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
$$n=\overline{a b c}=100 a+10 b+c$$

Here $a, b, c \in\{0,1,2, \cdots, 9\}$, 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,
$$n=f(n),$$

i.e.,
$$100 a+10 b+c=a+b+c+a b+b c+c a+a b c,$$

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
$$b=9 \text {. }$$

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: $199,299,399,499,599$, $699,799,899,999$.