Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Prove it Romania

Let $k$ and $n$ be integer numbers with $2 \le k \le n - 1$ . Consider a set $A$ of $n$ real numbers such that the sum of any $k$ distinct elements of $A$ is a rational number. Prove that all elements of the set $A$ are rational numbers.

Solution

The difference of any two elements from $A$ is a rational number. To show this, let $x \ne y \in A$ and choose other $k-1$ elements of $A$ – the choice can be made, for $k-1 \le n-2$ . Denote $s$ the sum of the $k-1$ elements and apply the hypothesis to infer that $x+s$ and $y+s$ are both rational numbers. Subtracting we get $x-y \in \mathbb{Q}$ , as claimed.

Now let $\alpha \in A$ . Rewrite the elements of $A$ as $\alpha + q_i$ , $q_i \in \mathbb{Q}$ . The sum of $k$ distinct elements is equal to $k\alpha + \sum q_i \in \mathbb{Q}$ , hence $\alpha \in \mathbb{Q}$ , as needed.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.