+1)!!*1 (24 points) 721 = (1+1+1)!!+1 (24 points) 1000 = 1/.1/.1/.1 (4.5 points) 1100 = 11/.1/.1 (3 points) I can't see how √,  or gamma can be useful. Is that a clue as to how to generate the missing values in the 0100 range? 
20070924, 06:00  #11  
Aug 2002
Ann Arbor, MI
433 Posts 
Quote:
Also, gamma can be useful if you wanted to get (n1)! instead of n! without wasting a 1. Last fiddled with by Kevin on 20070924 at 06:12 

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