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

 
using Primes
 
function satisfy1(x::Integer)
    prmslt100 = primes(100)
    for i in 2:(x ÷ 2)
        if i ∈ prmslt100 && x - i ∈ prmslt100
            return false
        end
    end
    return true
end
 
function satisfy2(x::Integer)
    once = false
    for i in 2:isqrt(x)
        if x % i == 0
            j = x ÷ i
            if 2 < j < 100 && satisfy1(i + j)
                if once return false end
                once = true
            end
        end
    end
    return once
end
 
function satisfyboth(x::Integer)
    if !satisfy1(x) return 0 end
    found = 0
    for i in 2:(x ÷ 2)
        if satisfy2(i * (x - i))
            if found > 0 return 0 end
            found = i
        end
    end
    return found
end
 
for i in 2:99
    if (j = satisfyboth(i)) > 0
        println("Solution: ($j, $(i - j))")
    end
end