2005 AMC 10A Problem 14

The first and last digits must have the same parity so their average is a digit. Both odd gives $5 \cdot 5 = 25$ pairs. Both even, with a nonzero leading digit, gives $4 \cdot 5 = 20$ pairs. Each pair fixes the middle digit, for a total of $25 + 20 = 45$ numbers. Thus, the correct answer is E.