2005 AMC 10A Problem 14

Call the digits $a$, $b$, $c$, so the number is $abc$ and $a$ cannot be $0$. The middle digit must be the average of the other two:
$$b = \frac{a + c}{2}.$$

1. The middle digit is not a free choice. Once $a$ and $c$ are picked, $b$ is fixed. So we only need to count good pairs $(a, c)$.
2. When is the average a whole number? When $a + c$ is even. That happens when $a$ and $c$ are both odd or both even. For example, $3$ and $7$ average to $5$, but $3$ and $6$ average to $4.5$, which is not a digit.
3. Is $b$ always a real digit? Yes. The average of two digits is between $0$ and $9$, so it is never too big.
4. Count, both odd. $a$ can be $1, 3, 5, 7, 9$ ($5$ choices) and $c$ can be $1, 3, 5, 7, 9$ ($5$ choices): $5 \times 5 = 25$ pairs.
5. Count, both even. $a$ can be $2, 4, 6, 8$ ($4$ choices, since $a$ cannot be $0$) and $c$ can be $0, 2, 4, 6, 8$ ($5$ choices): $4 \times 5 = 20$ pairs.
6. Total. $25 + 20 = 45$.

The answer is $\boxed{\textbf{(E)}\ 45}$.

"The average is a whole number" is really a question about odd and even: two whole numbers have a whole-number average exactly when they are both odd or both even. Noticing that the middle digit is decided for you turns a three-digit search into a quick count of pairs. A shortcut: for each of the $9$ first digits, exactly $5$ of the $10$ last digits match it, giving $9 \times 5 = 45$.

Go through the middle digit. For each $b$, count pairs $(a, c)$ with $a + c = 2b$ and $a \ge 1$:

For example, $b = 1$ allows $a + c = 2$, which is $1{+}1$ or $2{+}0$. The row adds up to $45$.