[Example 3.2.6] Find a three-digit number such that the ratio of the number to the sum of its digits is minimized.
Let the three-digit number $\overline{a b c}=100 a+10 b+c, a \in\{1,2, \cdots, 9\}, b, c \in\{0,1,2, \cdots, 9\}$, and the sum of its digits is $a+b+c$.
Also, let
$$
k=\frac{100 a+10 b+c}{a+b+c} .
$$

Our goal is to find a three-digit number that minimizes the value of $k$.
Transform the expression for $k$ to
$$
k=100-\frac{90 b+99 c}{a+b+c} .
$$

In this fraction, the numerator contains only the variables $b$ and $c$, while the denominator contains the variables $a$, $b$, and $c$.

First, we treat $b$ and $c$ as fixed numbers and $a$ as a variable. Clearly, when $a=1$, $\frac{90 b+99 c}{a+b+c}$ is maximized, thus minimizing $k$. Substituting $a=1$ yields
$$
\begin{aligned}
k & =100-\frac{90 b+99 c}{1+b+c}=100-90-\frac{9 c-90}{1+b+c} \\
& =10+\frac{90-9 c}{1+b+c}
\end{aligned}
$$

Next, we fix $c$ and treat $b$ as a variable. At this point, when $b=9$, $\frac{90-9 c}{1+b+c}$ is minimized, thus minimizing $k$. Substituting $b=9$ yields
$$
k=10+\frac{90-9 c}{1+9+c}=1+\frac{180}{10+c}
$$

Now, there is only one variable $c$. Clearly, when $c=9$, $k$ is minimized, and the value of $k$ at this point is $\frac{199}{19}$.
In solving this problem, we adopted the method of first freezing $b$ and $c$, and then sequentially unfreezing $b$ and $c$ to solve the problem.
The entire transformation process can be written as
$$
\begin{aligned}
k & =\frac{100 a+10 b+c}{a+b+c}=100-\frac{90 b+99 c}{a+b+c} \geqslant 100-\frac{90 b+99 c}{1+b+c} \\
& =10+\frac{90-9 c}{1+b+c} \geqslant 10+\frac{90-9 c}{10+c}=1+\frac{180}{10+c} \\
& \geqslant 1+\frac{180}{19}=\frac{199}{19}
\end{aligned}
$$

Thus, the desired three-digit number is 199.