Consider this problem:
A connected graph has the same number of vertices of degree 1 and 4. There is 1 vertex of degree 8, and no vertices of any other degree. The graph has size 24. Determine the order of the graph.
My attempt:
Since there's a vertex of degree $8$, there's at least $9$ vertices. Since a graph must have even number of odd vertices, and the number of vertex with degree $1$ is the same as the number of vertex of degree $4$, the size of $G = 9, 13, 17, ...$, hence it must increase $4$ at a time.
Since we already have a vertex of degree $8$, we can eliminate the possibility of $|G| > 25$, since $|G| = 25$ represents the case when the graph is minimally connected.
We have a pool of $\{9, 13, 17, 21, 25\}$. $25$ is impossible since it's when $G$ is minimal, so our pool is $\{9, 13, 17, 21\}$
Struggle:
From this point on I can't think of a better way to do this other than trying to brute force everything. Any help would be greatly appreciated.
A more general arithmetic/approach that solves this types of problems (if it exists) would be even better.
By the Handshaking Lemma, the sum of the vertex degrees is twice the number of edges. Suppose there are $k$ vertices of degree $1$. So:
$$k(1 + 4) + 8 = 24 \cdot 2$$
Thus, $5k = 40$, so $k = 8$. Thus, there are $8$ vertices of degree $1$, $8$ vertices of degree $4$, and $1$ vertex of degree $8$.
So $|G| = 17$.