tough logic problem

[thylacine]

I didn't look through the argument, but note that the sum also could not be of the form p^2+p for prime p, since p^3 factorises uniquely into factors bigger than one.

[Rasputin]

[ QUOTE ]
So that is one solution. This does not discount the possibility of other solutions. Basically there are two stipulations: m*n must be a number such that there must be a unique set of two of its factors that adds up to the summation of two non-primes. Additionally, m+n must equal some z where the following is true: given all pairs of numbers between 2 and 100 that add up to z, there must be a unique set of two numbers such that any two factors of the product of the two numbers can only add up to the summation of two non-prime numbers in a single way.

[/ QUOTE ]

I’d like to bump this to ask why.

Going back to the problem, we know the following facts:

1) There is a pair of distinct numbers from 2-100 inclusive.
2) One mathematician (P) was given the product of the two numbers.
3) Another mathematician (S) was given the sum of the two numbers.
4) P cannot determine what the sum of the numbers is
5) S knows that P cannot determine what the sum of the numbers is
6) When P finds out that S knows that P cannot determine what the sum of the numbers is, it provides him with enough information to determine what the sum of the numbers is
7) When S finds out that the above is true, it gives him enough information to determine what the product is.
8) From #4 we can deduce that there is more than one pair of numbers in the given range that would produce the product. Crunching the numbers gives me 2880 distinct products, 1793 of which only have one pair of factors in the required range leaving 1087 with more than one potential pair.
9) From #5 we can deduce that the sum is such that all of the pairs of integers in the range that would sum to this number would, when multiplied, result in a product which has more than one potential pair of factors in the given range. That is there must be at least two pairs of integers that sum to the sum and all of them must produce a product that is among the 1087 from #8. Crunching those numbers gives me ten possibilities for the sum, 11, 17, 23, 27, 29, 35, 37, 41, 47, and 53.

I completely understand that if both the numbers are prime, the guy given the product will be able to figure out what they are pretty easily. However, if one of the pair of integers is prime and one is not the guy given the product won’t be able to determine which pair it is.

For example, if the pair were to be 7, 10 the product is 70 and the sum is 17. The guy given the product doesn’t know if the pair is 7,10 or 2,35.

The guy given the sum doesn’t know if the pair is 2,15 or 3,14 or 5, 12 or 6,11 or 7,10.

2,15 multiplies to 30 which could also result from 3,10
3,14 multiplies to 42 which could also result from 2,21
5,12 multiplies to 60 which could also result from 2,30
6,11 multiplies to 66 which could also result from 2,33
7,10 multiplies to 70 which could also result from 2,35

If the pair were to be 5,6 the product is 30 and the sum is 11. The guy given the product doesn’t know if it’s 2,15 or 5,6.
The guy given the sum doesn’t know if the pair is 2,9 or 3, 8 or 4,7 or 5,6

2,9 multiplies to 18 which could also result from 3,6
3,8 multiplies to 24 which could also result from 2,12
4,7 multiples to 28 which could also result from 2,14
5,6 multiplies to 30 which could also result from 3,10

In both of these situations the man who knows the sum knows that the man who knows the product can’t pin down the pair because every possible combination of integers multiplies to a product that could also be the result of another pair of integers within the given range.

Now if you eliminate 2 from the range of integers I can see where it would work (though I haven’t crunched the numbers) but anytime the pair of integers is an even number and an odd number the man who knows the product won’t know if the pair is that pair or half the even number and twice the odd one.

[NiceCatch]

K is the set of all numbers that are not the sum of two prime numbers. k is any number in this set.

I'll try to answer your argument:

~~~
The guy given the sum doesn’t know if the pair is 2,15 or 3,14 or 5, 12 or 6,11 or 7,10.

2,15 multiplies to 30 which could also result from 3,10
*** It can't be 2,15 because 30 could also result from 5,6, which adds up to 11, another k number. The product (30) CANNOT potentially be the product of two sets of numbers that add up to two k numbers. If it were, P couldn't EXACTLY figure out m and n, just from knowing they were both not prime.
3,14 multiplies to 42 which could also result from 2,21
*** Once again, 2+21=23 is a k number. Same reason as above.
5,12 multiplies to 60 which could also result from 2,30
*** 60 is also the product of 20 and 3. Once again, 20+3=23 is a k number. Same reason as above.
6,11 multiplies to 66 which could also result from 2,33
*** Once again, 2+33=35 is a k number. Same reason as above.
7,10 multiplies to 70 which could also result from 2,35
*** Once again, 2+35=37 is a k number. Same reason as above.

4,13 multiplies to 52, which is also the product of 2 and 26 (4,13 and 2,26 are the ONLY integer pairs in the range from 2 to 100 that multiply to 52). 2+26 DOES NOT equal a k number, so 4,13 alone from the (m+n=17) numbers works.

I think your analysis missed all possible combinations of numbers that multiply to some of the numbers listed below. Look over my explanation for the (m+n=11) numbers in a previous post... I think it was fairly clear.