Let $A_1, A_2, \ldots , A_m$ be subsets of the set $\{ 1,2, \ldots , n \}$, such that the cardinal of each subset $A_i$, such $1 \le i \le m$ is not divisible by $30$, while the cardinal of each of the subsets $A_i \cap A_j$ for $1 \le i,j \le m$, $i \neq j$ is divisible by $30$. Prove \begin{align*} 2m - \left \lfloor \frac{m}{30} \right \rfloor \le 3n \end{align*}
Problem
Source: Balkan MO ShortList 2009 C2
Tags: combinatorics, linear algebra, set theory
04.11.2021 01:02
Bumping!
04.11.2021 19:15
Can someone post the solution?!
06.11.2021 13:53
Does anybody have any ideas?
12.11.2021 22:18
Anyone…?
12.12.2021 13:15
Bump again
02.08.2022 05:05
I learnt about this intriguing problem while listening to Teacher Yinghua Ai (艾颖华)'s lecture this morning. This is hard (just notice the AoPS thread hasn't got a full solution for 13 years). Ai told us the problem has the background of Oddtown Theorem, and the proof of the theorem requires using linear algebra (or, there is a combinatorial proof but is literally the 'translation' of linear algebra proof). In other words, we can generalize this problem to the following: Generalization wrote: Let $A_1, \cdots, A_m \subseteq \{1, 2, \cdots, n\}$ such that $|A_i|\not\equiv 0 \pmod l$ and $|A_i \cap A_j|\equiv 0 \pmod l$. Determine the upper bound for $m$. I'll introduce the beautiful results: $\quad \CIRCLE$ When $l$ is a prime, the answer is $\boxed {m\leqslant l}$. $\quad \CIRCLE$ When $l$ is composite, let $\omega(l)$ denote the number of prime factors of $l$. Then the answer is $\boxed {m\leqslant n\cdot\omega(l)}$. The first can be deduced also by linear algebra, and combining China MO 2021/2 we have the second result. But we may ask, is those the best bound? In fact, NO. In 1988(1989?), someone optimized the bound specially for $l=6$ and $l=30$. (Mem. The original thesis is hard to look up) He proved for $\pmod 6$ we obtain $\boxed{m\leqslant 2n-2\log_{2}n}$ and for $\pmod {30}$ we obtain $\boxed{m\leqslant 3n-2\log_{2}(2n)}$. Later on in 2018, a famous professor in combinatorics held a lecture specifically talked about the above newest bounds and claimed that "it cannot be optimized further". Looking back at what we want -- WAY TOO STRONG. So Ai questions the correctness of the problem statement. A short time ago, Ai gave the problem to Cheng Jiang (江城), IMO 2022 China team member, and Jiang hadn't solved it. So I'm afraid to tell you that this problem is still OPEN.
12.08.2022 05:12
JG666 wrote: When $l$ is a prime, the answer is $\boxed {m\leqslant l}$. Minor Mistake: Actually, when $l$ is a prime, the answer is $\boxed {m\leqslant nl}$. It seems that you copied Ai's answer(maybe I should say, his whiteboard ) without thinking twice.
12.08.2022 07:27
So this problem is really still open ?