Task A-3.2. (4 points)

What is the maximum possible ratio of a three-digit number to the sum of its digits?
## First solution.

We claim that this ratio is 100. For example, for the number 300, we have $\frac{300}{3+0+0}=100$.

(Of course, the same holds for all numbers of the form $\overline{a 00}$.)

For a three-digit number $\overline{a b c}$, the observed ratio is $\frac{100 a+10 b+c}{a+b+c}$.

Now we need to prove that this ratio cannot be greater than 100.

Clearly,

$$
\frac{100 a+10 b+c}{a+b+c} \leqslant \frac{100 a+100 b+100 c}{a+b+c}=100
$$

i.e., the observed ratio cannot be greater than 100.