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China National Olympiad

Number theory Difficulty 6.5 National Olympiad Find the answer

Let m,nm,n be positive integers. Find the minimum positive integer NN which satisfies the following condition. If there exists a set SS of integers that contains a complete residue system module mm such that ∣S∣=N| S | = N, then there exists a nonempty set A⊆SA \subseteq S so that n∣∑x∈Axn\mid {\sum\limits_{x \in A} x }.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let m m and n n be positive integers. We aim to find the minimum positive integer N N which satisfies the following condition: If there exists a set S S of integers that contains a complete residue system modulo m m such that ∣S∣=N |S| = N , then there exists a nonempty set A⊆S A \subseteq S so that n∣∑x∈Ax n \mid \sum_{x \in A} x .

First, let d=gcd⁡(m,n) d = \gcd(m, n) , and write m=ad m = ad and n=bd n = bd . The answer depends on the relationship between bd bd and ad(d+1)2 \frac{ad(d+1)}{2} .

The minimum positive integer N N is given by:
N={1if bd≤ad(d+1)2,bd−ad(d−1)2otherwise. N = \begin{cases} 1 & \text{if } bd \leq \frac{ad(d+1)}{2}, \\ bd - \frac{ad(d-1)}{2} & \text{otherwise}. \end{cases}

The answer is: \boxed{}{1if bd≤ad(d+1)2,bd−ad(d−1)2otherwise.\begin{cases} 1 & \text{if } bd \leq \frac{ad(d+1)}{2}, \\ bd - \frac{ad(d-1)}{2} & \text{otherwise}. \end{cases}}.

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Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.