[Limitedness, monotonicity]

For what value of $K$ is the quantity $A_{\mathrm{k}}=\frac{19^{k}+66^{k}}{k!}$ maximal?
Answer: For $k=65$. Let $B_{\mathrm{k}}=\frac{19^{k}}{k!}, C_{\mathrm{k}}=\frac{66^{k}}{k!}$. Then $A_{\mathrm{k}}=B_{\mathrm{k}}+C_{\mathrm{k}}, \frac{B_{k+1}}{B_{k}}=\frac{19}{k+1}, \frac{C_{k+1}}{C_{k}}=\frac{66}{k+1}$.

Therefore, for $k \leq 19$ both sequences are non-decreasing, and for $k \geq 65$ both sequences are non-increasing, i.e., the maximum value is achieved for some $k \in[19,65]$. Note that for $k \in[19,64]$ the inequalities $\frac{C_{\boldsymbol{h}+1}}{C_{\boldsymbol{k}}} \geq \frac{66}{65}$ and $\frac{B_{\boldsymbol{h}}}{C_{\boldsymbol{k}}}=\left(\frac{19}{66}\right)^{k}0
$$

i.e., for $k \in[19,65]$ the sequence $A_{\mathrm{k}}$ is increasing. Therefore, the value of $A_{\mathrm{k}}$ is maximal for $k=65$.