2007 AMC 10B Problem 8

The receipt looks like $b\,b\,c\,a\,c$. Once we choose $a$ and $c$, the middle digit $b$ is their average, so there is nothing left to choose. We only need to count the good choices of $a$ and $c$.

1. Try a few examples. $a = 1, c = 5$ gives $b = 3$, a digit. $a = 1, c = 4$ gives $b = 2.5$, not a digit.
2. The average of two whole numbers is whole exactly when their sum is even, which happens when both are even or both are odd.
3. When $b$ is whole, it automatically lies between $a$ and $c$, so $a < b < c$ holds. Also $b$ is at least $1$, so the receipt really starts with a nonzero digit.
4. Count pairs with $a$ smaller than $c$:
- Both even, from $0, 2, 4, 6, 8$: choosing $2$ of these $5$ digits gives $\frac{5 \times 4}{2} = 10$ pairs.
- Both odd, from $1, 3, 5, 7, 9$: again $10$ pairs.
5. Total: $10 + 10 = 20$.

The answer is $\boxed{\textbf{(D)}\ 20}$.

When one digit is decided by the others, count only the free choices. "The average is a whole number" is really an even/odd condition, and the digits split evenly into five even and five odd ones, so each group gives the same simple count.

The number is fixed once $a$ and $c$ are chosen, because $b = \frac{a + c}{2}$. For $b$ to be a digit it must be an integer, which happens exactly when $a$ and $c$ have the same parity. When it is an integer, $a < c$ automatically places $b$ strictly between them, and $b \ge 1$ so the number really has five digits.

So we count pairs $a < c$ of the same parity among the digits $0$ through $9$:

• both even, from $\{0, 2, 4, 6, 8\}$: $\binom{5}{2} = 10$ pairs;
• both odd, from $\{1, 3, 5, 7, 9\}$: $\binom{5}{2} = 10$ pairs.

Total $10 + 10 = 20$ receipts.

The answer is $\boxed{\textbf{(D)}\ 20}$.