10.43 Find the maximum value of the ratio of a three-digit number to the sum of its digits.
(17th Moscow Mathematical Olympiad, 1954)
[Solution] For the numbers $100, 200, 300, \cdots, 900$, the ratio of the number to the sum of its digits is 100.
We prove that 100 is the maximum value of this ratio.
Let $abc$ be any three-digit number different from $100, 200, \cdots, 900$, where $a, b, c \in \{0,1,2, \cdots, 9\}, a \neq 0$, and at least one of $b$ and $c$ is not equal to 0.
Then the sum of the digits of $\overline{abc}$ is $a+b+c \geqslant a+1$.
And $\overline{abc} < (a+1) \cdot 100$.
Thus, the ratio of $\overline{abc}$ to the sum of its digits is
$$
\frac{\overline{abc}}{a+b+c} < \frac{(a+1) \cdot 100}{a+1} = 100.
$$

Therefore, the maximum value of the ratio of a three-digit number to the sum of its digits is 100, and this maximum is achieved when the number is $100, 200, \cdots, 900$.