7. For a positive integer $n$, let the product of its digits be denoted as $a(n)$. Then the positive integer solution to the equation $n^{2}-17 n+56=a(n)$ is
$\qquad$ .
7. 4 Detailed Explanation: First, prove $a(n) \leqslant n$. Let $n=\overline{b_{k} b_{k-1} \cdots b_{1} b_{0}}$, then $n \geqslant b_{k} \times 10^{k} \geqslant b_{k} \prod_{i=1}^{k-1} b_{i}=\prod_{i=1}^{k} b_{i}=a(n)$. $\therefore n^{2}-17 n+56 \leqslant n$, which means $n^{2}-18 n+56 \leqslant 0$, solving this yields $n=4,5,6, \cdots, 14$. Upon verification, $n=4$.