Two distinct positive integers are called "relatively consistent" if the larger one can be written as a sum of some distinct positive divisors of the other one. Show that there exist 2018 positive integers such that any two of them are "relatively consistent"
Problem
Source: 2018 JBMO TST - Turkey, P2
Tags: number theory
p_tripa_c
29.03.2020 06:10
We claim that for any positive integer $n$, there exist $n$ positive integers which are pairwise relatively consistent.
The proof is by induction on $n$. The base case $n=1$ is trivial. Now, assume we have $k$ integers $a_1 < a_2 < \dots < a_k$ which are pairwise relatively consistent. We claim that the $k+1$ integers $2a_1, 2a_2, \dots, 2a_k, 2a_k + 1$ are pairwise relatively consistent.
Indeed, for each $1 \leq i < j \leq k$, let $S_{ij}$ be some set of factors of $a_i$ that sum to $a_j$. Then $2S_{ij}$ is a set of factors of $2a_i$ that sum to $2a_j$, so $2a_i$ and $2a_j$ are relatively consistent. Also, for each $1 \leq i \leq k$, we have that $2S_{ik} \cup \{1\}$ is a set of factors of $2a_i$ that sum to $2a_k + 1$, so $2a_i$ and $2a_k + 1$ are relatively consistent. Hence, $2a_1, 2a_2, \dots, 2a_k, 2a_k + 1$ are pairwise relatively consistent, as desired.
By induction, our initial claim holds for $n=2018$, which is the desired result.
BarisKoyuncu
29.03.2020 17:41
Take $2^{2018}, 2^{2018}+2^{2017}, 2^{2018}+2^{2017}+2^{2016} , ... , 2^{2018}+2^{2017}+2^{2016} + ...+ 2^{3}+2^{2}+2^{1}$ Let's show any 2 of these numbers are "relatively consistent". It is enough to prove that we can write $2^{2018}+2^{2017}+ ... + 2^{i}+2^{i-1}+ ... 2^j$ with sum of some distinct positive divisors of $2^{2018}+2^{2017}+ ... + 2^{i}$ for all $1\leq j<i\leq 2018$ For all k positive integers which $j\leq k\leq i-1$; $2^k|2^{2018}+2^{2017}+ ... + 2^{i}$ thus $2^j, 2^{j+1}, ... , 2^{i-1}$ are distinct divisors of $2^{2018}+2^{2017}+ ... + 2^{i}$ $2^{2018}+2^{2017}+ ... + 2^{i}$ also divisor of $2^{2018}+2^{2017}+ ... + 2^{i}$ Thus $2^j, 2^{j+1}, ... , 2^{i-1}, 2^{2018}+2^{2017}+ ... + 2^{i}$ are all distinct divisors of $2^{2018}+2^{2017}+ ... + 2^{i}$ and their sum equal to $2^{2018}+2^{2017}+ ... + 2^{i}+2^{i-1}+ ... 2^j$ so we can write $2^{2018}+2^{2017}+ ... + 2^{i}+2^{i-1}+ ... 2^j$ with sum of some distinct positive divisors of $2^{2018}+2^{2017}+ ... + 2^{i}$ for all $1\leq j<i\leq 2018$
felixsg
22.02.2022 14:19
A1=2018!, A2=2018!+2017!, A3=2018!+2017!+2016!, ...A2018=2018!+2017!+..+1!
wizixez
01.01.2025 15:35
For all $n$ there exist $n$ r.c numbers as mentioned @above we can give the example as $a_i =\sum (2019-k)!$