Solution:

Given that the three-digit number is greater than 100 and increases by 198 when the digits are reversed.

Let the number be of the form ABC, where A, B, and C are digits.

When the digits are reversed, the number becomes CBA.

Therefore, the number increases by (CBA - ABC) = (100C + 10B + A) - (100A + 10B + C) = 99C - 99A = 99(C - A).

We know that the number increases by 198, which is a multiple of 99.

Therefore, (C - A) must be either 2 or -2.

Case 1: (C - A) = 2

In this case, C can be any digit from 5 to 9, and A can be any digit from 0 to 4.

For each value of C, there is only one possible value of A that satisfies the condition.

Therefore, the total number of three-digit numbers in this case is 5.

Case 2: (C - A) = -2

In this case, A can be any digit from 5 to 9, and C can be any digit from 0 to 4.

For each value of A, there is only one possible value of C that satisfies the condition.

Therefore, the total number of three-digit numbers in this case is 5.

Hence, the total number of three-digit numbers that satisfy the condition is 5 + 5 = 10. However, we need to exclude the number 198, which is less than 100.

Therefore, the final answer is 10 - 1 = 9.

However, we need to consider the fact that the digits cannot be repeated.

Case 1: (C - A) = 2

In this case, there are 5 possible values of C and 5 possible values of A that satisfy the condition.

Therefore, the total number of three-digit numbers in this case is 5 x 5 = 25.

However, we need to exclude the cases where the digits are repeated.

When C = 9, A can only be 7.

When C = 8, A can only be 6 or 8.

When C = 7, A can only be 5 or 7.

Therefore, the total number of three-digit numbers in this case is 25 - 4 = 21.

Case 2: (C - A) = -2

In this case, there are 5 possible values of A and 5 possible values of C that satisfy the condition.

Therefore, the total number of three-digit numbers in this case is 5 x 5 = 25.

However, we need to exclude the cases where the digits are repeated.

When A = 9, C can only be 1.

When A = 8, C can only be 0 or 2.

When A = 7, C can only be 3 or 5.

When A = 6, C can only be 4 or 6.

Therefore, the total number of three-digit numbers in this case is 25 - 6 = 19.

Hence, the total number of three-digit numbers that satisfy the condition and do not have repeated digits is 21 + 19 = 40.

However, we need to exclude the number 198

Let the numbers be of the form 100a+10b+c, where a, b, and c represent single digits.