2025 AMC 12A Problem 15

Attempt Problem 15 of the 2025 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

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15.

A set of numbers is called sum-free if whenever xx and yy are (not necessarily distinct) elements of the set, x+yx + y is not an element of the set. For example, {1,4,6}\{1, 4, 6\} and the empty set are sum-free, but {2,4,5}\{2, 4, 5\} is not. What is the greatest possible number of elements in a sum-free subset of {1,2,3,,20}?\{1, 2, 3, \ldots, 20\}?

88

99

1010

1111

1212

Answer: C
Concepts:subsetsextremal argument
Difficulty rating: 1800
Solution:

The set {11,12,,20}\{11, 12, \ldots, 20\} has 1010 elements and is sum-free, since any two elements sum to at least 22>20.22 \gt 20.

For the upper bound, let a1<a2<<aka_1 \lt a_2 \lt \cdots \lt a_k be a sum-free subset. Each difference akaia_k - a_i for i<ki \lt k cannot lie in S,S, because (akai)+ai=akS(a_k - a_i) + a_i = a_k \in S would violate sum-freeness.

These k1k - 1 differences are distinct, lie in {1,,19},\{1, \ldots, 19\}, and are disjoint from the kk elements of S.S. So k+(k1)20,k + (k - 1) \le 20, giving k10.k \le 10.

Thus, the correct answer is C.

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