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Puzzle I: Uisng standard set theory $\mathsf{ZFC}$ (i.e., the Zermelo-Frenkel axioms, including the Asiom of Choice), find a set $x_0$ such that $\phi(x_0)$ holds, where $$\phi(x)\equiv (\mathsf{CH}\leftrightarrow x=\emptyset). $$ Here, $\mathsf{CH}$ is short for the continuum hypothesis. You cannot solve this puzzle, i.e., you cannot exhibit any concrete ...


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I'm interested in the community's take on this trivial example using Tapa rules: 2 3 2 _ _ _ _ _ _ _ _ _ 2 3 2 Basic rules force the second and fourth rows to be all shaded. For either side of the middle row, one could argue that shading that square leads to multiple solutions, but you cannot make that argument for the middle square. So there is a ...


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Yes. I have a sphere with a hole drilled through the centre. The length of this hole is 1 metre. What is the remaining volume of the sphere. ie. the 'Donut' If you decide I'm honest and not sending you on a wild goose chase, then you can solve this without calculus in your head.


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Whereas the question has already been answered, here is a corroboration on answering it with "yes", by providing a minimalistic mathematical example. Many non-linear relations contain one or multiple points with unique characteristics or singularities, which you can exploit in such a riddle. Consider for example a parabola $y=x^2$. Find me the ...


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Find any integer $n$ such that $n$ is the number of solutions to this puzzle.


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Is it possible to construct a puzzle that is: Solvable if you assume there is a single unique solution Not solvable if you do not make this assumption How about this one: Given a function on the natural numbers, $F: \mathbb N \Rightarrow \{0,1\}$, such that the probability that $F(n)=1$ is $2^{-n}$, find all $n$ such that $F(n)=1$. If you assume there is ...


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I know this question has already been answered (and in a better way than I could ever have!) but I feel like it is worth bringing up the generic category of puzzle "Knowledge Puzzle" or "Induction Puzzle", which kind of fits the original description. (This category includes problems like the "coloured hats" puzzle group, and ...


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No. If there is exactly one way to fill out the grid in a valid way, then that can be found without using any logic based on uniqueness. (It may be very painful, and require brute-forcing a lot, but it won't be impossible to find.) If there are exactly two valid ways to fill out the grid, then uniqueness logic cannot help you. You can never make a deduction ...


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At least not if the solution space is a countable infinite set, or smaller. In that case, we can compute the finite time person 2) will need to find the solution found by person 1) by enumeration, making the strategy of 2) a computable function and therefore violating the constraints of the question. Response to edit: The "does not have to be perfect-...


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