Why are polyhedra related to the prime numbers 2, 3 and 5, but not to the prime number 7? Just take a quick glance at all the numbers in these Wikipedia pages on polyhedra:


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*http://en.wikipedia.org/wiki/Platonic_solid

*http://en.wikipedia.org/wiki/Archimedean_solid

*http://en.wikipedia.org/wiki/Kepler-Poinsot_polyhedra

*http://en.wikipedia.org/wiki/Catalan_solids
One thing is very curious: there are a lot of numbers divisible by 2, by 3, and by 5, but there are almost none divisible by 7 (or any prime number greater than 7).
Therefore it seems that 2, 3 and 5 are particularly important in geometry. While 7 is not.
Is this really true? And if yes, why are 2, 3 and 5 very common, but not 7?
 A: As a comment notes, regular $n$-gons start to get too wide-angled to build into these sorts of highly-symmetric polyhedra. The internal angles of a regular $n$-gon are $\frac{n-2}{n}\pi$; if you have, say, two of these at each vertex, that fills up $\frac{2n-4}{n}\pi$. Since the sum of all the angles at a vertex must be less than $2\pi$, the remaining faces at a vertex must have angles less than $\frac 4 n \pi$. The smallest angle of a regular polygon is $\frac 1 3 \pi$, for a triangle, so $\frac 4 n > \frac 1 3$ or $12 > n$.
An alternative is to have just one $n$-gon at each vertex, and this is always possible: you can have one $n$-gon and two triangles, which will give you a $n$-gonal pyramid, which isn't symmetric enough for these categories; 
or an $n$-gon and three triangles, which gives you an $n$-gonal antiprism;
or an $n$-gon and two squares, which gives you an $n$-gonal prism.
These prisms and antiprisms are vertex-transitive and have regular faces, so they'd qualify as Archimedean solids, except that they're excluded by convention for no real reason.
So here's one answer:
Highly symmetric ("Archimedean-type") polyhedra with $p$-fold symmetry actually do occur for every $p$.
But if we decide, for whatever reason, that prisms and antiprisms are second-class citizens, why don't we have anything with 7-gons or 11-gons?
Well, suppose a vertex is in 2 heptagons and a triangle. These polyhedra are vertex-transitive, so every vertex is the same. (Well, the Catalan solids aren't, but they are dual to the Archimedean solids, which are.) 
One of the three edges at the vertex is between two heptagons, and the other two edges are between a heptagon and a triangle.
So consider one particular heptagon face at the vertex.
Across one edge is another heptagon.
Across the next edge is a triangle.
The vertex at the end of this edge is also incident to one triangle and two heptagons; so across the next edge is a heptagon again.
In this way, the edges must alternate between heptagon-heptagon and heptagon-triangle.
But you can't alternate between two types of edges on a 7-gon!
The exact same argument shows that you can't have any polyhedron where each vertex is incident to two $n$-gons and one $r$-gon where $n$ is odd and $r \neq n$.
Thus you can't have two or more $n$-gons at each vertex of a regular-faced polyhedron if $n =7$, $n = 11$, or $n \geq 12$.
I haven't dealt with all the cases where each vertex is incident with exactly one $n$-gon.  If you think about the above argument, though, it also shows you can't have each vertex incident to one $n$-gon, one $b$-gon, and one $c$-gon, where 
$n$ is odd and $b$, $c$, and $n$ are all different.
So the only possibility really remaining is having each vertex incident to one $n$-gon ($n \in \{7,11\}$), two triangles, and one square. But we can discount this by a similar sort of argument (draw pictures and consider what happens around a triangle.)
Also, you can't have any regular-faced polyhedron with a 7-valent vertex,
since the smallest angle you can get is on a triangle, and $7 \cdot \frac 1 3 \pi > 2 \pi$.
So, all this kind of stuff shows you why you don't have $n$-gon faces (or $n$-valent vertices) appearing in vertex-transitive regular-faced polyhedra for big $n$, except for prisms and antiprisms. A somewhat "deeper" answer is this:
The symmetry group of any polyhedron is a finite Coxeter group.
These are all the finite Coxeter groups.
The labels on the edges indicate the orders of rotations in the symmetry group; blank edges are order 3. Notice that the only numbers that appear are 3, 4, 5, and 6. (And $n$, but only on the diagram $I_n$, which is—you guessed it—the symmetry group of those ugly-stepchild prisms and antiprisms.)
In any highly symmetric polyhedron, all those numbers that you see in their descriptions will probably be reflected in the order of some symmetry. And the primes you mention are the only ones that show up in finite Coxeter groups; any other numbers force the group to be infinite.
A: You may be slightly misled by your focus on prime numbers. It is not so much that 2,3 and 5 are important, 3,4 and 5 are (and 6), at least for polyhedra with certain kinds of regularity like those you link to.
In every solid bounded by planar polygonal faces, at each corner three or more faces meet. How many more? Even with regular triangles, if six of them meet in one point, they must in the same plane (6*60° is a full angle), so the point is not corner of a 3d solid.
And which kind of regular polygons can form a 3d solid? Three regular hexagons meeting at a common point must again be in the same plane (full at that point, 3*120°), so the point cannot be a corner of a 3d solid.
One can construct solids with regular hexagons sharing edges with n-gons for n<6, and the Archimedean solid list you link even includes some with octagons.
Another interesting thing about the Archimedean solids named "truncated somethings" (but also others, possibly all) is that one could produce them by taking one of the five regular polyhedra and cutting of all corners in a regular way, possibly multiple times. And since there are only five regular polyhedra, with corners where 3, 4 or 5 faces meet, this cutting will only produce triangles, quadrangles or pentagons, but nothing with heretofore not present prime numbers.
Of course, there are prisms and pyramids containing two or one n-gon, resp., for any n, also 7, 11, 13..., but these are not as regular as the solids you were looking at.
