This is a homework question that has me stumped:
Let $G$ be a directed graph with $\boldsymbol n$ vertices and $\boldsymbol k$ strongly connected components (with $1 < k \leq \left\lfloor\dfrac{n}{3}\right\rfloor$). We also know that each strongly connected component has at least $3$ vertices. What is the minimum number of edges that $G$ may have?
I know that if $G$ is a directed graph with $n$ vertices and $k$ strongly connected components (with $k > 1$), the minimum number of edges that $G$ may have can be calculated with: $\boldsymbol{n - k + 1}$.
My initial hunch is that the general formula needs to be modified to account for the strongly connected component with at least $3$ vertices, but I'm not sure which approach to take and, most importantly, what properties to use.