Example 1. Find the minimum value of the quotient of a three-digit number divided by the sum of its digits, and write down this three-digit number.



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Let the three-digit number be $100x + 10y + z$, and the quotient of the number and the sum of its digits be $w$, where $1 \leqslant x \leqslant 9, 0 \leqslant y, z \leqslant 9$.
$$
w = \frac{100x + 10y + z}{x + y + z} = 1 + \frac{9(11x + y)}{x + y + z}.
$$

In equation (1), if $x$ and $y$ are fixed and $z$ varies, then $w$ reaches its minimum value when $z = 9$, so $z = 9$. Thus,
$$
\begin{aligned}
w & = 1 + \frac{9(11x + y)}{x + y + 9} \\
& = 1 + \frac{9[(x + y + 9) + 10x - 9]}{x + y + 9} \\
& = 1 + 9\left(1 + \frac{10x - 9}{x + y + 9}\right) \\
& = 15 + \frac{9(10x + 9)}{x + y + 3}.
\end{aligned}
$$

In the above equation, if $x$ is fixed and $y$ varies, then $w$ reaches its minimum value when $y = 9$, so $y = 9$. Therefore,
$$
\begin{aligned}
w & = 10 + \frac{9(10x - 9)}{x + 18} \\
& = 10 + \frac{9[(x + 18) + 9x - 27]}{x + 18} \\
& = 10 + 9\left(1 + \frac{9x - 27}{x + 18}\right) \\
& = 19 + 9\left[\frac{9(x + 18) - 189}{x + 18}\right] \\
& = 19 + 9\left(9 - \frac{189}{x + 18}\right) = 100 - \frac{1701}{x + 18}.
\end{aligned}
$$

In equation (3), to make $w$ reach its minimum value, $x$ must take its minimum value 1. Hence, the three-digit number is 199,
$$
w_{\mathrm{min}} = \frac{199}{1 + 9 + 9} = 10 \frac{9}{19}.
$$