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Puzzling

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    $\begingroup$ I think you can improve the fraction of remaining value per question to $1/\phi$ (where $\phi$ is the golden ratio) by setting the cutoff point that fraction away from the upper end of the range. That way either one question shrinks the range to $1/ \phi$ its previous length, or two questions shrink to $1-1/\phi$, which equals $(1/ \phi)^2$ and so is like you shrank by a ratio of $1/ \phi$ twice in succession. This lets you improve to about $\log_ \phi (10^{300}) \approx 1435$ questions. $\endgroup$
    – xnor
    Commented Jul 22, 2023 at 7:13
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    $\begingroup$ I think a significant flaw is saying that $10^{300}$ possibilities for Bob and $10^{300}$ for Charlie make $2*10^{300}$ total possibilities. I believe it makes $10^{600}$, and that this changes all your math. $\endgroup$
    – David G.
    Commented Jul 22, 2023 at 14:56
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    $\begingroup$ @xnor : the same conclusion (1436 rounded up) if using graphic representation using square of $10^{300} * 10^{300}$ possibilities when pivot point is chosen so that "no" and "yes" reduces the solution equally. Of course this is equivalent to your and Tim C's explanation. $\endgroup$
    – z100
    Commented Jul 22, 2023 at 19:10
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    $\begingroup$ @DavidG., the possibilities mentioned in the answer is not about what both numbers are, but about the final answer: what the maximum number is (100^300 possibilities), and from whom (2 possibilities: aB and aC). So if the number range is 1-4, the possibilities are 1B, 2B, 3B, 4B, 1C, 2C, 3C, 4C, 4x2=8 possibilities, and not referring to the possibilities of their number combinations (1-1, 1-2, 1-3, 1-4, 2-1, 2-2, 2-3, 2-4, 3-1, 3-2, 3-3, 3-4, 4-1, 4-2, 4-3, 4-4, a total of 4x4 combinations) $\endgroup$
    – justhalf
    Commented Jul 23, 2023 at 0:12
  • $\begingroup$ This is the same idea I came up with. But still don't see how it can be extended to solve the puzzle. Certainly a nice improvement. Thanks for the write up. $\endgroup$
    – Dmitry Kamenetsky
    Commented Jul 23, 2023 at 0:13