Problem

Source: 69 Polish MO 2018 Second Round - Problem 2

Tags: number theory, combinatorics, Divisors, Poland, inequalities



Let $n$ be a positive integer, which gives remainder $4$ of dividing by $8$. Numbers $1 = k_1 < k_2 < ... < k_m = n$ are all positive diivisors of $n$. Show that if $i \in \{ 1, 2, ..., m - 1 \}$ isn't divisible by $3$, then $k_{i + 1} \le 2k_{i}$.