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Alibaba Global Mathematics Competition

Combinatorics Difficulty 7.0 National Olympiad Find the answer

Professor Ma has formulated n different but equivalent statements A1,A2,…,An.A_{1}, A_{2}, \ldots, A_{n}. Every semester, he advises a student to prove an implication Ai⇒Aj,i≠A_{i} \Rightarrow A_{j}, i \neq j. This is the dissertation topic of this student. Every semester, he has only one student, and we assume that this student finishes her/his dissertation within the semester. No dissertation should be a direct logical consequence of previously given ones. For example, if Ai⇒AjA_{i} \Rightarrow A_{j} and Aj⇒AkA_{j} \Rightarrow A_{k} have already been used as dissertation topics, Professor Ma cannot use Ai⇒AkA_{i} \Rightarrow A_{k} as a new dissertation topic, as the implication follows from the previous dissertations. What is the maximal number of students that Professor Ma can advise?

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

Solution

We will first construct an answer with 12(n+2)(n−1)\frac{1}{2}(n+2)(n-1) students. Then, we will show this is the best possible answer. Construction: First, (n-1) students sequentially prove A1⇒AiA_{1} \Rightarrow A_{i} for i=2, …,\ldots, n. Then, (n-2) students sequentially prove A2⇒AiA_{2} \Rightarrow A_{i} for i=3, …,\ldots, n. Continue this until 1 student proves An−1⇒An.A_{n-1} \Rightarrow A_{n}. Note that all implications proven so far are valid these and have the form Ai⇒AjA_{i} \Rightarrow A_{j} for i<j. Next, (n-1) students sequentially prove An⇒An−1,An−1⇒An−2,⋯ ,A2⇒A1,A_{n} \Rightarrow A_{n-1}, A_{n-1} \Rightarrow A_{n-2}, \cdots, A_{2} \Rightarrow A_{1}, which are also valid theses. The total number of theses is ((n−1)+(n−2)+⋯+1)+(n−1)=12n(n−1)+(n−1)=12(n+2)(n−1).((n-1)+(n-2)+\cdots+1)+(n-1)=\frac{1}{2}n(n-1)+(n-1)=\frac{1}{2}(n+2)(n-1).

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