Mathematicians - Treat for the weekend / might drive you crazy
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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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(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
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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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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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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(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
6 and 7 The answer to everything. :-) Regardz Colin J Davies The most LinkedIn CPian (that I know of anyhow) :-)
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6 and 7 The answer to everything. :-) Regardz Colin J Davies The most LinkedIn CPian (that I know of anyhow) :-)
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The answer to everything but this one :) No explanation ??? The solution is so simple that it will explain everything and everything fits into the picture perfectly like a jigsaw puzzle "Not everything that counts can be counted..." -Albert Einstein
explanations are for sissys Regardz Colin J Davies The most LinkedIn CPian (that I know of anyhow) :-)
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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
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) :-)
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explanations are for sissys Regardz Colin J Davies The most LinkedIn CPian (that I know of anyhow) :-)
Guessing two numbers between 2 and 100, gives a probability of 0.02020202020202020202020202020201 chances of getting it correct so it is not. As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.-Albert Einstein
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Guessing two numbers between 2 and 100, gives a probability of 0.02020202020202020202020202020201 chances of getting it correct so it is not. As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.-Albert Einstein
I believe its abit less than that. Regardz Colin J Davies The most LinkedIn CPian (that I know of anyhow) :-)
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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
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; -
(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
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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I believe its abit less than that. Regardz Colin J Davies The most LinkedIn CPian (that I know of anyhow) :-)
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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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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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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;Check the clue below that can filter the range of numbers its better to try the other way (backward :) ) I don't like to keep you waiting but some people are still trying. I will post the solution tonight. May be before that you might hit the light bulb moment :) "Not everything that counts can be counted..." -Albert Einstein
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For those who are still trying for the solution here is a clue to drive you more crazy 3=> B: I already knew that. i will post the solution by tonight "Not everything that counts can be counted..." -Albert Einstein
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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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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(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
Is there more than one answer?
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Is there more than one answer?