1. 205 Starting from any three-digit number $n$, we get a new number $f(n)$, which is the sum of the three digits of $n$, the pairwise products of the three digits, and the product of the three digits.
（1）When $n=625$, find $\frac{n}{f(n)}$ (take the integer value).
（2）Find all three-digit numbers $n$ such that $\frac{n}{f(n)}=1$.
［Solution］Let
$$
n=\overline{a b c}=100 a+10 b+c,
$$

where $a, b, c \in\{0,1,2, \cdots, 9\}$, and $a \geqslant 1$. By the problem statement,
$$
f(n)=a+b+c+a b+b c+c a+a b c .
$$
(1) When $n=625$,
$$
\begin{aligned}
f(625) & =6+2+5+6 \times 2+2 \times 5+5 \times 6+6 \times 2 \times 5 \\
& =125, \\
\frac{625}{f(625)} & =5 .
\end{aligned}
$$
(2) By the problem statement,
$$
n=f(n),
$$

which means
$$
100 a+10 b+c=a+b+c+a b+b c+c a+a b c,
$$

solving for $c$, we get
$$
c=\frac{99 a+9 b-a b}{a+b+a b}
$$

From $c \leqslant 9$, we have
$$
99 a+9 b-a b \leqslant 9(a+b+a b),
$$

simplifying the above, we get
$$
90 a \leqslant 10 a b
$$

Since $a \geqslant 1$, therefore
$$
9 \leqslant b,
$$

which gives
$$
b=9 .
$$

Substituting into (1), we get
$$
c=\frac{90 a+81}{10 a+9}=9 .
$$

Thus,
$$
n=\overline{a 99}
$$

where $a \in\{1,2, \cdots, 9\}$.
It is easy to verify that when $n=\overline{a 99}, a \in\{1,2, \cdots, 9\}$, the equation
$$
\frac{\overline{a 99}}{f(\overline{a 99})}=1
$$

holds true.
Therefore, there are 9 three-digit numbers $n$ that satisfy the problem's requirements: 199, 299, 399, 499, 599, 699, 799, 899, 999.