Question 205, For a composite number $n$, let $f(n)$ be the sum of its smallest three positive divisors, and $g(n)$ be the sum of its largest two positive divisors. Find all composite numbers $\mathrm{n}$ such that $g(n)$ equals some positive integer power of $\mathrm{f}(\mathrm{n})$.
Question 205, Answer: If $n$ is odd, then all divisors of $n$ are odd, making $f(n)$ odd and $g(n)$ even. In this case, $g(n)$ cannot equal any positive integer power of $f(n)$. Therefore, $n$ must be even, let $n=2m$.

Let the smallest divisor of $n$ other than 1 and 2 be $d$. Then $f(n)=1+2+d=3+d$, and $g(n)=n+\frac{n}{2}=\frac{3n}{2}=3m$. Since $3m$ can be expressed as a positive integer power of $3+d$, it follows that $3|(3+d) \Rightarrow 3|d$. Considering the minimality of $d$, we have $d=3$. Thus, there exists a positive integer $\alpha$ such that $3m=6^{\alpha}$, i.e., $m=2^{\alpha} \cdot 3^{\alpha-1}$, and hence $n=2^{\alpha+1} \cdot 3^{\alpha-1}$. Noting that $3 \mid n$, we have $\alpha \geq 2$.
On the other hand, when $n=2^{\beta+2} \cdot 3^{\beta}\left(\beta \in \mathbb{N}^{+}\right)$, it is verified to satisfy the conditions.
In summary, all $n$ that satisfy the conditions are $n=2^{\beta+2} \cdot 3^{\beta}\left(\beta \in \mathbb{N}^{+}\right)$.