CAT 2021 Slot 1 QA Question 17: How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Question

How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Accepted Answer

Let the three-digit number be $100a + 10b + c$, where $a$ is from 1 to 9.
After reversing the digits, the number becomes $100c + 10b + a$. The increase is 198.
$(100c + 10b + a) - (100a + 10b + c) = 198$
$99(c-a) = 198$
$c-a = 2$
So $a$ can be 1 to 7, and then $c$ is fixed as $a + 2$. The middle dig