Credited with solving this puzzle:
A three-dimensional cube has eight vertices, twelve edges, and six faces. Let’s call them 0-D, 1-D, and 2-D faces, respectively.
Denote f(d,k) as the number of k-dimensional faces of an d-dimensional hyper cube, so f(3,1)=12.
Find three cubes (with different dimensions d1, d2, and d3) such that the number of k1, k2, and k3 dimension faces are the same, i.e. f(d1,k1) = f(d2,k2) = f(d3,k3).
We are looking for nontrivial solutions, so k1 should be less then d1.
Bonus: Find more than three.
