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Both \(k\) and \(f\) must be positive integers. Note that \(\frac{6f}{4+f}\) is an increasing function for positive \(f\text{,}\) bounded above by a horizontal asymptote at \(k=6\text{.}\) Thus the only possible values for \(k\) are 3, 4, and 5. Each of these are possible. To get \(k = 3\text{,}\) we need \(f = 4\) (this is the tetrahedron). For \(k = 4\) we take \(f = 8\) (the octahedron). For \(k = 5\) take \(f = 20\) (the icosahedron). Thus there are exactly three regular polyhedra with triangles for faces.