Question

Let $abc$ be a three digit number. Then $abc+bca+cab$ is not divisible by

• A. $a+b+c$
• B. $3$
• C. $37$
• D. $9$

Answer 1

The correct option is D

$9$

Explanation for the correct option:

Expanding form of number:

The general form of given numbers is as follows,

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

Now,

$$\begin{gathered} \Rightarrow abc+bca+cab=100a+10b+c+100b+10c+a+100c+10a+b \\ \Rightarrow abc+bca+cab=111a+111b+111c \\ \Rightarrow abc+bca+cab=111(a+b+c) \end{gathered}$$

$abc+bca+cab$ is divisible by $111$and its factors, $3$ and $37$ . It is also divisible by $(a+b+c)$.

Hence, option(D) is correct.