Let $G=(X,U)$ be a digraph with $n$ vertices, outwardly regular of degree $r$. It is known that by removing $5$ edges from $G$ an arborescence is obtained, indicate the number of vertices $G$.
Firstly, we have that $G$ is a digraph with $n$ vertices, outwardly regular, follows that $$\displaystyle\sum_{x\in X} d^-(x)=\sum_{x\in X} d^+(x)=n\cdot r=m,$$ i.e., in $G$ we have $m=nr$ edges. Knowing that removing $5$ edges of $G$ ($\overline{m}=m-5=nr-5$), we obtain an arborescence, which is a tree with a root. It is also known that the number of edges in a tree is $n-1$. Therefore, we have that $nr-5=n-1$, concluding that $n=\dfrac{4}{r-1}$.
I think the fact that arborescence has a root influences the number of vertices, but I don't know how. Can anyone give me some help?
Clearly $n$ must be an integer, so the only possibilities are $r=5$ and $n=1$, $r=3$ and $n=2$, and $r=2$ and $n=4$. Assuming that your definition of digraph does not allow multiple edges with the same souce and target vertices, only the last of these is actually possible, so $n=4$.