Icosahedron
I’ve been staring at my icosahedron, trying to solve the following puzzle by Konstantin Knop.
Puzzle. One face of an icosahedron is special. The numbers 2, 3, and 5 are written at its vertices in some order. All other vertices of the icosahedron are labeled with 1. In one query, we may ask an oracle for the product of the numbers assigned to any subset of icosahedron’s vertices. What is the minimum number of queries needed to determine the special face?
However, I misread the problem. I ended up solving a different puzzle instead—and had quite a bit of fun doing it.
Puzzle. One face of an icosahedron is special. The numbers 2, 3, and 5 are written at its vertices in some order. All other vertices of the icosahedron are labeled with 1. In one query, we may ask an oracle for the product of the numbers assigned to vertices of any one face of the icosahedron. What is the minimum number of queries needed to determine the special face?
I won’t post the solutions just yet, but let me begin with a simple observation: one question cannot possibly be enough. Indeed, with one question, the oracle’s answer must be a divisor of 30, and 30 has only 8 positive divisors. But an icosahedron has 20 faces, so a single question cannot distinguish among all possible choices for the special face.
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