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2007 AMC 10B Problem 8

2007 AMC 10B Problem 8AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 2 of 5 · about 3 min · position P1-10

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?

A)
$12$
B)
$16$
C)
$18$
D)
$20$
E)
$25$

Answer

Key insight

b is determined by a and c, and b is an integer exactly when a and c have the same parity, so count same-parity pairs a < c.

Solution

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

Why this works

When one variable is a function of the others, count the free variables only. The "average must be an integer" condition is a parity statement, and parity splits the digits into two equal-sized classes, each contributing a simple combination count.

The trap

Forgetting that b = (a + c)/2 must be a whole digit, and counting all 45 pairs a < c or a subset of them.

Common mistakes

Forgetting that b = (a + c)/2 must be a whole digit, and counting all 45 pairs a < c or a subset of them.
Excluding $a = 0$ because "digits can't start with $0$ ," even though $a$ is not the leading digit.

Techniques

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