2025 CMO 第 1 题

Let $a_1, a_2, \ldots, a_n$ be integers such that $a_1 > a_2 > \cdots > a_n > 1$. Let $M = \operatorname{lcm} \left( a_1, a_2, \ldots, a_n \right)$. For any finite nonempty set $X$ of positive integers, define $$f(X) = \min_{1 \leq i \leq n} \sum_{x \in X} \left\{ \frac{x}{a_i} \right\}.$$ Such a set $X$ is called *minimal* if for every proper subset $Y$ of it, $f(Y) < f(X)$ always holds.

Suppose $X$ is minimal and $f(X) \geq \frac{2}{a_n}$ . Prove that $$|X| \leq f(X) \cdot M.$$

令 $a_1, a_2, \ldots, a_n$ 为整数，使得 $a_1 > a_2 > \cdots > a_n > 1$。令 $M = \operatorname{lcm} \left( a_1, a_2, \ldots, a_n \right)$。对于任何正整数的有限非空集 $X$，定义 $$f(X) = \min_{1 \leq i \leq n} \sum_{x \in X} \left\{ \frac{x}{a_i} \right\}。$$ 这样的集合 $X$ 被称为*最小*，如果对于它的每个真子集 $Y$，$f(Y) < f(X)$ 始终成立。

假设 $X$ 是最小的并且 $f(X) \geq \frac{2}{a_n}$ 。证明 $$|X| \leq f(X) \cdot M.$$