MathDB

Problem 4

Part of 2024 Kyiv City MO Round 1

Problems(3)

Make S nonnegative

Source: Kyiv City MO 2024 Round 1, Problem 7.4

1/28/2024
For real numbers a1,a2,…,a200a_1, a_2, \ldots, a_{200}, we consider the value S=a1a2+a2a3+…+a199a200+a200a1S = a_1a_2 + a_2a_3 + \ldots + a_{199}a_{200} + a_{200}a_1. In one operation, you can change the sign of any number (that is, change aia_i to −ai-a_i), and then calculate the value of SS for the new numbers again. What is the smallest number of operations needed to always be able to make SS nonnegative?
Proposed by Oleksii Masalitin
algebracombinatoricsOperation
Maestro saves algebra

Source: Kyiv City MO 2024 Round 1, Problem 8.4

1/28/2024
Positive real numbers a1,a2,…,a2024a_1, a_2, \ldots, a_{2024} are arranged in a circle. It turned out that for any i=1,2,…,2024i = 1, 2, \ldots, 2024, the following condition holds: aiai+1<ai+2a_ia_{i+1} < a_{i+2}. (Here we assume that a2025=a1a_{2025} = a_1 and a2026=a2a_{2026} = a_2). What largest number of positive integers could there be among these numbers a1,a2,…,a2024a_1, a_2, \ldots, a_{2024}?
Proposed by Mykhailo Shtandenko
algebra
Fantastic polynomial problem

Source: Kyiv City MO 2024 Round 1, Problem 10.4

1/28/2024
For a positive integer nn, does there exist a permutation of all its positive integer divisors (d1,d2,…,dk)(d_1 , d_2 , \ldots, d_k) such that the equation dkxk−1+…+d2x+d1=0d_kx^{k-1} + \ldots + d_2x + d_1 = 0 has a rational root, if:
a) n=2024n = 2024; b) n=2025n = 2025?
Proposed by Mykyta Kharin
algebrapolynomialnumber theory