The eighth question, given a simple connected graph $\mathrm{G}(\mathrm{V}, \mathrm{E})$ of order 101, where each edge is contained in a cycle of length at most $\mathrm{c}(\mathrm{G})$, find the minimum possible value of $c(G)+|E(G)|$.
The eighth question, solution: We recursively define cycles $C_{1}, C_{2}, \ldots, C_{m}$: First, arbitrarily choose an edge $e_{1}$, and take a smallest cycle $C_{1}$ containing $e_{1}$; assuming $C_{1}, C_{2}, \ldots, C_{i-1}$ have been chosen, arbitrarily choose an edge $e_{i}$ not contained in any of the cycles $C_{1}, C_{2}, \ldots, C_{i-1}$ (if no such edge $e_{i}$ can be found, then terminate), and then take a smallest cycle $C_{i}$ containing $e_{i}$; suppose $C_{1}, C_{2}, \ldots, C_{m}$ have been chosen, and the process cannot continue further.

Now, sequentially delete $e_{m}, e_{m-1}, \cdots, e_{1}$ from $G$, this process will not disrupt the connectivity of $G$, so the number of edges remaining is at least $|\mathrm{V}|-1=100$. This indicates:
$|E(G)|-m \geq 100$
Furthermore, the cycles $C_{1}, C_{2}, \ldots, C_{m}$ include all the edges of $G$, and $|C_{i}| \leq c(G) (1 \leq i \leq m)$. Hence, $m \cdot c(G) \geq \sum_{i=1}^{m}|C_{i}| \geq |E(G)|$
Combining (1) and (2):
$$
c(G)+|E(G)| \geq \frac{|E(G)|}{m}+|E(G)| \geq \frac{100+m}{m}+(100+m)=101+\left(\frac{100}{m}+m\right) \geq 121
$$

On the other hand, if $G$ is 10 cycles of length 11 sharing a common vertex, then $|E(G)|=110$, and $c(G)=11$, so $c(G)+|E(G)|=121$.
In conclusion, the minimum value of $c(G)+|E(G)|$ is 121.