Euler has been mentioned a few times in this thread regarding the use of hexagons to cover a sphere, so let me go into more detail. All polyhedra have an
Euler characteristic, which is a constant that relates the number of vertices, edges, and faces. The Euler characteristic is represented with the Greek letter chi, but it looks like an x so I'll just use that. This is the Euler formula:
x = V - E + F
It has been proven that all convex polyhedra (that is, polyhedra without any "dents" or "spikes", and therefore suitable for representing a sphere) have an Euler characteristic of 2.
Here is a list of proofs.
Now, let's assume for the moment that you can tessellate, or tile, a sphere with only hexagons. Each edge is shared between two hexes, and each vertex is shared between three. Therefore we have this set of equations:
E = 6F/2 = 3F
V = 6F/3 = 2F
Substituting back into the Euler formula gives us this:
x = 2F - 3F + F
x = 0
Therefore the Euler characteristic of a hexagon tiling is 0. However, it has been proven that all convex polyhedra have an Euler characteristic of 2. Therefore you cannot tile a sphere with only hexagons.
This site has a proof that if you use pentagons and hexagons, there will always be 12 pentagons.
Now, that doesn't mean you
have to use pentagons. You could use some really stupid shape instead. If you take the standard Civ map and put hexes on it, you can think of it as a polyhedron with however many hexes and two whatever-gons at the poles. Pentagons make for a more uniform grid, but then you're back to the problem of representing the grid on a plane, which I discussed earlier.