The difference between 3 digit number and reversing its digit is always divisible by

Question

The difference between a 3 digit number and a number fomed by reversing its digit is always divisible by?

a.6

b.9

c.12

d.19

How to approach this question?

Answer 1

The number thus formed is divisible by 9

Answer 2

Let $100a + 10b + c$ be your number; its reverse is $100c + 10b + a$.

Subtracting: $$\begin{aligned} (100a + 10b + c) - (100c + 10b + a) &= (100-1)a + (10-10)b + (1-100)c \\ &= 99a - 99c\\ &= 99(a-c) \end{aligned}$$

So it is always divisible by $99$, and in particular, $9$.