There are three regular tilings on a plane, equilateral triangles, squares and hexagons, note that six equilateral triangles can be arranged as a regular hexagon. In any case, the number of sides of these polygons, this also being the number of corners, is always 3-smooth.
There are five platonic solids, and I have noticed that the number of edges, faces and corners of a platonic solid is always 5-smooth,* specifically, it is a factor of 5! or twelfty as Wendy calls it.
All three regular planar tilings and all five platonic solids, consist of regular polygons constructible with a compass and straight-edge.
So 5-smooth numbers seem to have some significance in geometry.
*All 3-smooth numbers are also 5-smooth, but some 5-smooth numbers are not 3-smooth.
This has to do with the fact that the symmetry group in 3 dimensions with the largest order is the icosahedral group, which has exactly order 120.
You really canâ€™t expect any primes larger than 5 in the order of a symmetry group, since its associated polyhedron should have a face with a multiple of that number of sides. Three heptagons canâ€™t meet at a vertex in spherical space, so there canâ€™t be a regular polytope of any dimension greater than 2 with heptagons for faces, which would mean that there shouldnâ€™t be any polyhedral symmetry group with 7 as a factor. Only in six dimensions and onward do we see the factor 7 in the order of a point group, but thatâ€™s because the six-dimensional simplex has 7 vertices. The same argument applies for all higher primes: there are no uniform polytope symmetry groups with prime p in their prime factorization from dimension 3 to p âˆ’ 1.
I donâ€™t know enough about this to prove that there canâ€™t be a 3D symmetry group with order, say, 1440, but at least I tried to explain why this sort of numbers have to be 5-smooth.