Graph theory : Trees
Suppose $G = (V, E)$ is a tree with 25 vertices such that there exist some constants $c \geq 0$ and $m \geq 2$ such that $\operatorname{deg} v \equiv_m c$ for all $v$. Note that we can equivalently write $\operatorname{deg} v = a_v m + c$ for some $a_v \geq 0$.

Since $G$ is connected by definition, we get that it must have 24 edges in total, giving us
$$ \sum_{v \in V} \operatorname{deg} v = 2 \cdot |E| = 48.$$
We can express this sum using the aforementioned decomposition of the degrees in $G$ to get
$$
\begin{align*}
 \sum_{v \in V} \operatorname{deg} v 
 &amp;= \sum_{v \in V} a_v m + c \\
 &amp;= c |V| + m \sum_{v \in V} a_v \\
 &amp;= 48.
\end{align*}
$$
Due to the both terms in the rewritten sum being non-negative, we get that $c$ must be 1 if we are to find any solutions.
Hence we get that $m$ must divide $48 - c|V| = 23$, which implies $m = 23$ as we required $m \geq 2$. 

At this point, we know that any such $G$ satisfying your conditions does so with $c = 1$ and $m = 23$.
By using our new formula again, we find that the sum of our $a_v$ must also equal 1 and therefore, $G$ must be isomorphic to a tree with 1 internal node and 24 leaves, which is also known as the star graph $S_{24}$.