2005 AMC 12A Problem 11

The middle digit is an integer only when the first and last digits are both odd or both even. Each such pair determines the middle digit uniquely. There are $5 \cdot 5 = 25$ odd-odd choices for the first and last digits. For even-even, the first digit cannot be $0,$ giving $4 \cdot 5 = 20$ choices. The total is $25 + 20 = 45.$ Thus, the correct answer is E.