Question

Let $abc$ be a three digit number. Then $abc-cba$ is not divisible by

• A. $9$
• B. $11$
• C. $18$
• D. $33$

Answer 1

The correct option is C

$18$

Explanation for the correct option:

Expanding form of number:

The general form of given numbers is as follows,

$$\begin{gathered} abc=100a+10b+c \\ cba=100c+10b+a \end{gathered}$$

Now,

$$\begin{gathered} \Rightarrow abc-cba=100a+10b+c-100c-10b-a \\ \Rightarrow abc-cba=99a-99c \\ \Rightarrow abc-cba=99(a-c) \\ \Rightarrow abc-cba=(3\times 3\times 11)(a-c) \end{gathered}$$

$abc-cba$ is divisible by $99$ and its factors, $3,9,11$ and $33$ . It is also divisible by $(a-c)$.

Hence, option(C) is correct.