10.115 Starting from any three-digit number $n$, we get a new number $f(n)$, which is the sum of the products of the three digits of $n$, taken two at a time, and the product of the three digits.
(1) When $n=625$, find $\frac{n}{f(n)}$.
(2) Find all three-digit numbers such that $\frac{n}{f(n)}=1$.
(UK Mathematical Olympiad, 1994)
[Solution] Let $n=\overline{a b c}=100 a+10 b+c$, where $a, b, c \in\{0,1,2, \cdots, 9\}$ and $a \geqslant 1$.
$$
f(n)=(a+b+c)+(a b+b c+c a)+a b c \text {. }
$$
(1) When $n=625$, $a=6, b=2, c=5$.
$$
\begin{array}{c}
f(625)=(6+2+5)+(12+10+30)+60=125, \\
\frac{625}{f(625)}=\frac{625}{125}=5 .
\end{array}
$$
(2) According to the problem, we have $n=f(n)$.
$$
\begin{aligned}
100 a+10 b+c & =(a+b+c)+(a b+b c+c a)+a b c, \\
c & =\frac{99 a+9 b-a b}{a+b+a b} \leqslant 9,
\end{aligned}
$$

which means
$$
99 a+9 b-a b \leqslant 9(a+b+a b),
$$

or $90 a \leqslant 10 a b$, hence $b \geqslant 9$.
On the other hand, $b \leqslant 9$, so $b=9$.
And $c=\frac{99 a+81-9 a}{a+9+9 a}=\frac{90 a+81}{10 a+9}=9$

Therefore, $n=a \overline{99}$,
Since $a$ can take $1,2, \cdots, 9$, we have
$$
n=199,299,399,499,599,699,799,899,999 .
$$