Art of Problem Solving

2025 AMC 10A Problems/Problem 21: Difference between revisions

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~metrixgo
~metrixgo
It’s 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


==See also==
==See also==

Revision as of 16:39, 6 November 2025

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

It’s 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

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.