2022 AMC 10B Problem 14

Construction. Take $S = \{13, 14, \dots, 25\}$, which has $13$ elements. The smallest possible sum of two elements is $13 + 13 = 26 > 25$, so no sum lands in $S$.

Upper bound. Let $M$ be the largest element of $S$. Every other element lies in $\{1, \dots, M-1\}$. Group these numbers into pairs $\{k, M-k\}$ with $k < M - k$. If both members of a pair were in $S$, their sum $M$ would be in $S$, forbidden. Also, if $M$ is even, the number $M/2$ cannot be in $S$, since $M/2 + M/2 = M$.

• If $M$ is odd, $\{1, \dots, M-1\}$ splits into $\frac{M-1}{2}$ pairs, each contributing at most one element, so $|S| \le \frac{M-1}{2} + 1 = \frac{M+1}{2} \le \frac{26}{2} = 13$.
• If $M$ is even, there are $\frac{M-2}{2}$ pairs plus the excluded $M/2$, so $|S| \le \frac{M-2}{2} + 1 = \frac{M}{2} \le 12$.

Either way $|S| \le 13$, and $13$ is achieved.

The answer is $\boxed{\textbf{(B)}\ 13}$.

An extremal problem has two halves: an example showing the value is reachable and an argument that nothing larger works. The natural example is "all large numbers," and the bound comes from pairing numbers that sum to the maximum element, a pigeonhole argument that is the standard tool for sum-free sets. Note that "not necessarily distinct" matters: it is what excludes $M/2$.

The odd numbers $\{1, 3, \dots, 25\}$ also form a valid set (odd + odd is even), and it also has $13$ elements. Seeing two different constructions of size $13$ and none of size $14$ supports the answer, but the pairing argument is what proves it.

• Stopping at the construction and assuming a larger set must exist because 13 is the second-smallest choice.
• Forgetting that $x + x$ counts, which would allow $M/2$ into the set and suggest a bound of $14$ for even $M$.