119. What is the greatest possible value of the ratio of a three-digit number to the sum of its digits?
119. Let \( N = \overline{a b} c \), where \( a, b, c \) are the digits of the number; it is clear that 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 \( a+b+c \geqslant a+1 \), and since the leading digit of \( N \) 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.

Note. In the same way, it can be proven that the maximum value of the ratio of a \( k \)-digit number \( N = \overline{a_{k} a_{k-1} \ldots a_{1}} \) to the sum \( a_{k} + a_{k-1} + \ldots + a_{1} \) of its digits is \( 10^{k-1} \); this value is achieved only for "round" numbers, where the last \( k-1 \) digits are zeros.