You are given some $12$ non-zero, not necessarily distinct real numbers. Find all positive integers $k$ from $1$ to $12$, such that among these numbers you can always choose $k$ numbers whose sum has the same sign as their product, that is, either both the sum and the product are positive, or both are negative. Proposed by Anton Trygub
Problem
Source: Ukrainian Mathematical Olympiad 2024. Day 2, Problem 11.5
Tags: algebra
21.03.2024 10:56
Answer: All odd $1\leq k \leq 12$ except $11$. Let $S$ be the set with $12$ non-zero real numbers. If $k$ is even, then the product must be positive and the sum must be negative in the set $\{-1,-1,...,-1\}$ which gives contradiction. Claim: $k=1,3,5$ is achievable. Assume that there exists a set where at least one of $k=1,3,5$ is not achievable. If there exists $k$ negative numbers, then that subset has negative product and sum which is not possible because of our assumption. If there exists $k$ positive numbers, then that subset has positive product and sum which is not possible because of our assumption, again. So $12\leq 2k\iff 6\leq k$ which gives a contradiction. For $k=7$ Assume that there exists a set where $k=7$ is not achievable. There must be at most $6$ positive and $6$ negative real numbers. Since there are $12$ numbers, there are exactly $6$ positive and $6$ negative numbers. If we take a set $T$ which contains $4$ negative and $3$ positive numbers, since the product is positive, the sum must be negative. By changing a positive number in $T$ with a negative number which is not in $T$, we have negative sum and negative product which gives a contradiction. For $k=9$ Assume that there exists a set where $k=9$ is not achievable. There must be at most $8$ positive, $8$ negative real numbers. If there exists exactly $4$ positive and $8$ negative real numbers, then we can choose $4$ positive and $5$ negative numbers. The product is negative so the sum must be positive. After changing a positive in the subset with a negative which is not in the subset, the product becomes positive and the sum stays positive which gives a contradiction. If there are at least $5$ positive numbers, then we take a subset which contains $4$ positive and $5$ negative numbers. After changing a negative in the subset with a positive which is not in the subset, the product becomes positive and the sum stays positive which contradicts. For $k=11$ The set $(-19,-19,2,2,2,2,2,2,2,2,2,2)$ satisfies. Thanks, Below.
20.05.2024 19:14
For k=11 @above how about the set (-19,-19, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2) Choose any 11 of them. You either get 1 or 2 -19s. If you have 1 the sum is 1 while the product is negative. If you have 2 the sum is negative but the sum is clearly positive. The other parts of the solution are correct