2025 AMC 10A Problem 21

Ten is attainable. Two different sets of size $10$ work.

• The top half $\{11, 12, \ldots, 20\}$: any two of its elements (equal or not) sum to at least $22$, which is outside $\{1, \ldots, 20\}$ altogether.
• The odd numbers $\{1, 3, 5, \ldots, 19\}$: a sum of two odd numbers is even, so it is never one of them.

Eleven is impossible. Let $S$ be sum-free with $|S| = k$ and let $m$ be its largest element, so $m \le 20$.

For any $x \in S$ we cannot have $m - x \in S$, since then $x + (m-x) = m$ would be an element of $S$. So the "mirror" map $x \mapsto m - x$ sends $S$ into the complement of $S$.

Apply this to the $k-1$ elements of $S \setminus \{m\}$. Each such $x$ satisfies $1 \le x \le m-1$, so its mirror $m - x$ also lies in $\{1, 2, \ldots, m-1\}$, and the map is injective. We therefore have two sets inside $\{1, 2, \ldots, m-1\}$:
$$S \setminus \{m\} \quad \text{and} \quad \{\,m - x : x \in S \setminus \{m\}\,\},$$
each of size $k-1$, and by the previous paragraph they are disjoint. Counting,
$$2(k-1) \le m - 1 \le 19,$$
so $k - 1 \le 9.5$, and since $k$ is an integer, $k \le 10$.

The greatest possible size is $10$.

The answer is $\boxed{\textbf{(C)}\ 10}$.

The extremal principle — name the largest element and ask what it forbids — converts a global condition into a statement about a single number. Here $m$ splits $\{1, \ldots, m-1\}$ into pairs $\{x, m-x\}$ (with $m/2$ alone if $m$ is even), and sum-freeness says at most one member of each pair is in $S$; the disjointness argument above is just that pairing counted in one line, which also disposes of the even case since $m/2 + m/2 = m$ rules out $m/2$ automatically. The same bound shows that a sum-free subset of $\{1, \ldots, n\}$ has at most $\lceil n/2 \rceil$ elements, and the top half always attains it.

One can also bound using the largest element without mirroring the whole set. List the elements of $S$ in increasing order as $x_1 < x_2 < \cdots < x_k = m$. The $k-1$ differences
$$m - x_1 > m - x_2 > \cdots > m - x_{k-1} \ge 1$$
are distinct, strictly positive, and none of them is in $S$. Together with the $k$ elements of $S$ itself, that is $(k-1) + k = 2k-1$ distinct numbers in $\{1, \ldots, 20\}$, so $2k - 1 \le 20$ and $k \le 10$.

• Taking {10, 11, ..., 20}, whose 11 elements have no two distinct members summing into the set, and forgetting that x = y is allowed so 10 + 10 = 20 is a violation; that lands on choice (D).
• Finding one $10$-element example and stopping, without any argument that $11$ is impossible; with the choices ranging up to $12$ that is a guess rather than a solution.
• Assuming the optimum must be the odd numbers or the top half and concluding that a mixed set could never be as large; the bound, not the examples, is what rules out $11$.