A number consists of three digits of which the middle one is zero and their sum is 4. If the number formed by interchanging the first and last digits is greater than the number itself by 198, then the difference between the first and last digits is

(a) 1
(b) 2 ✓ UPSC's answer
(c) 3
(d) 4

Why the answer is (b)

• Let the three-digit number be represented as $100a + 10b + c$, where $a$ is the first digit, $b$ is the middle digit, and $c$ is the last digit.

• The problem states the middle digit is zero, so $b = 0$, making the number $100a + c$.

• The sum of the digits is 4, so $a + 0 + c = 4$, which simplifies to $a + c = 4$.

• Interchanging the first and last digits gives the number $100c + a$.

• The new number is greater than the original by 198, so $(100c + a) - (100a + c) = 198$.

• Simplifying the equation: $99c - 99a = 198$, which divides to $c - a = 2$.

• Since $c - a = 2$, the difference between the first and last digits is 2, corresponding to option (b).

Why the other options are wrong

(a) 1: The difference is 2, not 1, as derived from the equation $c - a = 2$.
(c) 3: The difference is 2, not 3, which would imply a sum of digits different from 4 or a different difference value.
(d) 4: The difference is 2, not 4, which would require the digits to be 0 and 4, but the first digit cannot be 0 in a three-digit number.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2018, held on 3 June 2018. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.