[byearthithatius]

"Which means we're one of these [2A, A, A]; [3A, 2A, A]; [4A, 3A, A]; [3A, A, 2A]; [4A, A, 3A]; [5A, 2A, 3A]; [8A, 3A, 5A]"

Why? Couldn't it be an infinite number of 3 size arrays comprised of A where two elements sum to the third? [24A, 13A, 11A]? How did we deduce this set of arrays?

EDIT: Solved from another reddit comment. Tuples without a common factor like the one above are considered as a=1.

"They're not eliminated; they correspond to a = 1."

[jhhh]

I think that answer was poorly phrased because those possibilities are eliminated in a sense. There is a better answer further in the thread that explains "If the solution was not one of the flipped triplets, then the first player would not have worked out the solution." Thus if it was one of your other infinite triplets (eg. 65, 12, 53) then round 2 player 1 would've still answered 'I don't know'. Since they did respond with a definitive answer it had to be one of the formula solutions, since those were the only solutions they could prove. And since the only formula with a factor in 65 is 5 the correct formula must be [5A, 2A, 3A] and thus [65, 26, 39].

You should be able to generate an infinite number of these problems just by multiplying the first formula factor by a prime number. Like the same question but the person answers '52' restricts you to either [4a, 3a, a] or [4a, a, 3a]. Since the question only asks for the product of all the terms the answer is 4 * 13 + 3 * 13 + 13 = 104.

[WithinReason]

Look at it this way: Person 1 sees the numbers 26 and 39, and has to guess his own number. It must be one of only 2 possibilities: 13 or 65. All he has to do is eliminate one of those possibilities.