Which is the greatest possible value of n

Question

In a number with at least two digits, the last digit was deleted. The resulting number was n times smaller than the previous one. Which is the greatest possible value of n?

A).10 B).11 C).19 D).20

Answer 1

Hint: $n=\frac{10a+b}{a}$ with $a\ge1 $ and $b\le 9$

Answer 2

Say the number is $(10^ra_r+...+10^2a_2 + 10a_1 + a_0)$. New number after deleting last digit becomes $(10^{r-1}a_r+...+10a_2 + a_1)$. By the problem, $$n\cdot (10^{r-1}a_r+...+10a_2 + a_1) = (10^ra_r+...+10^2a_2 + 10a_1 + a_0)$$ or, $$n = \frac{(10^ra_r+...+10^2a_2 + 10a_1 + a_0)}{(10^{r-1}a_r+...+10a_2 + a_1)}$$ $$=\frac{10(10^{r-1}a_r+...+10a_2 + a_1 +\frac{a_0}{10})}{(10^{r-1}a_r+...+10a_2 + a_1)}$$ or, $$n = 10 + \frac{a_0}{(10^{r-1}a_r+...+10a_2 + a_1)}$$
Now $0 \le a_i\le 9$ $\forall$ $i$ and $a_r \not = 0$ for any $r$
Hence max value of n follows as $n = 19$ when $a_0=9,a_1=1,a_r=0$ $\forall$ $r\ge 2$