How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

To solve this problem, we need to find the three-digit numbers that increase by 198 when their digits are reversed.

Step 1: Let's denote the three-digit number as ABC, where A, B, and C are its digits. When reversed, it becomes CBA.

Step 2: According to the problem, CBA = ABC + 198.

Step 3: We can express ABC and CBA in terms of A, B, and C: ABC = 100A + 10B + C and CBA = 100C + 10B + A.

Step 4: Substituting these expressions into the equation from step 2, we get 100C + 10B + A = 100A + 10B + C + 198.

Step 5: Simplifying this equation, we get 99C - 99A = 198, or C - A = 2.

Step 6: Since A and C are digits, they must be integers between 0 and 9. Given that C - A = 2, the possible pairs (A, C) are (0, 2), (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), and (7, 9).

Step 7: However, since we're looking for three-digit numbers, A cannot be 0. So, the valid pairs are (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), and (7, 9).

Step 8: For each pair, there are 10 choices for B (0 through 9).

Step 9: Therefore, there are 7 pairs * 10 choices = 70 such numbers.

So, 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.