ChatGPT 4.5:

Let’s carefully analyze this puzzle step-by-step:

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Step 1: Define the Problem Clearly

We have three people in a circle. Let’s call them:
• Person A (first person)
• Person B (second person)
• Person C (third person)

Each has a positive integer above their head. Each sees the other two numbers but not their own. The numbers satisfy:
A + B = C \quad \text{or} \quad B + C = A \quad \text{or} \quad C + A = B

Given the puzzle scenario, we need to find out the exact numbers given the clues, and ultimately find the product of the three numbers.

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Step 2: Logic of the Puzzle (Step-by-Step Reasoning)

Initial conditions:
The numbers are positive integers. The puzzle implies a situation of ambiguity at first, but after multiple rounds of “I don’t know,” Person A finally knows their number is 65.

Let’s denote clearly the conditions again:
• Exactly one of the following three equations is true:
• A + B = C, or
• B + C = A, or
• C + A = B.

Person A sees B and C, but not A. Similarly for Persons B and C.

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Step 3: Logical deductions from the repeated “I don’t know”

First round deductions:
• Person A initially doesn’t know their number. This means, from Person A’s perspective, both scenarios A = B + C and A + B = C (or A + C = B) are still possible. If Person A had seen identical numbers, for instance, Person A could have instantly known their number. Thus, Person A sees two different numbers.
• Similarly, Person B doesn’t know either, implying that Person B also sees ambiguity in their own number. Thus, Person B also sees two numbers that could result in at least two possible solutions.
• Person C also doesn’t know, same logic as above.

Thus, after one full round, we understand that initially, all three numbers must create ambiguity. No trivial case (like two identical numbers) is present.

Second round deductions:

After hearing that no one knows their number, each person gains information: they now know the scenario is one in which none of the other two people could have immediately solved it.

Crucially, in the second round:
• Person A now knows their number is 65 after considering the previous statements. This implies that Person A sees numbers B and C, and from these numbers plus the statements of uncertainty, Person A deduces that their own number can only be 65.

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Step 4: Deriving the Numbers (Critical deduction)

Let’s denote clearly again: Person A’s number is now known to be 65.

We have three possible equations. Let’s test each carefully given A = 65:
1. Case 1: A + B = C
Then we have:
• 65 + B = C. Person A sees numbers B and C, so they would see B and 65 + B. Person A sees numbers B and 65+B. If Person A sees these two numbers, they might initially consider their number could be either the sum or difference of these two.
• However, if Person A sees two numbers, and one is exactly 2× the other, Person A could immediately guess their number is the difference (since if you see numbers X and 2X, your number is clearly X). So this scenario of B and 65+B is okay only if these two numbers don’t exhibit an immediate trivial scenario.
• Let’s hold this scenario and continue examining others first.
2. Case 2: B + C = A (So, B + C = 65)
• Then Person A sees numbers B and C, adding exactly to 65.
• Initially, Person A sees two numbers summing exactly to 65. Could Person A know their number immediately at first glance? No, since their number could either be the sum (65) or the difference of the two numbers. Ambiguity here initially makes sense.
• Initially, Person B and Person C also see ambiguous scenarios. They both don’t know their numbers either at first round. This scenario is plausible.
3. Case 3: C + A = B (So, C + 65 = B)
• Then Person A sees numbers B = C + 65 and C.
• Person A sees C and C+65. Similar reasoning as Case 1.

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Step 5: Identifying the Correct Scenario

The puzzle’s logic typically aligns best with Case 2: B + C = A = 65.

Here’s why clearly stated:
• Initially, Person A sees two numbers adding to exactly 65. At first glance, Person A’s number could either be 65 (sum) or the difference between the two numbers.
• Person B also initially sees two numbers and cannot distinguish immediately their own number, nor can Person C.
• After everyone states “I don’t know,” Person A gains critical insight: if the difference between the two numbers (that A sees) allowed Person B or Person C to know their number immediately, they’d have known right away. The fact that B and C didn’t know means the difference is not immediately obvious to B or C. Thus, the difference cannot be trivial (such as identical numbers or easy multiples).

Therefore, the two numbers must be such that their difference doesn’t immediately clue in the other persons. This occurs most clearly with pairs of numbers which have multiple factorization possibilities or aren’t trivial multiples.

But we must carefully test quickly to find a suitable, plausible integer pair whose sum is 65 and whose ambiguity would logically survive through multiple rounds. The puzzle is a classic logic puzzle with a known structure:

A common scenario in these classic puzzles is the use of Fibonacci-like or similar “golden” sets. Let’s quickly check the following well-known ambiguous triple:
• Numbers: 65, 39, 26
• Check quickly: indeed, 39 + 26 = 65.

Checking logic quickly:
• Person A sees 39 and 26. Initially ambiguous (number could be 65 or 13).
• Person B sees 65 and 26. Could be either 39 or 91 at first glance. Initially ambiguous.
• Person C sees 65 and 39. Could be either 26 or 104 initially. Initially ambiguous.

After each person says they don’t know:
• Now, Person A considers: If my number was 13, then Person B would see numbers 13 and 26. Person B would then immediately know their number is 39 (13+26=39), no ambiguity. But Person B didn’t know! So Person A’s number cannot be 13. Hence, Person A confidently concludes their number is 65.

Thus, the numbers 65, 39, and 26 satisfy the puzzle conditions perfectly!

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Step 6: Final Answer (Product of the Numbers)

The three numbers identified clearly are:
• A = 65
• B = 39
• C = 26

Then, the product is:
65 \times 39 \times 26
= 65 \times 1014
= 65,910

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Final Answer:

\boxed{65,910}