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How far down the rabbit hole of not knowing can we go? While working on the generalization of I don't know the two numbers... but now I do, I thought of these two extensions. Both have the same conditions, but a different conversation taking place:

Conditions:

Two perfect logicians, Summer and Proctor, are told that integers $x$ and $y$ have been chosen such that $1<x<y$ and $x+y<100$. Summer is given the value $x+y$ and Proctor is given the value $x⋅y$.

Conversation A:

  • Proctor: "I cannot determine the two numbers."
  • Summer: "I knew that."
  • Proctor: "I still cannot determine the two numbers."
  • Summer: "Now I can determine them."
  • Proctor: "Now I can, too."

Conversation B:

  • Proctor: "I cannot determine the two numbers."
  • Summer: "I knew that."
  • Proctor: "I still cannot determine the two numbers."
  • Summer: "Neither can I."
  • Proctor: "Now I can determine them."
  • Summer: "Now I can, too."

We could take this even farther, but based on the solution to the original problem each step will add a significant amount of complexity so I don't want to go too far down the rabbit hole (at least not yet).

Would either of these conversations be possible? If so, what are the values of $x$ and $y$? If there is no solution for $x+y<100$, I would accept a solution that violates that restriction.

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  • $\begingroup$ You need to clarify a bit. In conversation A, when Summer says "Now I can determine them", does he mean "I couldn't before, but NOW I can"? If so, there's no solution. If not, (5, 6) is the solution (a very trivial one however). $\endgroup$ – Taylor Brandstetter Oct 15 '14 at 17:35
  • $\begingroup$ @TaylorBrandstetter Yes, he means that he was only able to figure them out after Proctor's second statement. Go ahead and post an answer explaining why there is no solution for conversation A. Is that true for x+y not having a restriction? $\endgroup$ – Rob Watts Oct 15 '14 at 18:46
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I actually misspoke in my comment earlier. Here's a complete answer:

Stepping through conversation A:

Proctor: "I cannot determine the two numbers."

Condition A: The product is non-unique.

Summer: "I knew that."

Condition B: The sum can only be made from pairs that satisfy condition A (form non-unique products). For example, 11 could be formed from (2,9), (3,8), (4,7), or (5,6), none of which have unique products.

Proctor: "I still cannot determine the two numbers."

Condition C: The product can be made from more than one pair that satisfies condition B. This rules out (2,9), for example: the only other pair that has the same product is (3,6), which has a sum of 9, which can be formed from (2,7), which has a unique product.

Summer: "Now I can determine them."

Condition D: The sum could be formed from more than one pair that satisfies condition A, but only one that satisfies condition C. Only possibility at this point is (5,6).

Proctor: "Now I can, too."

Condition E: The product could be formed from more than one pair that satisfies condition B, but only one that satisfies condition D. (5,6) obviously still works here.

For Conversation B, we have:

Summer: "Neither can I."

Condition D: The sum could be formed from more than one pair that satisfies condition C. 58 possible pairs remain at this point.

Proctor: "Now I can determine them."

Condition E becomes: The product could be formed from more than one pair that satisfies condition B, but only one that satisfies condition D. Only remaining pair at this point is (2, 15).

Summer: "Now I can, too."

Finally Condition F: The sum could be formed from more than one pair that satisfies condition C, but only one that satisfies condition E. (2, 15) still works.

However, add more "neither can I"s than that and you'll be stuck with 57 possible pairs.

It's interesting that these still work. However I don't think you can beat the original puzzle in elegance. In the original, an outside observer only knows the solution AFTER Summer says "Now I can too". But notice that in these two conversations, we knew the answer before the final statements. I wonder how else you could vary the puzzle, besides lengthening it; could be a interesting exercise.

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