Is there a polyhedron with an odd number of faces, each of which is a polygon with an odd number of sides?

Let us prove that such a figure does not exist.

To do this, count the number of edges (sides) in this figure.

Let’s denote by S – the number of edges.

Obviously, S must be a natural number.

Consider another value A, which is equal to the sum of the sides on each face.

Then the relation S = A / 2 must be satisfied (since we counted each edge twice).

But the number A is odd because it contains an odd number of odd terms (an odd number of faces that have an odd number of sides).

Hence from the relation S is non-integer, which is impossible.