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

defmodule Puzzle do
  def sum_and_product do
    s1 = for x <- 2..49, y <- x+1..99, x+y<100, do: {x,y}
    s2 = Enum.filter(s1, fn p ->
      Enum.all?(sumEq(s1,p), fn q -> length(mulEq(s1,q)) != 1 end)
    end)
    s3 = Enum.filter(s2, fn p -> only1?(mulEq(s1,p), s2) end)
    Enum.filter(s3, fn p -> only1?(sumEq(s1,p), s3) end) |> IO.inspect 
  end
 
  defp add({x,y}), do: x + y
 
  defp mul({x,y}), do: x * y
 
  defp sumEq(s, p), do: Enum.filter(s, fn q -> add(p) == add(q) end)
 
  defp mulEq(s, p), do: Enum.filter(s, fn q -> mul(p) == mul(q) end)
 
  defp only1?(a, b) do
    MapSet.size(MapSet.intersection(MapSet.new(a), MapSet.new(b))) == 1
  end
end
 
Puzzle.sum_and_product