The 29th question: Given an integer $n \geq 2$, for any pairwise coprime positive integers $a_{1}, a_{2}, \ldots, a_{n}$, let $A=a_{1}+a_{2}+\ldots+$ $a_{n}$. Suppose the greatest common divisor of $A$ and $a_{i} (i=1, 2, \ldots, n)$ is $d_{i}$, and the greatest common divisor of the remaining $n-1$ numbers after removing $a_{i}$ from $a_{1}, a_{2}, \ldots, a_{n}$ is $D_{i}$. Find the minimum value of $\prod_{i=1}^{n} \frac{A-a_{i}}{d_{i} D_{i}}$.
Question 29:
Solution: Consider $D_{1}=\left(a_{2}, a_{3}, \ldots, a_{n}\right), d_{2}=\left(a_{1}+a_{2}+\ldots+a_{n}, a_{2}\right)$, it is clear that $D_{1}\left|a_{2}, d_{2}\right| a_{2}$. Let $\left(D_{1}, d_{2}\right)=d$, then $d\left|a_{2}, d\right| a_{3}, \cdots, d \mid a_{n}$, and $d \mid a_{1}+a_{2}+\ldots+a_{n}$, thus we know $d \mid\left(a_{1}, a_{2}, \ldots, a_{n}\right)$. Noting that $\left(a_{1}, a_{2}, \ldots, a_{n}\right)=1$, hence $d=1$, so $\left(D_{1}, d_{2}\right)=1$. Therefore, $D_{1} d_{2} \leq a_{2}$.

Similarly, we can see that $\mathrm{D}_{2} \mathrm{~d}_{3} \leq \mathrm{a}_{3}, \mathrm{D}_{3} \mathrm{~d}_{4} \leq \mathrm{a}_{4}, \cdots, \mathrm{D}_{\mathrm{n}-1} \mathrm{~d}_{\mathrm{n}} \leq \mathrm{a}_{\mathrm{n}}, \mathrm{D}_{\mathrm{n}} \mathrm{d}_{1} \leq \mathrm{a}_{1}$. Thus, $\Pi_{\mathrm{i}=1}^{\mathrm{n}}\left(\mathrm{d}_{\mathrm{i}} \mathrm{D}_{\mathrm{i}}\right)=$ $\prod_{i=1}^{n}\left(D_{i} d_{i+1}\right) \leq \prod_{i=1}^{n} a_{i+1}=\prod_{i=1}^{n} a_{i}$, where $d_{n+1}=d_{1}, D_{n+1}=D_{1}, \quad a_{n+1}=a_{n}$.

Furthermore, according to the AM-GM inequality, we have $\prod_{i=1}^{n}\left(A-a_{i}\right)=\prod_{i=1}^{n}\left(\sum_{j \neq i} a_{j}\right) \geq \prod_{i=1}^{n}\left[(n-1)\left(\prod_{j \neq 1} a_{j}\right)^{\frac{1}{n-1}}\right]=(n-$ 1) ${ }^{\mathrm{n}} \cdot \prod_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{i}}$. Therefore,
$$
\prod_{i=1}^{n} \frac{A-a_{i}}{d_{i} D_{i}}=\frac{\prod_{i=1}^{n}\left(A-a_{i}\right)}{\prod_{i=1}^{n}\left(d_{i} D_{i}\right)} \geq \frac{(n-1)^{n} \cdot \prod_{i=1}^{n} a_{i}}{\prod_{i=1}^{n} a_{i}}=(n-1)^{n}
$$

On the other hand, when $a_{1}=a_{2}=\ldots=a_{n}=1$, then $\prod_{i=1}^{n} \frac{A-a_{i}}{d_{1} D_{i}}=(n-1)^{n}$.
In conclusion, the minimum value of $\prod_{\mathrm{i}=1}^{\mathrm{n}} \frac{\mathrm{A}-\mathrm{a}_{1}}{\mathrm{~d}_{\mathrm{d}} \mathrm{D}_{\mathrm{i}}}$ is $(\mathrm{n}-1)^{\mathrm{n}}$.