2007 AMC 10B Problem 8

P1-10~3 minPrint

Basic Counting Number Properties Set up the equation/formula and compute; no special trick needed Use an invariant, parity, or coloring argument

On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form $bbcac,$ where $0 \le a < b < c \le 9,$ and $b$ was the average of $a$ and $c.$ How many different five-digit numbers satisfy all these properties?

Show solution without answering

Key insight

Once a and c are picked, b is their average, so it is fixed. It is a whole number exactly when a and c are both even or both odd.

Solution

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 $\boxed{\textbf{(D)}\ 20}$ .

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(D)}\ 20}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Algebra, Chapter 20: Special Functions (§20.3)