2025 AMC 10A Problems/Problem 21

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

$\textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12$

Solution 1

Let our subset be $\{11,12,13,...,20\}.$ If we add any one element from the set $\{1,2,3,...,10\},$ we will have to remove at least one element from our current subset. Hence, the size of our set cannot exceed $\boxed{\text{(C) }10}.$

~Tacos_are_yummy_1

Solution 2

Let our subset be $\{1,3,5,...,19\}.$ Since odd numbers + odd numbers will always sum to an even number, this subset holds true. Leo, the addition of any even number will result in a violation of the rule, so the maximum number of elements is $\boxed{\text{(C) }10}.$

~KMS888

2025 AMC 10A (Problems • Answer Key • Resources)
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2025 AMC 12A (Problems • Answer Key • Resources)
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2025 AMC 10A Problems/Problem 21

Solution 1

Solution 2

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