Art of Problem Solving

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

Solution 3 (???)

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

See also

2025 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2025 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.