Let the original three-digit number be denoted by \( ABC \), where \( A \) is the hundreds digit, \( B \) is the tens digit, and \( C \) is the units digit. When the digits are reversed to form \( CBA \), the number increases by 198. The objective is to determine the count of such three-digit numbers that satisfy this condition.
The number \( ABC \) can be expressed as: \[ ABC = 100A + 10B + C \] The reversed number \( CBA \) can be expressed as: \[ CBA = 100C + 10B + A \] The problem states: \[ CBA = ABC + 198 \] Substituting the expanded forms: \[ 100C + 10B + A = 100A + 10B + C + 198 \]
Eliminate the \( 10B \) term from both sides: \[ 100C + A = 100A + C + 198 \] Group like terms: \[ 100C - C = 100A - A + 198 \] Simplify the equation: \[ 99C - 99A = 198 \] Divide both sides by 99: \[ C - A = 2 \]
Given that \( A \) and \( C \) are single-digit integers, the pairs \( (A, C) \) satisfying \( C - A = 2 \) are: \[ (A, C) = (0, 2), (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9) \] As \( A \) must be non-zero for a three-digit number, the pair \( (0, 2) \) is invalid. Therefore, there are 7 valid pairs: \[ (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9) \]
The tens digit \( B \) can be any integer from 0 to 9, independently of the values of \( A \) and \( C \).
With 7 valid combinations for \( (A, C) \) and 10 possible values for \( B \), the total number of valid three-digit numbers is: \[ 7 \times 10 = 70 \]
The total count of such three-digit numbers is \( \boxed{70} \).