Let's share the thought of dividing every three-digit (natural) number by the sum of its digits. What is the smallest quotient that results?

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I. solution. Let the consecutive digits of an arbitrary three-digit natural number be $A, B, C$, where $1 \leq A \leq 9$, and $0 \leq B, C \leq 9$. The quotient of the number and the sum of its digits can be transformed as follows:

$$
q=\frac{100 A+10 B+C}{A+B+C}=1+9 \cdot \frac{11 A+B}{A+B+C} \geq 1+9 \cdot \frac{11 A+B}{A+B+9}
$$

In the second form, we replaced the fraction with a smaller or equal value by substituting the largest possible value for $C$, thus increasing or leaving the denominator unchanged. We repeat this for $B$:

$$
\begin{aligned}
q & \geq 1+9 \cdot \frac{A+B+9+10 A-9}{A+B+9}=10+9 \cdot \frac{10 A-9}{A+B+9} \geq \\
& \geq 10+9 \cdot \frac{10 A-9}{A+18}
\end{aligned}
$$

since the numerator is positive. Continuing,

$$
\begin{aligned}
q & \geq 10+9 \cdot \frac{10(A+18)-189}{A+18}=100-\frac{9 \cdot 189}{A+18} \geq \\
& \geq 100-\frac{9 \cdot 189}{19}=\frac{199}{19}
\end{aligned}
$$

Here, we increased or left the subtrahend unchanged by decreasing or leaving the denominator unchanged, substituting the smallest possible value for $A$, which is 1.

According to this, the quotient in question is never less than

$$
\frac{199}{19}=\frac{199}{1+9+9}
$$

and this is achieved when starting from the number 199.

László Domokos (Tatabánya, Árpád g. IV. o. t.)

Remark. The following transformations show that the digits of the number giving the smallest $q$ can be determined in any order:

$$
q=100-\frac{90 B+99 C}{A+B+C}=10+\frac{90 A-9 C}{A+B+C}=1+\frac{99 A+9 B}{A+B+C}
$$

Imre Kloknicer (Budapest, Bláthy O. t. III. o. t.)

II. solution. When dividing a three-digit number $x=100 A+10 B-C(>100)$ by the sum of its digits $y(>1)$, the quotient decreases if we increase the tens or units digit by 1, or decrease the hundreds digit by 1 (leaving the other two digits unchanged in each case):

$$
\begin{aligned}
& \frac{x}{y}-\frac{x+10}{y+1}=\frac{x-10 y}{y(y+1)}=\frac{9(10 A-C)}{y(y+1)}>0 \\
& \frac{x}{y}-\frac{x+1}{y+1}=\frac{x-y}{y(y+1)}=\frac{99 A+9 B}{y(y+1)}>0 \\
& \frac{x}{y}-\frac{x-100}{y-1}=\frac{100 y-x}{y(y-1)}=\frac{90 B+99 C}{y(y-1)} \geq 0
\end{aligned}
$$

since $10 A \geq 10>C$. The tens and units digits can only be increased until $B=9$ and $C=9$ respectively, and the hundreds digit can only be decreased to $A=1$, and in (1) equality does not hold if $B, C>0$. By performing the changes one after the other, we arrive at the number 199, and the quotient 199/19 thus obtained cannot be decreased further, and in the case of $x=100, y=1$ not considered in the examination, we also get a larger quotient, so 199/19 is the smallest value of the quotient in question.

László Babai (Budapest, Fazekas M. gyak. g. II. o. t.)