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Title: A Logical Length Post by ThudanBlunder on Nov 30th, 2009, 4:22pm Four ranchers have the following conversation: SMITH: Down in Todd County, which is a 23-mile square, I own a rectangular ranch that measures a whole number of miles each way and parallels the borders of the county. JONES: Wait a minute, I happen to know the area of your ranch; let me see if I can figure its length. [Jones calculates furiously.] JONES: No, I need more information. Is the width more than half the length? [Smith answers the question.] JONES: I now know the length. BROWN: I also know the area and, although I did not hear your answer to Jones' question, I can also tell you the length. GREEN: I do not know the area but I can now tell you the length of the ranch. What is the length of the ranch? |
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Title: Re: A Logical Length Post by Immanuel_Bonfils on Nov 30th, 2009, 7:16pm [hide]If Smith answer must be affirmative or negative, the length is 16 miles. [/hide] |
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Title: Re: A Logical Length Post by ThudanBlunder on Dec 1st, 2009, 2:44am Sorry, 16 is not even a candidate length. |
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Title: Re: A Logical Length Post by Immanuel_Bonfils on Dec 3rd, 2009, 10:01am Wouldn't it be 23 square miles, rather than 23 mile square? [hide]for a ranch with 4 x 4, should we take length = 4 ?[/hide] |
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Title: Re: A Logical Length Post by R on Dec 22nd, 2009, 4:35am The area of the Todd County is 23-square miles. As Smith owns a rectangular ranch, whose sides are parallel to the border of the county, I assumed that Todd County is also rectangular. And if that is right, I wonder what are the sides of the Todd County, as 23 is a prime number. The only possibility is width = 1 and length = 23. And in that case, if someone knows the area, he knows the length too. |
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Title: Re: A Logical Length Post by SMQ on Dec 22nd, 2009, 4:43am on 12/22/09 at 04:35:48, R wrote:
No, it's a "23-mile square." i.e. a square 23-miles to a side. --SMQ |
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Title: Re: A Logical Length Post by Hippo on Dec 22nd, 2009, 7:47am I have got these lengths possible: [hide]4,5,6,7,8,10,12,14,18, and 20[/hide]. Oh, sorry [hide]There is Mr. Brown, and there is no solution at all[/hide]. [hide](12,20/144,180)[/hide] |
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Title: Re: A Logical Length Post by ThudanBlunder on Dec 22nd, 2009, 11:34am on 12/03/09 at 10:01:08, Immanuel_Bonfils wrote:
No. But if we consider areas less than 23 I see that 20 = 2 x 10 = 4 x 5 is the only area that would explain the conversation between Smith and Jones. Quote:
Although I suppose rectangles include squares, I would never describe a square as a rectangle. Pre-emptive strike on nitpickers: width > 1 |
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Title: Re: A Logical Length Post by SMQ on Dec 22nd, 2009, 1:14pm on 12/22/09 at 11:34:11, ThudanBlunder wrote:
What about 12 = 2 x 6 = 3 x 4? Quote:
Are you also defining width < length? I.e. the "width" is the lesser of the two dimensions and the "length" is the greater? --SMQ |
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Title: Re: A Logical Length Post by ThudanBlunder on Dec 22nd, 2009, 3:11pm on 12/22/09 at 13:14:16, SMQ wrote:
Oops, dunno how I missed that. :-[ Trouble is, Brown cannot uniquely identify the possible lengths on learning that Jones can. on 12/22/09 at 13:14:16, SMQ wrote:
As 'length' is the longest dimension of an object, is a further definition necessary? (Or are you just trying to wind me up?) :P |
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Title: Re: A Logical Length Post by towr on Dec 23rd, 2009, 3:03am on 12/22/09 at 15:11:57, ThudanBlunder wrote:
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Title: Re: A Logical Length Post by ThudanBlunder on Dec 23rd, 2009, 4:10am on 12/23/09 at 03:03:22, towr wrote:
This gives me hope that I will one day know as much as you. :) |
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Title: Re: A Logical Length Post by SMQ on Dec 23rd, 2009, 5:36am on 12/22/09 at 15:11:57, ThudanBlunder wrote:
Nope, just trying to be sure I'm interpreting the problem as intended, since I currently find 15 areas satisfying Smith's, Jones', and Brown's statements (13 with width > 1), and am having a difficult time understanding Green's statement unless he already had knowledge of some other parameter of the ranch's configuration. --SMQ |
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Title: Re: A Logical Length Post by ThudanBlunder on Dec 23rd, 2009, 6:24am on 12/23/09 at 05:36:10, SMQ wrote:
Is that so? Hmm, (with width > 1) I have only 7 areas for which Smith's answer is useful to Jones. :-/ [hide]180, 120, 60, 40, 24, 20, and 12[/hide] |
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Title: Re: A Logical Length Post by SMQ on Dec 26th, 2009, 5:54am Smith and Jones only, 37/39 candidate areas (candidates with width = 1 in italics, Smith's answer to Jones' question in parentheses): [hide]6 = 1 x 6 (no) = 2 x 3 (yes) 12 = 1 x 12 (no) = 2 x 6 (no) = 3 x 4 (yes) 15 = 1 x 15 (no) = 3 x 5 (yes) 20 = 1 x 20 (no) = 2 x 10 (no) = 4 x 5 (yes) 24 = 2 x 12 (no) = 3 x 8 (no) = 4 x 6 (yes) 28 = 2 x 14 (no) = 4 x 7 (yes) 30 = 2 x 15 (no) = 3 x 10 (no) = 5 x 6 (yes) 40 = 2 x 20 (no) = 3 x 10 (no) = 5 x 8 (yes) 42 = 2 x 21 (no) = 3 x 14 (no) = 6 x 7 (yes) 45 = 3 x 15 (no) = 5 x 9 (yes) 48 = 3 x 16 (no) = 4 x 12 (no) = 6 x 8 (yes) 54 = 3 x 18 (no) = 6 x 9 (yes) 56 = 4 x 14 (no) = 7 x 8 (yes) 60 = 3 x 20 (no) = 4 x 15 (no) = 5 x 12 (no) = 6 x 10 (yes) 63 = 3 x 21 (no) = 7 x 9 (yes) 66 = 3 x 22 (no) = 6 x 11 (yes) 70 = 5 x 14 (no) = 7 x 10 (yes) 72 = 4 x 18 (no) = 6 x 12 (no) = 8 x 9 (yes) 80 = 4 x 20 (no) = 5 x 16 (no) = 8 x 10 (yes) 84 = 4 x 21 (no) = 6 x 14 (no) = 7 x 12 (yes) 88 = 4 x 22 (no) = 8 x 11 (yes) 90 = 5 x 18 (no) = 6 x 15 (no) = 9 x 10 (yes) 96 = 6 x 16 (no) = 8 x 12 (yes) 108 = 6 x 18 (no) = 9 x 12 (yes) 110 = 5 x 22 (no) = 10 x 11 (yes) 112 = 7 x 16 (no) = 8 x 14 (yes) 120 = 6 x 20 (no) = 8 x 15 (yes) = 10 x 12 (yes) 126 = 6 x 21 (no) = 7 x 18 (no) = 9 x 14 (yes) 132 = 6 x 22 (no) = 11 x 12 (yes) 140 = 7 x 20 (no) = 10 x 14 (yes) 144 = 8 x 18 (no) = 9 x 16 (yes) 154 = 7 x 22 (no) = 11 x 14 (yes) 160 = 8 x 20 (no) = 10 x 16 (yes) 168 = 8 x 21 (no) = 12 x 14 (yes) 176 = 8 x 22 (no) = 11 x 16 (yes) 180 = 9 x 20 (no) = 10 x 18 (yes) = 12 x 15 (yes) 198 = 9 x 22 (no) = 11 x 18 (yes) 210 = 10 x 21 (no) = 14 x 15 (yes) 220 = 10 x 22 (no) = 11 x 20 (yes) [/hide] Smith, Jones and Brown, 13/15 candidate areas (dimensions Brown deduces in bold): [hide]12 = 1 x 12 (no) = 2 x 6 (no) = 3 x 4 (yes) 20 = 1 x 20 (no) = 2 x 10 (no) = 4 x 5 (yes) 24 = 2 x 12 (no) = 3 x 8 (no) = 4 x 6 (yes) 30 = 2 x 15 (no) = 3 x 10 (no) = 5 x 6 (yes) 40 = 2 x 20 (no) = 4 x 10 (no) = 5 x 8 (yes) 42 = 2 x 21 (no) = 3 x 14 (no) = 6 x 7 (yes) 48 = 3 x 16 (no) = 4 x 12 (no) = 6 x 8 (yes) 60 = 3 x 20 (no) = 4 x 15 (no) = 5 x 12 = 6 x 10 (yes) 72 = 4 x 18 (no) = 6 x 12 (no) = 8 x 9 (yes) 80 = 4 x 20 (no) = 5 x 16 (no) = 8 x 10 (yes) 84 = 4 x 21 (no) = 6 x 14 (no) = 7 x 12 (yes) 90 = 5 x 18 (no) = 6 x 15 (no) = 9 x 10 (yes) 120 = 6 x 20 (no) = 8 x 15 (yes) = 10 x 12 (yes) 126 = 6 x 21 (no) = 7 x 18 (no) = 9 x 14 (yes) 180 = 9 x 20 (no) = 10 x 18 (yes) = 12 x 15 (yes) [/hide] --SMQ |
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Title: Re: A Logical Length Post by ThudanBlunder on Dec 26th, 2009, 7:43am Nice work, SMQ. I thought at first that you had cooked another of 'my' puzzles, but perhaps not. [hide] From your group of 13 candidate areas, Brown can conclude that the length must be 4, 5, 6 (twice), 7, 8, (twice), 9, 10 (twice), 14, or 20 (twice). So 6, 8, and 10 occur twice when W > L/2 and only 20 occurs twice when W < L/2. Since we are told that Green, who could also deduce the 13 candidate areas, knew the length but not the area when he heard the answer to Jones' question, we can conclude that the answer must have been 'No.' Hence length is established as 20, the width as 6 or 9, and the area as 120 or 180.[/hide] |
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Title: Re: A Logical Length Post by SMQ on Dec 26th, 2009, 7:59am Ahh, so given that Green was able to deduce the length of the ranch, we must additionally conclude he [hide]heard the answer to Jones' question[/hide]. That was the inference I had been missing, and it [hide]removes the need for the restriction that width > 1 as well (although it does still require that square ranches not be considered, otherwise 144 = 8 x 18 (no) = 9 x 16 (yes) = 12 x 12 (yes) would be a candidate as well.)[/hide] Definitely an interesting puzzle! --SMQ |
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Title: Re: A Logical Length Post by ThudanBlunder on Dec 26th, 2009, 8:50am on 12/26/09 at 07:59:36, SMQ wrote:
But surely, as with any other problem, they can all hear each other by default, unless otherwise stated? |
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Title: Re: A Logical Length Post by SMQ on Dec 26th, 2009, 10:48am on 12/26/09 at 08:50:43, ThudanBlunder wrote:
I was...unfamiliar with that convention. Rather--and especially with a problem as tightly-worded as this one--I tend to use an approach of "assume nothing relevant unless explicitly stated" and then attempt to determine the minimum information needed to reach a solution. Since Jones heard an answer that Brown didn't, I didn't want to make an assumption as to whether or not Green had heard it without evidence. Since I now see that his having heard the answer explains both his statement and the meta question of why the problem asks for the length rather than the area, I am content that the expected and most logical answer has been found. :) --SMQ |
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Title: Re: A Logical Length Post by ThudanBlunder on Dec 26th, 2009, 11:46am on 12/26/09 at 10:48:43, SMQ wrote:
:) As for 'rectangular' admitting squares, I suppose a better word would have been oblong (http://www.merriam-webster.com/dictionary/oblong). |
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Title: Re: A Logical Length Post by Hippo on Dec 26th, 2009, 1:31pm on 12/22/09 at 07:47:08, Hippo wrote:
If I would consider square is not rectangle may be I would find the solution :) ... |
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Title: Re: A Logical Length Post by ThudanBlunder on Dec 26th, 2009, 2:39pm on 12/26/09 at 13:31:46, Hippo wrote:
Sorry Hippo, you were light-years ahead of us all the time, haha. ;) But this puzzle is sorta similar to one (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs;action=display;num=1259685054) you objected to a few weeks ago. :P |
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Title: Re: A Logical Length Post by Hippo on Dec 27th, 2009, 3:16pm on 12/26/09 at 14:39:51, ThudanBlunder wrote:
:) No. Current one was stated much better :) and the square not being rectangle is the only problem of it. |
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