Introduction:
To solve this problem, we need to find the three-digit numbers that are greater than 100 and increase by 198 when the three digits are arranged in the reverse order.

Approach:
We can break down the problem into smaller steps to find the solution:

Step 1: Analyzing the numbers:
Let's consider a three-digit number, ABC, where A, B, and C represent the hundreds, tens, and units digits, respectively. The reverse of this number is CBA.
We can write the given condition as:
ABC - CBA = 198

Step 2: Analyzing the equation:
We need to analyze the equation ABC - CBA = 198 further to find a pattern. Let's write the equation in terms of place values:
(100A + 10B + C) - (100C + 10B + A) = 198

Simplifying the equation, we get:
99A - 99C = 198
Dividing both sides by 99, we have:
A - C = 2

Step 3: Finding the possible values:
Since A and C are digits, they can take values from 1 to 9. We need to find the values of A and C that satisfy the equation A - C = 2.

Step 4: Analyzing the possible values:
Let's examine the possible values of A and C that satisfy the equation A - C = 2:

- When A = 3, C = 1
- When A = 4, C = 2
- When A = 5, C = 3
- When A = 6, C = 4
- When A = 7, C = 5
- When A = 8, C = 6
- When A = 9, C = 7

Step 5: Counting the numbers:
Now that we have found the possible values of A and C, we can find the corresponding numbers. Since B can be any digit from 0 to 9, we have 10 possibilities for B.

Therefore, the total number of three-digit numbers that satisfy the given condition is 7 (values of A) * 10 (values of B) = 70.

Conclusion:
There are 70 three-digit numbers that are greater than 100 and increase by 198 when the three digits are arranged in the reverse order.