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Let G be a connected 5-regular embedded planar graph, in which every face has the same degree. Find the number of faces of G.

So let there be $e$ edges, $f$ faces with each degree $d$ and $v$ vertices each of degree $5$ so using the handshaking lemma for vertices and faces we get equations:

$f \cdot d = 2e = 5v$

Euler's formula states $v + f - e = 2$

$$\implies fd/5 + f - fd/2 = 2$$

$$\implies 10f - 3df = 20$$

Also euler's $e \le 3v - 6$

But now I am stuck, any ideas?

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You got $(10-3d)f=20$. What are possible values for $d$? In the end you will obtain a graph which should be familiar to you.

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  • $\begingroup$ I am not sure how to deduce $d$ in this case. Do you mean that we should be checking for which integer values $f$ and $(10 - 3d)$ multiply to produce $20$? In this case $d$ can be $3$ or $2$. $\endgroup$ – asdfghjk Apr 5 '18 at 23:38

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