2025 AMC 10A Problem 21

This is a "greatest possible" question, so it has two halves: find a set that works, then show nothing bigger can work.

Part 1: ten numbers is possible.

1. Take the top half, $11, 12, \ldots, 20$. The smallest sum you can make is $11 + 11 = 22$, which is bigger than $20$. So no sum lands back in the set. That is $10$ numbers.
2. (Another set that works: the odd numbers $1, 3, 5, \ldots, 19$. Odd plus odd is always even, so a sum is never one of them.)

Part 2: eleven numbers is impossible.

1. The largest number in the set is the one that limits everything, so look at it first. Suppose the largest number is $20$.
2. Pair up the numbers below $20$ so that each pair adds to $20$:

The number $10$ is left over without a partner.

1. Because $20$ is in the set, the set cannot contain both numbers of any pair. For example, if it had $3$ and $17$, then $3 + 17 = 20$ would be in the set, which breaks the rule.
2. The set cannot contain $10$ either, because $10 + 10 = 20$. (The problem says the two numbers may be the same.)
3. So below $20$ the set has at most one number from each of the $9$ pairs. Adding $20$ itself gives at most $9 + 1 = 10$ numbers.
4. If the largest number is smaller than $20$, the same pairing works, with even fewer numbers below it to choose from. For example, if the largest is $19$, the pairs $1{+}18, 2{+}17, \ldots, 9{+}10$ give at most $9$ numbers below it, and again at most $10$ in total.

Ten is possible and eleven is not, so the greatest size is $10$.

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

When a question asks how many numbers you can pick without breaking a rule, look at one special number (here the largest) and see what it forbids. Pairing the other numbers so that each pair "fights" over one spot turns the question into counting pairs. The same idea shows that from $1$ to $n$ you can never pick more than about half the numbers, and the top half always works.

Count forbidden numbers instead of pairs. Say the set has $k$ numbers and the largest is $m$. For each of the other $k - 1$ numbers $x$, the number $m - x$ cannot be in the set (otherwise $x + (m - x) = m$). These $k - 1$ "forbidden" numbers are all different from each other, because different values of $x$ give different values of $m - x$. So there are $k$ numbers in the set, plus $k - 1$ different numbers that are not in the set, all between $1$ and $20$. That is $2k - 1 \le 20$, so $k \le 10$.

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