How to Solve the Sum and Product Puzzle

There are two unique integers x and y. Both of them are greater than 1 and their sum is less than or equal to 100. Mathematician P knows their product p and mathematician S knows the sum s. The following dialogue takes place:

P: I don't know what x and y are.
S: I knew you wouldn't know.
P: But now I do know.
S: And so do I!

Here's how to solve the sum and product puzzle.

Difficulty:

Challenging

Instructions

1. Determine the possible values of s. Because x and y are unique and both greater than 1, we know that s is at least 5 (2+3=5). From P's first statement we know that x and y cannot both be prime.

2. Eliminate all integers between 5 and 100 inclusive that are not the sum of unique prime numbers. Eliminate 6 because 2 and 4 are the only integers greater than 1 which will sum to 6. The product of 2 and 4 is 8, which has unique factors of 2 and 4. Further eliminate all integers greater than the smallest prime over 50. This gives us the following set S for values for s: 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53.

3. Observe that P's second statement shows that p has one and only one (x, y) pair such that x+y is an element of S. Because all members of S are odd, this means that one member of (x, y) is even and the other is odd. Therefore, p=q(2^k) where q is prime and k>1. S's second statement tells us that s has one and only one (x, y) pair that has a good value for p.

4. Show that s=17 implies p=52. We will eliminate all (x, y) pairs for s=17 where p is not equal to 52:

   2 x 15 = 30 = 5 x 6; 5+6=11
   3 x 14 = 42 = 2 x 21; 2+21=23
   5 x 12 = 60 = 3 x 20; 3+20=23
   6 x 11 = 66 = 2 x 33; 2+33=35
   7 x 10 = 70 = 2 x 35; 2+35=37
   8 x 9 = 72 = 3 x 24; 3+24=27

   Note the sums 11, 23, 27, 35 and 37 are elements of S.

5. Establish that all elements of S except 17 have at least two good products.

   11=4+7=8+3
   23=4+19=16+7
   27=4+23=8+19
   29=16+13
   35=4+31=16+19
   37=8+29=32+5
   41=4+37
   47=4+43=16+31
   51=4+47=8+43
   53=16+37

   Establish a second good product for 29, 41, 53, 59, 65, 89, 97.

   29=2+27; 2x27=54=3x18=6x9; 3+18=21, 6+9=15; 15, 21 not elements of S
   41=7+34; 7x34=288=14x17; 14+17=31; 31 not an element of S
   53=10+43; 10x43=430=5x86; 5+86=91; 91 not an element of S

   Because all values of S have more than one good value for p except 17 and since the good product of 17 is 52, then the solution is (x, y)=4,13.

Resources

• Read a rigorous treatment of this problem.