118. What is the largest and the smallest value that the ratio of a two-digit number to the sum of its digits can take?
S o l u t i o n. Let $x, y$ be the first and second digits of a two-digit number, respectively. Since $x \neq 0$, denoting the considered ratio by $t$, we have:

$$
t=\frac{10 x+y}{x+y}=1+\frac{9 x}{x+y}=1+\frac{9}{1+\frac{y}{x}}
$$

From this, it is clear that $t$ is greater the smaller the ratio $\frac{y}{x}$. Therefore, the largest value of $t=10$ is obtained when $y=0$ and for any $x$, while the smallest $t=1.9-$ is obtained when $\frac{y}{x}=9$, i.e., when $x=1$, $y=9$.

A n s w e r: $10 ; 1.9$