4. Given a set of identical regular pentagons, with natural numbers from 1 to 5 written at each vertex, as shown in the figure. The pentagons can be rotated and flipped. They were stacked (vertex to vertex), and it turned out that the sums of the numbers at each of the five vertices are the same. How many pentagons could there be in this stack?

![](https://cdn.mathpix.com/cropped/2024_05_06_903bafdb6e65c3ac820dg-1.jpg?height=208&width=229&top_left_y=1986&top_left_x=959)
Solution. Stack two pentagons, flipping one of them, and align their numbers as follows: $1-5, 2-4, 3-3, 4-2$, and 5-1. Then we get a stack that meets the condition. Let's call it a "pair." Stack five pentagons, rotating them relative to each other so that the numbers align as 1-2-3-4-5, 2-3-4-5-1, and so on. Then we get a stack that also meets the condition. Let's call it a "quintet." Any even number of pentagons in a stack can be obtained by sequentially stacking "pairs." Any odd number of pentagons, greater than 3, can be obtained from one "quintet" and the necessary number of "pairs." Note that a stack of one pentagon does not meet the condition. We will prove that it is impossible to form a stack that meets the condition from three pentagons. In each pentagon, the numbers written at adjacent vertices differ by either 1 or 4. Suppose that in a stack of three pentagons, the sums of the numbers written at two adjacent vertices are equal, that is, $x_{1}+x_{2}+x_{3}=\left(x_{1} \pm a_{1}\right)+\left(x_{2} \pm a_{2}\right)+\left(x_{3} \pm a_{3}\right)$, where on the left is the sum of the numbers at one vertex, and on the right is the sum of the numbers at the adjacent vertex ($a_{i}$ can only be 1 or 4). By enumeration, we can verify that this equality cannot be true (the enumeration can be shortened by using the idea of parity).

Answer: any natural number except 1 and 3.