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Topic: A Logical Length (Read 884 times) |
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ThudnBlunder
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A Logical Length
« on: Nov 30th, 2009, 4:22pm » |
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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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| « Last Edit: May 28th, 2011, 6:19am by ThudnBlunder » |
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Immanuel_Bonfils
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Re: A Logical Length
« Reply #1 on: Nov 30th, 2009, 7:16pm » |
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If Smith answer must be affirmative or negative, the length is 16 miles.
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ThudnBlunder
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Re: A Logical Length
« Reply #2 on: Dec 1st, 2009, 2:44am » |
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Sorry, 16 is not even a candidate length.
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Immanuel_Bonfils
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Re: A Logical Length
« Reply #3 on: Dec 3rd, 2009, 10:01am » |
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Wouldn't it be 23 square miles, rather than 23 mile square?
for a ranch with 4 x 4, should we take length = 4 ?
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R
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Re: A Logical Length
« Reply #4 on: Dec 22nd, 2009, 4:35am » |
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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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SMQ
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Re: A Logical Length
« Reply #5 on: Dec 22nd, 2009, 4:43am » |
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on Dec 22nd, 2009, 4:35am, R wrote:| The area of the Todd County is 23-square miles. |
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No, it's a "23-mile square." i.e. a square 23-miles to a side.
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Hippo
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Re: A Logical Length
« Reply #6 on: Dec 22nd, 2009, 7:47am » |
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I have got these lengths possible: 4,5,6,7,8,10,12,14,18, and 20. Oh, sorry There is Mr. Brown, and there is no solution at all. (12,20/144,180)
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| « Last Edit: Dec 22nd, 2009, 7:55am by Hippo » |
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ThudnBlunder
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Re: A Logical Length
« Reply #7 on: Dec 22nd, 2009, 11:34am » |
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on Dec 3rd, 2009, 10:01am, Immanuel_Bonfils wrote:Wouldn't it be 23 square miles, rather than 23 mile square?
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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:| for a ranch with 4 x 4, should we take length = 4 ? |
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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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| « Last Edit: Dec 22nd, 2009, 3:43pm by ThudnBlunder » |
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SMQ
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Re: A Logical Length
« Reply #8 on: Dec 22nd, 2009, 1:14pm » |
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on Dec 22nd, 2009, 11:34am, ThudanBlunder wrote:| 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. |
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What about 12 = 2 x 6 = 3 x 4?
Quote:| Pre-emptive strike on nitpickers: width > 1 |
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Are you also defining width < length? I.e. the "width" is the lesser of the two dimensions and the "length" is the greater?
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ThudnBlunder
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Re: A Logical Length
« Reply #9 on: Dec 22nd, 2009, 3:11pm » |
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on Dec 22nd, 2009, 1:14pm, SMQ wrote:
What about 12 = 2 x 6 = 3 x 4?
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Oops, dunno how I missed that. :-[
Trouble is, Brown cannot uniquely identify the possible lengths on learning that Jones can.
on Dec 22nd, 2009, 1:14pm, SMQ wrote:
Are you also defining width < length?
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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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| « Last Edit: May 28th, 2011, 6:22am by ThudnBlunder » |
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towr
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Re: A Logical Length
« Reply #10 on: Dec 23rd, 2009, 3:03am » |
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on Dec 22nd, 2009, 3:11pm, ThudanBlunder wrote:| As 'length' is the longest dimension of an object, is a further definition necessary? |
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I can't say I knew that that was how length was defined.
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ThudnBlunder
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Re: A Logical Length
« Reply #11 on: Dec 23rd, 2009, 4:10am » |
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on Dec 23rd, 2009, 3:03am, towr wrote:
I can't say I knew that that was how length was defined. |
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This gives me hope that I will one day know as much as you. :)
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SMQ
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Re: A Logical Length
« Reply #12 on: Dec 23rd, 2009, 5:36am » |
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on Dec 22nd, 2009, 3:11pm, ThudanBlunder 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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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.
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ThudnBlunder
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Re: A Logical Length
« Reply #13 on: Dec 23rd, 2009, 6:24am » |
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on Dec 23rd, 2009, 5:36am, SMQ 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),
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Is that so? Hmm, (with width > 1) I have only 7 areas for which Smith's answer is useful to Jones. :-/ 180, 120, 60, 40, 24, 20, and 12
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| « Last Edit: Dec 23rd, 2009, 6:39am by ThudnBlunder » |
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SMQ
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Re: A Logical Length
« Reply #14 on: Dec 26th, 2009, 5:54am » |
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Smith and Jones only, 37/39 candidate areas (candidates with width = 1 in italics, Smith's answer to Jones' question in parentheses):
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)
Smith, Jones and Brown, 13/15 candidate areas (dimensions Brown deduces in bold):
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)
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ThudnBlunder
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Re: A Logical Length
« Reply #15 on: Dec 26th, 2009, 7:43am » |
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Nice work, SMQ. I thought at first that you had cooked another of 'my' puzzles, but perhaps not.
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.
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| « Last Edit: Dec 26th, 2009, 12:39pm by ThudnBlunder » |
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SMQ
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Re: A Logical Length
« Reply #16 on: Dec 26th, 2009, 7:59am » |
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Ahh, so given that Green was able to deduce the length of the ranch, we must additionally conclude he heard the answer to Jones' question. That was the inference I had been missing, and it 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.) Definitely an interesting puzzle!
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ThudnBlunder
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Re: A Logical Length
« Reply #17 on: Dec 26th, 2009, 8:50am » |
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on Dec 26th, 2009, 7:59am, SMQ wrote:Ahh, so given that Green was able to deduce the length of the ranch, we must additionally conclude he heard the answer to Jones' question. That was the inference I had been missing...
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But surely, as with any other problem, they can all hear each other by default, unless otherwise stated?
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Re: A Logical Length
« Reply #18 on: Dec 26th, 2009, 10:48am » |
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on Dec 26th, 2009, 8:50am, ThudanBlunder wrote:| But surely, as with any other problem, they can all hear each other by default, unless otherwise stated? |
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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. 
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ThudnBlunder
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Re: A Logical Length
« Reply #19 on: Dec 26th, 2009, 11:46am » |
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on Dec 26th, 2009, 10:48am, SMQ wrote:
As for 'rectangular' admitting squares, I suppose a better word would have been oblong.
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Hippo
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Re: A Logical Length
« Reply #20 on: Dec 26th, 2009, 1:31pm » |
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on Dec 22nd, 2009, 7:47am, Hippo wrote:| I have got these lengths possible: 4,5,6,7,8,10,12,14,18, and 20. Oh, sorry There is Mr. Brown, and there is no solution at all. (12,20/144,180) |
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If I would consider square is not rectangle may be I would find the solution ...
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ThudnBlunder
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Re: A Logical Length
« Reply #21 on: Dec 26th, 2009, 2:39pm » |
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on Dec 26th, 2009, 1:31pm, Hippo wrote:
If I would consider square is not rectangle may be I would find the solution ... |
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Sorry Hippo, you were light-years ahead of us all the time, haha. 
But this puzzle is sorta similar to one you objected to a few weeks ago. 
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| « Last Edit: Dec 26th, 2009, 2:41pm by ThudnBlunder » |
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Hippo
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Re: A Logical Length
« Reply #22 on: Dec 27th, 2009, 3:16pm » |
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on Dec 26th, 2009, 2:39pm, ThudanBlunder wrote:
Sorry Hippo, you were light-years ahead of us all the time, haha.
But this puzzle is sorta similar to one you objected to a few weeks ago.
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No. Current one was stated much better and the square not being rectangle is the only problem of it.
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| « Last Edit: Dec 27th, 2009, 3:16pm by Hippo » |
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