Finding the number of edges that connect to a single vertex in a dodecahedron

Please note my geometry background is very weak (high school geometry is all I have), so I would appreciate it if someone could explain it in very layman terms how to do this.

I am trying to solve the following problem

Prove that it is not possible to assign the integers $1,2,3,\ldots,20$ to the 20 vertices of a regular dodecahedron so that the five numbers at the vertices of each of the 12 pentagonal faces have the same sum.

I recognize how to solve this, but to do this I need to know how many times a single vertex is counted when adding all the faces up i.e. the number of edges that connect to a single vertex.

I also remember reading a side note somewhere in my Geometry course about something that looked like this: $v+e-f=2$ or something like that - can anybody help me figure out what this was and if I am remembering it right?

Source: Art of Problem Solving Volume 2, Ch.13 Equations and Expressions

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For a regular solid there must be the same number of faces, which is the same as the number of edges, meeting at each vertex. If you have $N$ faces of $S$ sides each, there are $NS/2$ total edges as each edge is counted twice. If there are $n$ faces at each vertex, there are $NS/n$ vertices. So the Euler formula becomes $NS/n-NS/2+N=2$ (note the sign changes from the way you quoted it). For a dodecagon, there are 12 faces of 5 sides each, so this becomes $60/n-30+12=2$ which shows $n=3$. There are three edges and three faces at each vertex.

Another way to see it is that for a convex polyhedron, the sum of the angles at a vertex must be less than $360^{\circ}$. As the angle of a pentagon is $108^{\circ}$, you can only have three at a vertex.

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It is high time that you provide yourself with a picture of a dodecahedron. You will then see that exactly three faces (and three edges) meet at each vertex.

Assume that the described allocation problem can be solved with a certain sum $s$ per face. Then $12 s$ would count every number between $1$ and $20$ exactly three times.

Euler showed that $v-e+f=2$, where $v$, $e$, and $f$ are respectively the number of vertices, edges, and faces. See this article.