# Logic problem

• Dec 3rd 2010, 07:37 AM
wonderboy1953
Logic problem
Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.
P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.
Given that the above statements are true, what are the two numbers? (computer assistance is allowed.)

Moderator edit: Apporved Challenge question.
• Dec 3rd 2010, 01:25 PM
CaptainBlack
Quote:

Originally Posted by wonderboy1953
Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.
P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.
Given that the above statements are true, what are the two numbers? (computer assistance is allowed.)

Moderator edit: Apporved Challenge question.

P: I cannot determine the two numbers. - the number I have is not the product of two primes

S: I knew that. - the sum I have cannot be the sum of two primes

P: Now I can determine them. - there is only one factorisation such that the sum cannot be the sum of two primes

S: So can I. - there is only one decomposition of my number into a sum such that the sum of a factorisation of their product cannot be the sum of two primes

CB
• Dec 3rd 2010, 06:31 PM
Wilmer
Quite a golden oldie:

• Dec 3rd 2010, 08:07 PM
CaptainBlack
Quote:

Originally Posted by Wilmer
Quite a golden oldie:

I am reasonably sure that I have solved this (or a number of problems remarkably like this) before as a NewScientist Enigma.

CB
• Dec 4th 2010, 08:05 AM
MSM
found a solution
13 and 4 (Smirk)
• Dec 4th 2010, 11:28 AM
wonderboy1953
Quote:

Originally Posted by MSM
13 and 4 (Smirk)

You're the winner (good job).
• Dec 4th 2010, 12:23 PM
wonderboy1953
Quote:

Originally Posted by Wilmer
Quite a golden oldie: