4. Determine the smallest positive integer $C$ such that $\frac{6^{n}}{n!} \leq C$ for all positive integers $n$.
4. Answer: 65
Define $f(n)=\frac{6^{n}}{n!}$ for $n=1,2,3, \cdots$. It is clear that $f(1)=6, f(2)=18, f(3)=36, f(4)=54$ and $f(5)=64.8$. For all $n \geq 6$,
$$
f(n)=\frac{6^{n}}{n!} \leq \frac{6}{1} \times \frac{6}{2} \times \frac{6}{3} \times \frac{6}{4} \times \frac{6}{5} \times \frac{6}{6}=6 \times 3 \times 2 \times 1.8=64.8
$$