2005 AMC 10A Problem 14

Call the digits $a$, $b$, $c$ with $a \neq 0$. The condition $b = \frac{a+c}{2}$ has two consequences: $a + c$ must be even, and once $a$ and $c$ are chosen, $b$ is fixed. Since $a$ and $c$ are digits, $b = \frac{a+c}{2}$ is automatically between $0$ and $9$, so it is always a legal digit.

So we just count pairs $(a, c)$ with $a + c$ even, i.e. with $a$ and $c$ of the same parity.

• $a$ odd ($5$ choices: $1,3,5,7,9$), $c$ odd ($5$ choices): $25$ pairs.
• $a$ even ($4$ choices: $2,4,6,8$), $c$ even ($5$ choices: $0,2,4,6,8$): $20$ pairs.

Total $25 + 20 = 45$.

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

An "average" condition between digits is really a divisibility condition: two integers have an integer average exactly when they share parity. Recognizing that the middle digit is determined (not chosen) turns a three-digit search into a two-digit parity count. Whichever parity $a$ has, exactly $5$ of the $10$ possible last digits match it, giving the slick $9 \cdot 5$.

Casework on the middle digit $b$: the pairs $(a, c)$ with $a + c = 2b$ number $2, 4, 6, 8, 9, 7, 5, 3, 1$ for $b = 1, \dots, 9$, again summing to $45$.

• Forgetting that 0 is an allowed (even) last digit, or allowing 0 as a first digit, which shifts the count by 4 or 5.
• Worrying that $b$ might exceed $9$ or need checking; the average of two digits is always a digit.