Maths is still after x and y
X and Y are two different whole numbers greater than 1. Their sum is no greater than 100, and Y is greater than X. S and P are two mathematicians (and consequently perfect logicians); S knows the sum X + Y and P knows the product X * Y. Both S and P know all the information in this paragraph.
The following conversation occurs:
S says "P does not know X and Y."
P says "Now I know X and Y."
S says "Now I also know X and Y."
What are X and Y?
instructions: enter the answer in a format of x0y, where x and y are the values of X and Y respectively and are separated by a 0 which represents a ",". Example X=1,Y=2 can be input as 102. Happy Solving!
The answer is 4013.
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1 solution
First of all, trivially, xy cannot be prime. It also cannot be the square of a prime, for that would imply x = y.
We now deduce as much as possible from each of the logicians' statements. We have only public information: the problem statement, the logicians' statements, and the knowledge that the logicians, being perfect, will always make correct and complete deductions. Each logician has, in addition, one piece of private information: sum or product.
P's statement implies that xy cannot have exactly two distinct proper factors whose sum is less than 100. Call such a pair of factors eligible.
For example, xy cannot be the product of two distinct primes, for then P could deduce the numbers. Likewise, xy cannot be the cube of a prime, such as 33 = 27, for then 3×9 would be a unique factorization; or the fourth power of a prime.
Other combinations are ruled out by the fact that the sum of the two factors must be less than 100. For example, xy cannot be 242 = 2×112, since 11×22 is the unique eligible factorization; 2×121 being ineligible. Similarly for xy = 318 = 2×3×53.
If S was sure that P could not deduce the numbers, then none of the possible summands of x+y can be such that their product has exactly one pair of eligible factors. For example, x+y could not be 51, since summands 17 and 34 produce xy = 578, which would permit P to deduce the numbers.
We can generate a list of values of x+y that are never the sum of precisely two eligible factors.
Eligible sums: 11, 17, 23, 27, 29, 35, 37, 41, 47, 53.
(We can use Goldbach's Conjecture, which states that every even integer greater than 2 can be expressed as the sum of two primes, to deduce that the above list can contain only odd numbers. Although the conjecture remains unproven, it has been verified empirically up to 4×1018.)
P now knows that x+y is one of the values listed above. If this enables P to deduce x and y, then, of the eligible factorizations of xy, there must be precisely one for which the sum of the factors is in the list. The table below, shows all such xy, together with the corresponding x, y, and x+y. The table is sorted by sum and then product.
Note that a product may be absent from the table for one of two reasons. Either none of its eligible factorizations appears in the above list of eligible sums (example: 12 = 2×6 and 3×4; sums 8 and 7), or more than one such factorization appears (example: 30 = 2×15 and 5×6; sums 17 and 11.)
If S can deduce the numbers from the table below, there must be a sum that appears exactly once in the table. Checking the table, we find just one such sum: 17.
Therefore, we are able to deduce that the numbers are x = 4 and y = 13.
