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2025 AMC 12A Problem 15

Problem 15 of 25IntermediateCombinatoricsProblem-Solving Techniques

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\}?

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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 ak−aia_k - a_i for i<ki \lt k cannot lie in S,S, because (ak−ai)+ai=ak∈S(a_k - a_i) + a_i = a_k \in S would violate sum-freeness. These k−1k - 1 differences are distinct, lie in {1,…,19},\{1, \ldots, 19\}, and are disjoint from the kk elements of S.S. So k+(k−1)≤20,k + (k - 1) \le 20, giving k≤10.k \le 10. Thus, the correct answer is C.
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Tagged: subsets · extremal argument

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