2025 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 24 Easy

Call a positive integer fair if no digit is used more than once, it has no 0s, and no digit ins adjacent to two greater digits. For example, 196, 23, and 12463 are fair, but 1546, 320, and 34321 are not fair. How many fair positive integers are there?

  • A.

    511

  • B.

    2584

  • C.

    9841

  • D.

    17711

  • E.

    19682

Answer:C

For k selected digits, let a_k count arrangements where no digit is adjacent to two larger digits.

Base cases: a_1 = 1, a_2 = 2.

For k \geq 3, the smallest digit must occupy either the first or last position.

Each placement corresponds bijectively to arranging k-1 digits with the same constraint, giving a_k = 2a_{k-1}, hence a_k = 2^{k-1}.

Total fair integers: \sum_{k=1}^{9}\binom{9}{k} \cdot 2^{k-1} = \frac{\sum_{k=0}^{9}\binom{9}{k} \cdot 2^k - 1}{2} = \frac{(2+1)^9 - 1}{2} = \frac{3^9 - 1}{2} = 9841

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