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Difference between revisions of "2013 OIM Problems/Problem 5"

(Created page with "== Problem == Let <math>A</math> and <math>B</math> be two sets such that: 1. <math>A \cup B</math> is the set of positive integers. 2. <math>A \cap B</math> is empty. 3. I...")
 
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== Solution ==
 
== Solution ==
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{{solution}}
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We know that <math>2027</math> and <math>2029</math> are primes greater than <math>2013</math> by testing. As a result, let some number <math>x</math> be in <math>A</math>. Then <math>x+2029</math> is in <math>B</math>, so <math>(x+2029)-2027=x+2</math> is in <math>A</math>. We can also perform this in the opposite direction; given some <math>x\in A</math>, <math>x+2027</math> is in <math>B</math>, so <math>(x+2027)-2029=x-2</math> is in <math>A</math>.
  
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We can repeat this process, and we can easily prove using induction that the set <math>A</math> contains all of the positive integers <math>x+2k</math> for integers <math>k</math>; therefore, <math>A</math> must contain either only the even integers or only the odd integers. Clearly, <math>B</math> must then contain only the odd integers or only the even integers, respectively. Both of these cases work (because in order for two positive integers to have a difference as such a prime, they must differ by an odd integer, and separating odds and evens facilitates this). As a result, the two solutions are
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<cmath>\boxed{A=\{x|x\equiv0\pmod{2}\},B=\{x|x\equiv1\pmod{2}\}}</cmath>
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<cmath>\text{or}</cmath>
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<cmath>\boxed{A=\{x|x\equiv1\pmod{2}\},B=\{x|x\equiv0\pmod{2}\}}</cmath>
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~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
 
== See also ==
 
== See also ==
 
[[OIM Problems and Solutions]]
 
[[OIM Problems and Solutions]]

Latest revision as of 19:32, 29 April 2025

Problem

Let $A$ and $B$ be two sets such that:

1. $A \cup B$ is the set of positive integers.

2. $A \cap B$ is empty.

3. If two positive integers have as difference a prime greater than 2013, then one of them is in $A$ and the other in $B$.

Find all the possibilities for sets $A$ and $B$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

We know that $2027$ and $2029$ are primes greater than $2013$ by testing. As a result, let some number $x$ be in $A$. Then $x+2029$ is in $B$, so $(x+2029)-2027=x+2$ is in $A$. We can also perform this in the opposite direction; given some $x\in A$, $x+2027$ is in $B$, so $(x+2027)-2029=x-2$ is in $A$.

We can repeat this process, and we can easily prove using induction that the set $A$ contains all of the positive integers $x+2k$ for integers $k$; therefore, $A$ must contain either only the even integers or only the odd integers. Clearly, $B$ must then contain only the odd integers or only the even integers, respectively. Both of these cases work (because in order for two positive integers to have a difference as such a prime, they must differ by an odd integer, and separating odds and evens facilitates this). As a result, the two solutions are \[\boxed{A=\{x|x\equiv0\pmod{2}\},B=\{x|x\equiv1\pmod{2}\}}\] \[\text{or}\] \[\boxed{A=\{x|x\equiv1\pmod{2}\},B=\{x|x\equiv0\pmod{2}\}}\] ~ eevee9406

See also

OIM Problems and Solutions

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