# Task 2. (20 points)

Find the maximum possible value of the ratio of a three-digit number to the sum of its digits.

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# Solution.

Let $N=\overline{a b c}$, where $a, b, c$ are the digits of the number. Clearly, for "round" numbers $N=$ $100, 200, \ldots, 900$, we have $\frac{N}{a+b+c}=100$. Furthermore, if the number $N$ is not round, then $b+c>0$ and $a+b+c \geq a+1$. Since the leading digit of the number $N$ is $a$, we have $N<(a+1) \cdot 100$ and

$$
\frac{N}{a+b+c}<\frac{(a+1) \cdot 100}{a+1}=100
$$

Thus, the maximum value of the considered ratio is 100. This value is achieved only for "round" numbers.