Mathematicians - Treat for the weekend / might drive you crazy
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Now you are making no sense, and even contradicting your earlier posts. In the first post you said:
2 <= a,b <= 100
Now you say
2 >= a,b >= 100
I believe you meant the first one. And a and b in BOTH of the solutions on the page I showed are between 2 and 100. David
Sorry it was the latter
David O`Neil wrote:
2 <= a,b <= 100
and you were right these makes no sense, i wrote that without thinking and ALSO it contradicts the problem it self becuse if there are two solutions then how come the the two persons are sure about the anwser they will still have to decide between the two solutions right ? Midnight jitters :) -- modified at 19:22 Sunday 13th November, 2005
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(Actually, This is an old problem but just to stop smartsy CPians from using the word "nanosecond" forever let's blast them with it! ). There are 2 integers a and b (Though programmers like i and j more as integers) And two CPian A and B. (The result will decide who are they) Rules of the game ------------------------------- I. 2 <= a,b <= 100 II. A is told the product of a and b III. B is told the sum of a and b. IV. Neither is told the values The cPians conversation ::: ------------------------------- 1=> B: I cannot determine a, b. 2=> A: I cannot determine a, b 3=> B: I already knew that. 4=> A: In that case, I now know them 5=> B: In that case, I too now know a, b. What are the numbers ??? ------------------------------- Is'nt it simple and beautiful ? I can hear you saying " huh! " cheers. "Not everything that counts can be counted..." -Albert Einstein
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Quartz... wrote:
where r u guys !
Thinking! Obsessivelly!:mad: "Thanks" for wrecking my weekend!:mad: X| Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)
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Quartz... wrote:
2 <= a,b <= 100
Do you mean 2 <= a <= 100 and 2 <= b <= 100 or 2 <= a and b <= 100 ?
Picture a huge catholic cathedral. In it there's many people, including a gregorian monk choir. You know, those who sing beautifully. Then they start singing, in latin, as they always do: "Ad hominem..." -Jörgen Sigvardsson
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OK, your post has been up for several hours now. How about posting your solution? Personally, I think the problem is indeterminate, at least without additional constraints.
Software Zen:
delete this; -
2 & 3 or vica versa My reasoning the lowest 2 numbers that are not 2 & 2. 1. B: (a x b) = 6 2. A: (a + b) = 5 3. B: a != b and a < 4 and b < 4 4. A: Can only be 2 & 3 5: B: Can only be 2 & 3 :confused: Probably very incorrect... :p [update] Similar to Rui A. Rebelo's solution [update] xacc.ide-0.1-rc3 released! Download and screenshots -- modified at 12:23 Saturday 12th November, 2005
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Now you are making no sense, and even contradicting your earlier posts. In the first post you said:
2 <= a,b <= 100
Now you say
2 >= a,b >= 100
I believe you meant the first one. And a and b in BOTH of the solutions on the page I showed are between 2 and 100. David
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Is A and B necesary CPians? Or can they be other kinds of people too? If the former, my original hunch was perhaps the user id of 2 specific users under 10000 that can be broken down to the requirements... (long shot :) ) xacc.ide-0.1-rc3 released! Download and screenshots
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The numbers are 3 and 4. Explanation: B is told the sum: 7. He knows that only 2 different pairs of numbers greater than 2 can give a sum of 7 : 3+4 and 2+5. Therefore:
Quartz... wrote:
1=> B: I cannot determine a, b.
A is told the product: 12. He knows that there are just 2 different pairs of numbers greater than 2 which give this result: 2*6 and 3*4. So:
Quartz... wrote:
2=> A: I cannot determine a, b
Now, B knows 2 things. He knows A's ambiguity if the numbers are 3 and 4. And he knows that if they were 2 and 5 (as B considered) then A wouldn't have an ambiguity; these 2 numbers would generate a unique product of 10. Facing A's ambiguity:
Quartz... wrote:
3=> B: I already knew that.
A perceives that B solved it's own ambiguity and the only possibility of this happening is that the numbers are 3 and 4. If they were 2 and 6 as he considered the sum would be 8, wich would give 3 diferent possibilities to B (4+4, 2+6, 3+5). So:
Quartz... wrote:
4=> A: In that case, I now know them
And B already knew the answer. Makes sense? Did I preserve my sanity? Rui A. Rebelo I don't smoke, don't gamble, don't sniff, don't drink and don't womanize. My only defect is that I lie just a little bit, sometimes. Tim Maia (brazilian pop singer)
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Interpreting your qu differently gives 4, 13. I think this used to be called "the impossible problem" Regardz Colin J Davies The most LinkedIn CPian (that I know of anyhow) :-)