Correct Option (2)
Let the three-digit number be represented as 100x + 10y + z, where x is the hundreds digit, y is the tens digit (middle digit), and z is the units digit (last digit).
- According to the problem statement, the middle digit is zero, so y = 0.
- The sum of the digits is 4: x + y + z = 4. Substituting y = 0, this equation simplifies to x + z = 4.
- The original number can therefore be expressed as 100x + z.
- When the first and last digits are interchanged, the new number becomes 100z + 10y + x. With y = 0, the new number is 100z + x.
- The problem states that the number formed by interchanging the first and last digits exceeds the original number by 198. This can be formulated as:
(100z + x) - (100x + z) = 198
- Simplifying the equation:
99z - 99x = 198
99(z - x) = 198
z - x = 198 / 99
z - x = 2
Thus, the difference between the first and last digits is 2.
Incorrect Options:
Options 1 (1), 3 (3), and 4 (4) are incorrect. The mathematical derivation, based on the problem's conditions, establishes that the difference between the first and last digits must be 2. Any other value for (z - x) would not satisfy the equation 99(z - x) = 198, which is a direct consequence of the problem's premise.