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That's true, I haven't proven that no other number pairs work. My reasoning was that if there were two possible answers, then the question would not have implied that there is a single solution. Therefore, I am assuming that there is only one solution. Since (3,8) works, that must be the solution.Quote:
Originally posted by gstercken
Nice idea. But are you sure that no other pair of numbers would fit to the same argumentation?
Not exactly. Of course the question only makes sense if there is a single solution. But if there would be other numbers fitting your argumentation, that would only mean that your argumentation cannot be the correct solution. In this case, there had to be a different kind of reasoning which would unambiguosy lead to a single pair of numbers. The fact that (3,8) "works" is true only in the context of your (possibly wrong) solution.Quote:
Originally posted by solarflare
Since (3,8) works, that must be the solution.
Don't misunderstand me, I am not saying that your solution is wrong. But to prove that it is right, it must be guaranteed that no other pair of numbers "works" in this reasoning.
There must be a flaw in the reasoning or in the question. Anyway, 3 and 8 was not the correct answer.
In the reasoning I have seen, there is one step I still don't understand, and I now think it might be wrong. But it was reducing the different possibilites to just one. It really seems that there are multiple solutions now. Or something is missing in the question, but I can't find what. :(
I guess you'll have to go for another question, and I'll try to find the missing information that implies there is only ONE solution.
If you say which part, I can either clarify it or realize my own mistake.Quote:
Originally posted by Elrond
In the reasoning I have seen, there is one step I still don't understand, and I now think it might be wrong.
That's not correctQuote:
Originally posted by solarflare
Commentary: S thinks If it were (2,9), whose product is 18, Mr. P could have thought the numbers were (3,6), which would also have fit his statements; since there are two possibilities, how can he know which one? So it can't be (2,9). The same logic holds for (4,7) and (5,6), leaving (3,8) as the only working possibility. Therefore, the numbers must be (3,8).
okay that's what i thought of....
if a+b = s and a*b = p
p is not a product of two prime numbers
therefore s cannot be a sum of two prime numbers otrewise S wouldn't have known for sure that P cannot know the numbers that means that 3+6=9 could not be it becaouse 9 can be 2+7
so s is one of these numbers and both men nknow it after the second statement 11, 17... etc nubers that cannot be made like a sum of two prime numbers
so suppose S knew 11 and P knew 24 p whould've guessed the numbers after S saying that he knew P wouldn't. but s would be wondering between 2*9 = 18, 3*8 =24 and 5*6 = 15 so the sum cannot be eleven
so s is not 17 becaouse p could be 52=3*14 or 66 = 11*6 and S won't know the answer again
so a is prime and b is ot the type 2*m where m is a prime number and s can has only one way of bein represented like a sum of this kind of numbers so s is not 11 and not 17 and not 23
so a is 13 and b is 22
So you're saying s can only be odd numbers (as each even number can be written as the sum of two primes) minus all odd numbers that are equal to a prime number + 2. And we already know that p cannot be the product of two prime numbers.Quote:
Originally posted by SeventhStar
therefore s cannot be a sum of two prime numbers...
That already excludes quite some possibilities.
taht's still not proven but it's true for all numbers less than 100 000Quote:
Originally posted by Simon666
each even number can be written as the sum of two primes
I think it is calculated just a little bit further than 100.000, according to GStercken 4*10^14. So I doubt it would not be true. :D:D:DQuote:
Originally posted by SeventhStar
taht's still not proven but it's true for all numbers less than 100 000
okay but it isn't proven for sure.Quote:
Originally posted by Simon666
I think it is calculated just a little bit further than 100.000, according to GStercken 4*10^14. So I doubt it would not be true. :D:D:D
Look, if it is proven for at least all numbers between 2 and 99 that is already enough for this riddle. :rolleyes:Quote:
Originally posted by SeventhStar
okay but it isn't proven for sure.
Yep! And I think that:Quote:
Originally posted by Simon666
Look, if it is proven for at least all numbers between 2 and 99 that is already enough for this riddle. :rolleyes:
Quote:
]Originally posted by me
A is 13 and B is 22