China Southeastern Mathematical Olympiad

By $2x \mid A$, note that $A = 4 \cdot p_2^{p_2} \cdots p_n^{p_n}$. We may suppose that $x = 2^{\alpha_1} p_2^{\alpha_2} \cdots p_n^{\alpha_n}$, where $0 \le \alpha_1 \le 1$, $0 \le \alpha_i \le p_i$ ($i = 2, 3, \dots, n$). Then, we have
$$\frac{A}{x} = 2^{2-\alpha_1} p_2^{p_2-\alpha_2} \cdots p_n^{p_n-\alpha_n}.$$
Hence, the number of different divisors of $\frac{A}{x}$ is
$$(3 - \alpha_1)(p_2 - \alpha_2 + 1)\cdots(p_n - \alpha_n + 1).$$
We know that
$$(3 - \alpha_1)(p_2 - \alpha_2 + 1)\cdots(p_n - \alpha_n + 1) = x = 2^{\alpha_1} p_2^{\alpha_2} \cdots p_n^{\alpha_n}. \qquad \textcircled{1}$$
By induction on $n$, we shall prove that the array $(\alpha_1, \alpha_2, \dots, \alpha_n)$ satisfying $\textcircled{1}$ is $(1, 1, \dots, 1)$ ($n \ge 2$).

(1) If $n=2$, then $\textcircled{1}$ becomes $(3 - \alpha_1)(4 - \alpha_2) = 2^{\alpha_1} 3^{\alpha_2}$, where $\alpha_1 \in \{0, 1\}$. If $\alpha_1 = 0$, then $3(4 - \alpha_2) = 3^{\alpha_2}$ which has no integer solution $\alpha_2$. If $\alpha_1 = 1$, then $2(4 - \alpha_2) = 2 \cdot 3^{\alpha_2}$. We have $\alpha_2 = 1$. Thus, $(\alpha_1, \alpha_2) = (1, 1)$. That is, the conclusion is true for $n=2$.

(2) Suppose that the conclusion is true for $n = k - 1$ ($k \ge 3$), then when $n = k$, $\textcircled{1}$ becomes
$$(3 - \alpha_1)(p_2 - \alpha_2 + 1)\cdots(p_{k-1} - \alpha_{k-1} + 1)(p_k - \alpha_k + 1) \qquad \textcircled{2} \\ = 2^{\alpha_1} p_2^{\alpha_2} \cdots p_{k-1}^{\alpha_{k-1}} p_k^{\alpha_k}.$$
If $\alpha_k \ge 2$, considering
$$0 < p_k - \alpha_k + 1 < p_k,$$
$$0 < p_i - \alpha_i + 1 \le p_i + 1 < p_k \quad (1 \le i \le k - 1),$$
we see that the left-hand side of $\textcircled{2}$ cannot be divided by $p_k$, but the right-hand side of $\textcircled{2}$ is a multiple of $p_k$, which is a contradiction.

If $\alpha_k = 0$, then $\textcircled{2}$ becomes
$$(3 - \alpha_1)(p_2 - \alpha_2 + 1)\cdots(p_{k-1} - \alpha_{k-1} + 1)(p_k + 1) \quad \textcircled{3} \\ = 2^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}.$$
Note that $p_2, p_3, \dots, p_k$ are odd primes, thus, on the one hand, $p_k + 1$ is even. So the left-hand side of $\textcircled{3}$ is even. On the other hand, the right side of $\textcircled{3}$ is odd. So $\alpha_1 = 1$. But then $3 - \alpha_1 = 2$, so the left-hand side of $\textcircled{3}$ is a multiple of $4$, but the right-hand side of $\textcircled{3}$ is not, which is a contradiction.

By the above argument, we must have $\alpha_k = 1$, and in $\textcircled{2}$,
$$p_k - \alpha_k + 1 = p_k^{\alpha_k} = p_k.$$
Thus,
$$(3 - \alpha_1)(p_2 - \alpha_2 + 1)\cdots(p_{k-1} - \alpha_{k-1} + 1) = 2^{\alpha_1} p_2^{\alpha_2} \cdots p_{k-1}^{\alpha_{k-1}}.$$
By the induction hypotheses, $\alpha_1 = \alpha_2 = \cdots = \alpha_{k-1} = 1$.
Thus, $\alpha_1 = \alpha_2 = \cdots = \alpha_{k-1} = \alpha_k = 1$, that is, the conclusion is true for $n = k$.

By (1) and (2), we conclude that $(\alpha_1, \alpha_2, \dots, \alpha_n) = (1, 1, \dots, 1)$, so the integer required is $x = 2p_2 \cdots p_n = p_1 p_2 \cdots p_n$.