For a natural number n , let n^ denote the number obtained by...

\because n^{\prime} \neq n$so atleast one digit of a must be zero Let$n=\overline{a b c}$Case I. If only$c=0 \Rightarrow \overline{a b} \mid \overline{a b c} \Rightarrow \overline{a b} \mid 10 \overline{a b}$(always true) Number of numbers$n=9 \times 9=81$Case II. If only$b=0 \Rightarrow \overline{a c} \mid \overline{a b c} $$\begin{aligned} & \Rightarrow 10 a+c \mid 100 a+c \\ & \Rightarrow 10 a+c \mid 10 a-8 c \end{aligned}$$$So either$10 a=8 c$or$8 c-10 a=\lambda(10 a+c)$for some natural number$\lambda$. When$10 a=8 c \Rightarrow 5 a=4 c \Rightarrow(a, c)=(4,5)$When$c(8-\lambda)=10 a(\lambda+1) \Rightarrow c=a$\left[\frac{90}{8-\lambda}-10\right]$ for $\lambda=1, c=a$\left(\frac{20}{7}\right)$(Not possible) for$\lambda=2, c=5 a \Rightarrow(a, c)=(1,5)$for$\lambda=3, c=8 a \Rightarrow(a, c)=(1,8)$for$\lambda=4, c=a$\left(\frac{25}{2}\right)$ (Not possible) and no further value of $\lambda$ is needed to be checked Number of numbers $n=3$, which are 405,105 and 108. Case III. If both $b=c=0$, then all values of $n$ are acceptable. Number of numbers $n=9$ So total number of possible values of $n=93$.