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
29.03.2020 06:10
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 $1\leq j<i\leq 2018$
22.02.2022 14:19
A1=2018!, A2=2018!+2017!, A3=2018!+2017!+2016!, ...A2018=2018!+2017!+..+1!
01.01.2025 15:35
For all $n$ there exist $n$ r.c numbers as mentioned @above we can give the example as $a_i =\sum (2019-k)!$