2011 AMC 10A Problem 13

The number is a three-digit integer. For it to lie between $200$ and $700$, the hundreds digit must be $2$ or $5$ (the only allowed digits in that range). For it to be even, the units digit must be $2$ or $8$.

Case 1: hundreds digit $2$. The units digit cannot also be $2$, so it is $8$. The tens digit is any of the remaining $4$ digits. That gives $4$ numbers.

Case 2: hundreds digit $5$. The units digit can be $2$ or $8$ ($2$ choices), then the tens digit is any of the remaining $4$ digits. That gives $2 \cdot 4 = 8$ numbers.

Total: $4 + 8 = 12$.

The answer is $\boxed{\textbf{(A)}\ 12}$.

Fill the most restricted positions first (hundreds digit for the range, units digit for parity), and split into cases whenever a choice for one position changes the options for another. The distinct-digit condition is what couples the hundreds and units places here.

• Multiplying 2 4 2 = 16 for the three positions without noticing that hundreds digit 2 removes 2 as a units digit option.
• Allowing the hundreds digit $7$ ($700$ to $799$ is not below $700$) or forgetting that digits must be distinct.