Difference between revisions of "2020 OIM Problems/Problem 3"
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An "''operation''" consists of choosing two elements <math>a_k</math> and <math>a_l</math> of a sequence, with <math>k \ne l</math>, | An "''operation''" consists of choosing two elements <math>a_k</math> and <math>a_l</math> of a sequence, with <math>k \ne l</math>, | ||
and replace <math>a_l</math> with <math>a'_l</math> Show that, given a collection of <math>2^n-1</math> ''Limeñan'' sequences, each formed by <math>n</math> integers numbers, there are two of them, say <math>\beta</math> and <math>\gamma</math>, such that it is possible to transform <math>\beta</math> into <math>\gamma</math> by a finite number of operations. | and replace <math>a_l</math> with <math>a'_l</math> Show that, given a collection of <math>2^n-1</math> ''Limeñan'' sequences, each formed by <math>n</math> integers numbers, there are two of them, say <math>\beta</math> and <math>\gamma</math>, such that it is possible to transform <math>\beta</math> into <math>\gamma</math> by a finite number of operations. | ||
| − | + | '''Clarification:''' If all the elements of a sequence are equal, then that sequence is not ''Limeña''. | |
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com | ||
Latest revision as of 09:38, 23 December 2023
Problem
Let 
be an integer. A sequence 
of 
integers is "Limeña" (from Lima, Perú) if
![\[gcd \left\{ a_i - a_j | a_i > a_j, 1 \le i, j \le n\right\} = 1\]](http://latex.artofproblemsolving.com/8/0/7/8076851153da59ac38cadc3d80430249a908f382.png)
An "operation" consists of choosing two elements 
and 
of a sequence, with 
,
and replace with 
Show that, given a collection of 
Limeñan sequences, each formed by integers numbers, there are two of them, say 
and 
, such that it is possible to transform into by a finite number of operations.
Clarification: If all the elements of a sequence are equal, then that sequence is not Limeña.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
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