Analysis of the provided information:
Let the original three-digit number be \( ABC \), 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.
This translates to the equation:
\[ CBA = ABC + 198 \]
Expressing the numbers in terms of their digits:
\[ ABC = 100A + 10B + C \]
\[ CBA = 100C + 10B + A \]
Substituting these into the equation:
\[ 100C + 10B + A = 100A + 10B + C + 198 \]
Simplifying the equation by combining like terms:
\[ 99C - 99A = 198 \]
Dividing both sides by 99:
\[ C - A = 2 \]
Given that A and C are single-digit integers, the possible pairs for (A, C) are:
\[ (A, C) = (0, 2), (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9) \]
Since the number must be a three-digit number, the hundreds digit A cannot be 0. Therefore, the pair (0, 2) is invalid.
This leaves 7 valid combinations for (A, C).
The tens digit B can be any digit from 0 to 9. For each of the 7 valid (A, C) pairs, there are 10 possible values for B.
The total number of three-digit numbers satisfying the given conditions is:
\[ 7 \times 10 = 70 \]
Thus, there are 70 such three-digit numbers.