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 $\boxed{\text{(C) }10}.$

~Kevin Wang

Solution 3 (Those who know)

I'm kinda surprised that this question is just the copy-and-paste version of 2022 10B Problem 14. This problem is easier yet it's at problem 21. Nice problem quality you got there, huh, just adding a fancy definition, and yay, you got a brand new problem!

https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_14

~metrixgo

I litterly remember the problem from 2022. If anything, this one's easier. Not complaining though.

Its literally just the number of elements if you only put the odd numbers in like what this should’ve been problem 10 not that semicircle problem ~grogg007

frfr I thought so too ~BOTNATE

Video Solution (In 1 Min)

https://youtu.be/V_zh78Ae8xw?si=D8dEsX4ST3JORj6x ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

2025 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2025 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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