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Difference between revisions of "2016 UMO Problems/Problem 3"

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==Problem ==
 
==Problem ==
  
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Can each positive integer <math>1,2,3,\ldots</math> be colored either red or blue, such that for all positive integers <math>a,b,c,d</math> (not necessarily distinct), if <math>a + b + c = d</math> then <math>a,b,c,d</math> are not all the same color?
  
  
 
== Solution ==
 
== Solution ==
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Solution We claim that no such coloring exists. For the sake of contradiction, assume that such a coloring exists. Without loss of generality, we may assume that 1 is colored red. In the string of steps below, if a number n is known to be colored red, we denote it as <math>n^R</math>, and if it is known to be colored blue, we denote it as <math>n^B</math>. Each of the implications below follow from the fact that if all but one of the distinct numbers in <math>a + b + c = d</math> are the same color, then the final number must be a different color. <math>1^R + 1^R + 1^R = 3 \Rightarrow 3</math> is blue. <math>3^B + 3^B + 3^B = 9 \Rightarrow 9</math> is red. <math>1^R + 4 + 4 = 9^R \Rightarrow 4</math> is blue. Now we have <math>1^R + 1^R + 9^R = 11 \Rightarrow 11</math> is blue AND <math>3^B + 4^B + 4^B = 11 \Rightarrow 11</math> is red. This is a contradiction because <math>11</math> can only have one color. Therefore, no such coloring exists.
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== See Also ==
 
== See Also ==
 
{{UMO box|year=2016|num-b=2|num-a=4}}
 
{{UMO box|year=2016|num-b=2|num-a=4}}
  
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[[Category:]]
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[[Category: Intermediate Combinatorics Problems]]

Latest revision as of 03:09, 22 January 2019

Problem

Can each positive integer $1,2,3,\ldots$ be colored either red or blue, such that for all positive integers $a,b,c,d$ (not necessarily distinct), if $a + b + c = d$ then $a,b,c,d$ are not all the same color?


Solution

Solution We claim that no such coloring exists. For the sake of contradiction, assume that such a coloring exists. Without loss of generality, we may assume that 1 is colored red. In the string of steps below, if a number n is known to be colored red, we denote it as $n^R$, and if it is known to be colored blue, we denote it as $n^B$. Each of the implications below follow from the fact that if all but one of the distinct numbers in $a + b + c = d$ are the same color, then the final number must be a different color. $1^R + 1^R + 1^R = 3 \Rightarrow 3$ is blue. $3^B + 3^B + 3^B = 9 \Rightarrow 9$ is red. $1^R + 4 + 4 = 9^R \Rightarrow 4$ is blue. Now we have $1^R + 1^R + 9^R = 11 \Rightarrow 11$ is blue AND $3^B + 4^B + 4^B = 11 \Rightarrow 11$ is red. This is a contradiction because $11$ can only have one color. Therefore, no such coloring exists.


See Also

2016 UMO (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All UMO Problems and Solutions
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