Question

Consider the set $A$ of natural numbers $n_r$ whose units digit is nonzero, such that if this units digit is erased, then the resulting number divides $n_r$. If K is the number of elements in the set A, then.

• A. K is infinite
• B. K is finite but $K>100$
• C. $25\le K\le 100$
• D. $K<25$

Answer 1

Answer: (D)

$K<25$

Let $A\in (n_1,n_2,\dots n_k)$

and $n_r$ be any natural number of set $A$.

$n_r=a+10b+100c\dots$

where $a,b$ and $c$ are unit's, ten's and ${100}^{\prime }s$ digit respectively of $n_r$.

$\Rightarrow 0\le a,b,c\le 9$ and $a\ne 0\dots \dots \left(1\right)$

Given, that if units digit $a$ is erased, then the resulting number divides $n_r$.

$\therefore \frac{a+10b+100c\dots }{b+10c+100d\dots }=$ natural number

$\Rightarrow \frac{a}{10b+100c\dots }+10=$ natural number

$\therefore a$ must be greater than $b+10c+\dots$

$\Rightarrow c=0$ and $1\le b\le a\le 9$ From $\left(1\right)$

Now we have, $\frac{a}{b}=$ integer natural number

Possible number of sets

$=9(\text{If}b=1)+4(\text{If}b=2)+3(\text{If}b=2)+2(b=4)+1(b=5)+1(b=6)+1(b=7)+1(b=8)+1(b=9)$

$=24$

$\therefore k<25$

Option $D$ is correct.