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Harvard-MIT Mathematics Tournament

Combinatorics Difficulty 5.6 AIME, harder Prove it United States

Problem:

How many ways are there to label the faces of a regular octahedron with the integers 11–88, using each exactly once, so that any two faces that share an edge have numbers that are relatively prime? Physically realizable rotations are considered indistinguishable, but physically unrealizable reflections are considered different.

Solution

Solution:

Well, instead of labeling the faces of a regular octahedron, we may label the vertices of a cube. Then, as no two even numbers may be adjacent, the even numbers better form a regular tetrahedron, which can be done in 22 ways (because rotations are indistinguishable but reflections are different). Then 33 must be opposite 66, and the remaining numbers – 1,5,71, 5, 7 – may be filled in at will, in 3!=63! = 6 ways. The answer is thus 2×6=122 \times 6 = 12.

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