1. (2005 Romanian Mathematical Olympiad) Given that $n$ is an integer, let $p(n)$ denote the product of its digits (in decimal representation).
（1）Prove that: $p(n) \leqslant n$;
（2）Find all $n$ such that $10 p(n)=n^{2}+$ $4 n-2005$ holds.
1. (1) Suppose $n$ has $k+1$ digits, $k \in \mathbf{N}$, then $n=10^{k} a_{k}+10^{k-1} a_{k-1}+\cdots+10 a_{1}+a_{0}$, where $a_{1}, a_{2}, \cdots, a_{k} \in\{1,2, \cdots, 9\}$.
Thus, we have $p(n)=a_{0} a_{1} \cdots a_{k} \leqslant a_{k} 9^{k} \leqslant a_{k} 10^{k} \leqslant n$.
Therefore, $p(n) \leqslant n$.
(2) First, from $n^{2}+4 n-2005 \geqslant 0$, we get $n \geqslant 43$.
Second, from $n^{2}+4 n-2005=10 p(n) \leqslant 10 n$, we get $n \leqslant 47$.
Thus, we deduce that $n \in\{43,44,45,46,47\}$.
Checking each one, we find $n=45$.