In the solution, the number of divisors of $n$ is denoted by $d(n)$. Obviously $x$ must be of the form $\prod_{i=1}^n p_i^{a_i}$ where $0 \le a_i \le p_i$. We have $v_2(A)=2$ and since $\frac{A}{x}$ is even, we have $v_2(x)=0$ or $1$. Case I: $v_2(x)=0 \implies d(A/x)$ is odd. So, $A/x$ must be a perfect square. Therefore, $a_i$ is odd $\forall i \in [2,n]$. Since, $v_2(A/x)=2 \implies 2+1=3|d(A/x)=x$. So, $v_3(A/x)=2$ or $0$. In the first case, we have $3|p_2-a_2+1$ and $3|p_1-a_1+1 \implies 3^2|d(A/x)=x \implies v_3(A/x) \le 1$, a contradiction. Therefore, $v_3(A/x)=0$ and as a consequence, $3^3|x$. Now, $d(A/x)=x$ gives \[3 \prod_{i=3}^n(p_i-a_i+1) = 3^3 \prod_{i=3}^np_i^{a_i}\] But, since $a_i$ is odd, $a_i \ge 1 \implies p_i^{a_i} \ge p_i \ge p_i-a_i+1$ and this gives a clear contradiction in the above equation. Case II: \[v_2(x)=1 \implies \frac{A}{x}=2 \prod_{i=2}^np_i^{p_i-a_i}\] \[ \implies d(A/x)= 2\prod_{i=2}^n(p_i-a_i+1) = x=2\prod_{i=2}^np_i^{a_i}\] \[ \implies \prod_{i=2}^n(p_i-a_i+1)=\prod_{i=2}^np_i^{a_i}\] Obviously, the $RHS$ is odd and so should be the $LHS$. Therefore $p_i-a_i+1$ is odd $\forall i \in [2,n]$. As a result, $a_i$ is odd $\forall i \in [2,n] \implies a_i \ge 1$. But then, $p_i^{a_i} \ge p_i \ge p_i-a_i+1 \forall i \in [2,n]$ with equality iff $a_i=1$. So, for the equality to hold, we need $a_i=1 \forall i \in [2,n]$. This gives $x=\prod_{i=1}^np_i$, and it is easy to verify that this is indeed a solution.