Task[edit]
Solve the "Impossible Puzzle":


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?


Guidance
It can be hard to wrap one's head around what the three lines of dialog between S (the "sum guy") and P (the "product guy") convey about the values of X and Y.

So for your convenience, here's a break-down:





 Quote

 Implied fact


 1)

 S says "P does not know X and Y."

 For every possible sum decomposition of the number X+Y, the product has in turn more than one product decomposition.


 2)

 P says "Now I know X and Y."

 The number X*Y has only one product decomposition for which fact 1 is true.


 3)

 S says "Now I also know X and Y."

 The number X+Y has only one sum decomposition for which fact 2 is true.

Terminology:

 "sum decomposition" of a number = Any pair of positive integers (A, B) so that A+B equals the number. Here, with the additional constraint 2 ≤ A < B.
 "product decomposition" of a number = Any pair of positive integers (A, B) so that A*B equals the number. Here, with the additional constraint 2 ≤ A < B.


Your program can solve the puzzle by considering all possible pairs (X, Y) in the range 2 ≤ X < Y ≤ 98, and then successively eliminating candidates based on the three facts. It turns out only one solution remains!

See the Python example for an implementation that uses this approach with a few optimizations.

See also
   Wikipedia:   Sum and Product Puzzle

object ImpossiblePuzzle extends App {
  type XY = (Int, Int)
  val step0 = for {
    x <- 1 to 100
    y <- 1 to 100
    if 1 < x && x < y && x + y < 100
  } yield (x, y)
 
  def sum(xy: XY) = xy._1 + xy._2
  def prod(xy: XY) = xy._1 * xy._2
  def sumEq(xy: XY) = step0 filter { sum(_) == sum(xy) }
  def prodEq(xy: XY) = step0 filter { prod(_) == prod(xy) }
 
  val step2 = step0 filter { sumEq(_) forall { prodEq(_).size != 1 }}
  val step3 = step2 filter { prodEq(_).intersect(step2).size == 1 }
  val step4 = step3 filter { sumEq(_).intersect(step3).size == 1 }
  println(step4)
}