To solve the problem of finding how many three-digit numbers exist such that the middle digit is the sum of the first and last digits, we can break down the problem as follows:

Let the three-digit number be represented by ABC, where A, B, and C are the digits of the number.

• A is the hundreds place digit, which ranges from 1 to 9 (it cannot be 0 for a three-digit number).
• B is the tens place digit.
• C is the units place digit, which ranges from 0 to 9.

According to the problem, the condition is that:

B = A + C

Let's determine the possible values of A and C to satisfy this equation.

1. If A = 1, then B = 1 + C and C can range from 0 to 8, giving us 9 possible numbers.
2. If A = 2, then B = 2 + C and C can range from 0 to 7, giving us 8 possible numbers.
3. If A = 3, then B = 3 + C and C can range from 0 to 6, giving us 7 possible numbers.
4. If A = 4, then B = 4 + C and C can range from 0 to 5, giving us 6 possible numbers.
5. If A = 5, then B = 5 + C and C can range from 0 to 4, giving us 5 possible numbers.
6. If A = 6, then B = 6 + C and C can range from 0 to 3, giving us 4 possible numbers.
7. If A = 7, then B = 7 + C and C can range from 0 to 2, giving us 3 possible numbers.
8. If A = 8, then B = 8 + C and C can range from 0 to 1, giving us 2 possible numbers.
9. If A = 9, then B = 9 + C and C can only be 0, giving us 1 possible number.

Adding all possibilities gives us:

9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45

Therefore, there are 45 three-digit numbers where the middle digit equals the sum of the first and last digits.