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2009 AMC 10B Problem 11

2009 AMC 10B Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 2 of 5 · about 2 min · position P11-15

How many $7$ -digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$ , $2$ , $3$ , $3$ , $5$ , $5$ , $5$ ?

A)
$6$
B)
$12$
C)
$24$
D)
$36$
E)
$48$

Answer

Key insight

The middle digit must be the odd-count 5, and the first three digits are one 2, one 3, one 5 in some order: 3! palindromes.

Solution

A $7$ -digit palindrome has the form $abcdcba$ : the last three digits mirror the first three, and the fourth digit $d$ stands alone.

Every digit other than $d$ is used an even number of times, so $d$ must be the digit with odd multiplicity. Among $2, 2, 3, 3, 5, 5, 5$ that is $5$ , so $d = 5$ . Removing one $5$ leaves $2, 2, 3, 3, 5, 5$ , which must split as a first half $abc$ and its mirror, so $\{a, b, c\} = \{2, 3, 5\}$ with each used once.

The first three digits are therefore a permutation of $2, 3, 5$ : there are $3! = 6$ of them, and each produces exactly one palindrome.

The answer is $\boxed{\textbf{(A)}\ 6}$ .

Why this works

A palindrome is completely determined by its first half plus the middle digit, so count those instead of the whole string. Multiplicity does the rest: the digit with odd count must occupy the center, and the remaining digits pair off across the middle.

The trap

Counting arrangements of all seven digits (7!/(2!2!3!) = 210) or forgetting that the middle digit is forced.

Common mistakes

Counting arrangements of all seven digits (7!/(2!2!3!) = 210) or forgetting that the middle digit is forced.
Allowing the first half to repeat a digit (for example $2, 2, 5$ ), which would require four $2$ 's in total.

Techniques

Map the objects to something easier to count