Question

The difference between a 3 digit number and the number formed by reversing its digits is not always divisible by

• A. 9
• B. 11
• C. 99
• D. 5

Answer 1

The correct option is D 5

Let $100a+10b+c$ be the original number.
Then, the number obtained by reversing its digits is $100c+10b+a$.

Assuming a>c, subtracting smaller number from larger
$(100a+10b+c)-(100c+10b+a)$
$=(100-1)a+(10-10)b+(1-100)c$
$=99a-99c=99(a-c)$
So, it is always divisible by $99$.

Now, $99=11\times 9$
Therefore, $99(a-c)=11\times 9(a-c)$
Hence, the difference obtained is always divisible by 99, 9 and 11.

So, the difference obtained is not divisible by 5 unless (a - c) is divisible by 5.