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Taiwan IMO Selection Camp

Number theory Difficulty 4.7 AIME Prove it Taiwan

Find all composite numbers nn such that the positive divisors of nn greater than 11 can be arranged on a circle so that no two adjacent ones are relatively prime.

Solution

When n=pα,α≥2n = p^{\alpha}, \alpha \ge 2, arranging them as p,p2,…,pαp, p^2, \dots, p^{\alpha} obviously satisfies the condition.

When n=p1α1p2α2⋯pkαk,k≥3n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}, k \ge 3, first arrange p1p2p3⋯pkp1p_1 p_2 p_3 \cdots p_k p_1 in order on the circle, then arrange the multiples of pjp_j between pj−1pjp_{j-1} p_j and pjpj+1p_j p_{j+1}. The resulting arrangement satisfies the condition.

Finally, when n=paqb,a>1n = p^a q^b, a > 1, first place pqpq and p2qp^2 q on the circle, put the multiples of pp on one arc, and the multiples of qq on the other arc.

The remaining case n=pqn = pq cannot be achieved.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.