A set $S$ of positive integers is said to be channeler if for any three distinct numbers $a,b,c \in S$, we have $a\mid bc$, $b\mid ca$, $c\mid ab$. a) Prove that for any finite set of positive integers $ \{ c_1, c_2, \ldots, c_n \} $ there exist infinitely many positive integers $k$, such that the set $ \{ kc_1, kc_2, \ldots, kc_n \} $ is a channeler set. b) Prove that for any integer $n \ge 3$ there is a channeler set who has exactly $n$ elements, and such that no integer greater than $1$ divides all of its elements.
Problem
Source: http://oim2013.opm.org.pa/pdfs/examen_pt.pdf
Tags: number theory, least common multiple, number theory proposed
professordad
13.08.2014 20:40
$k = m \cdot \text{lcm} \ (c_1, c_2, \cdots, c_n)$ where $m \in \mathbb{Z}^+$.
Davi Medeiros
13.08.2014 21:03
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bcp123
13.08.2014 21:05
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mavropnevma
13.08.2014 21:38
b) Consider $n$ distinct primes $p_i$, $1\leq i \leq n$, and take $c_i = \dfrac {1}{p_i}\prod_{j=1}^n p_j$.
jasperE3
30.03.2022 17:27
$(\text a)$ Let $k=d\cdot\prod_{i=1}^nc_i$ for any $d\ge1$. Clearly this works for $n=1$ and $n=2$, since the channeler condition disappears, so assume $n\ge3$. Then for any distinct $a,b,c\in\{1,2,\ldots,n\}$ we have: $$kc_a=dc_a\cdot\prod_{i=1}^nc_i\mid dc_a\cdot\prod_{i=1}^nc_i\cdot\frac{c_bc_c}{c_a}\cdot\prod_{i=1}^nc_i\mid d^2c_bc_c\cdot\left(\prod_{i=1}^nc_i\right)^2=k^2c_bc_c$$and by symmetry $\{kc_1,kc_2,\ldots,kc_n\}$ is channeler. (actually, all multiples of $\operatorname{lcm}(c_i)$ work) $(\text b)$ Let $S_n=\bigcup_{k=1}^n\left\{\frac{p_n\#}{p_k}\right\}$ for any $n\ge3$, where $p_i$ is the $i$th prime and $i\#$ is the product of primes less than or equal to $i$. We know that $S_n$ is a set of positive integers, $|S_n|=n$, and that $\gcd(S_n)=1$, so it suffices to show that $S_n$ is channeler. Indeed, for any $\frac{p_n\#}{p_a},\frac{p_n\#}{p_b},\frac{p_n\#}{p_c}\in S_n$ we have: $$\frac{p_n\#}{p_a}\mid p_n\#\mid p_n\#\cdot\frac{p_n\#}{p_a\cdot p_b}=\frac{p_n\#}{p_b}\cdot\frac{p_n\#}{p_c}$$so $S_n$ is channeler.