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

We need three-digit numbers that become 198 larger when we reverse their digits. For example, if we have a number like 234, its reverse would be 432, and we want 432 - 234 = 198.

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We represent any three-digit number using place values:

Units place = a

Tens place = b

Hundreds place = c

So our original number = 100c + 10b + a

In a number like 234, the digit 2 contributes 200 (2×100), digit 3 contributes 30 (3×10), and digit 4 contributes 4 (4×1).

When we reverse the digits, we get:

New number = 100a + 10b + c

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Since the reversed number is 198 more than the original:

$(100a + 10b + c) - (100c + 10b + a) = 198$

The problem states that the number "increases by 198" when digits are reversed, meaning the reversed number is larger.

$100a + 10b + c - 100c - 10b - a = 198$

$100a - a + 10b - 10b + c - 100c = 198$

$99a - 99c = 198$

We collected like terms: 100a - a = 99a, and c - 100c = -99c

$99(a - c) = 198$

$a - c = 2$

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Now we need to find all possible values of a and c such that:

$a - c = 2$ and the number is a valid three-digit number > 100

For a three-digit number: $c \geq 1$ (hundreds place can't be 0)

For any digit: $a \leq 9$ (digits can't exceed 9)

Valid pairs (a, c):

If c = 1, then a = 3 → (3,1)

If c = 2, then a = 4 → (4,2)

If c = 3, then a = 5 → (5,3)

If c = 4, then a = 6 → (6,4)

If c = 5, then a = 7 → (7,5)

If c = 6, then a = 8 → (8,6)

If c = 7, then a = 9 → (9,7)

We stop at c = 7 because if c = 8, then a would be 10, which isn't a valid digit.

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We have 7 valid combinations for (a, c)

For each combination, b can be any digit from 0 to 9 (10 choices)

Total numbers = 7 × 10 = 70

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We check with (a, c) = (3, 1) and b = 2:

Original number: 123

Reversed number: 321

Difference: 321 - 123 = 198

Therefore, there are 70 three-digit numbers that satisfy the given condition.