7. (10 points) The largest divisor of the natural number $A$ other than itself is $d$, and the largest divisor of the natural number $A+2$ other than itself is $d+2$. Then the value of $A$ is $\qquad$ .
【Analysis】First, prove that $A$ is a prime number, $d=1$, which leads to $d+2=3$. Since $d+2$ is the largest divisor of $A+2$ except for itself, we conclude that $A+2=9$. From this, the problem can be solved.

【Solution】If $A$ is even, then $d=\frac{A}{2}, d+2=\frac{A+2}{2}$, which is a contradiction. Therefore, $A$ is odd, and $A$ and $A+2$ are coprime, as are $d$ and $d+2$. If $A$ is not a prime number, let $A=md$, where $m$ is the smallest prime factor of $A$. Then $(d+2) \mid (md+2)$,
$$
\begin{array}{l}
\therefore (d+2) \mid (md+2m+2-2m), \\
\therefore (d+2) \mid (2m-2),
\end{array}
$$
$\therefore (d+2) \mid (m-1)$, which contradicts the fact that $m$ is the smallest prime factor.
$\therefore A$ is a prime number,
$$
\therefore d=1, d+2=3,
$$
$\because d+2$ is the largest divisor of $A+2$ except for itself,
$$
\begin{array}{l}
\therefore A+2=9, \\
\therefore A=7.
\end{array}
$$

The answer is 7.