
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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May 7th, 2018, 02:31 AM  #1  
Member Joined: Jan 2018 From: Belgrade Posts: 43 Thanks: 2  Painted numbers Quote:
That try of the solution assumed that equations in the problem are true for 3 fixed numbers, and then the answer is no. But, colored numbers in the equations of the problem can be different (but only painted with the same color). Last edited by lua; May 7th, 2018 at 02:52 AM.  
May 7th, 2018, 03:20 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 
Think "mod 3". That is, paint the numbers 1= biue, 2= red, 3= green, 4= blue, 5= red, 6= green, etc. Every number equal to 0 mod 3 is green, every number equal to 1 mod 3 is blue, and every number equal to 2 mod 3 is red.

May 7th, 2018, 03:34 AM  #3 
Member Joined: Jan 2018 From: Belgrade Posts: 43 Thanks: 2 
OK. In that case: $\displaystyle G\equiv 0 (mod 3)$, $\displaystyle B\equiv 1 (mod 3)$ and $\displaystyle R\equiv 2 (mod 3)$. Equations from problem become: $\displaystyle B+R\equiv 1+2\equiv 0 (mod 3) = G$, $\displaystyle B+G\equiv 1+0\equiv 1 (mod 3) = B$ (not $\displaystyle R$), and $\displaystyle R+G\equiv 2+0\equiv 2 (mod 3) = R$ (not $\displaystyle B$). 
May 7th, 2018, 04:14 AM  #4 
Member Joined: Jan 2018 From: Belgrade Posts: 43 Thanks: 2 
I'll try another way. It is obvious that $\displaystyle 1$, $\displaystyle 2$ and $\displaystyle 3$ must be painted with three different colors (because $\displaystyle 1+2=3$). Then, $\displaystyle 4$ must be painted with the same color as $\displaystyle 2$ ($\displaystyle 1+3=4$). $\displaystyle 5$ must be painted differently from $\displaystyle 2$ and from $\displaystyle 3$ ($\displaystyle 2+3=5$), but differently from $\displaystyle 1$ and $\displaystyle 4$ ($\displaystyle 1+4=5$), as well. To obtain that, we would have to have 4 different colors: $\displaystyle 1$ is painted with the first color, $\displaystyle 2$ (and $\displaystyle 4$) with the second, $\displaystyle 3$ with the third. $\displaystyle 5$ must be painted differently from all of these, but it is impossible because we have only three different colors. So, answer to the question in the problem is no. 
May 7th, 2018, 05:03 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,030 Thanks: 420 
In my opinion, it is easier to think as follows. Let r, b, and g be respectively the smallest red, blue, and green numbers and, without loss of generality, r < b < g. Thus 1 is a red number. Thus 1 + b = g. How about g + 1. Well it must be blue and equals b + 2. And b + 2 + 1 must be green and so on. Thus every number not less than b is either blue or green. And that means that the sum of any two numbers not less than b must be blue or green, including the sum of a blue and a green number. But the sum any pair of blue and green numbers must be red. Contradiction. 
May 7th, 2018, 05:25 AM  #6 
Member Joined: Jan 2018 From: Belgrade Posts: 43 Thanks: 2 
I understand. Thanks!


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