Each number from the set $\{1,2,3,4,5,6,7,8\}$ is either colored red or blue, following these rules: a) The number $4$ is colored red, and there is at least one number colored blue. b) If two numbers $x, y$ have different colors and $x + y \leq 8$, then the number $x + y$ is colored blue. c) If two numbers $x, y$ have different colors and $x \cdot y \leq 8$, then the number $x \cdot y$ is colored red. Find all possible ways the numbers from this set can be colored.
Problem
Source: Argentina TST 2011, Problem 1
Tags: combinatorics proposed, combinatorics
Blitzkrieg97
22.09.2014 19:36
it's easy solution by considering variants :
" 1 " is colored with blue,because if it was red,by given "a" condition we have 1 blue number ( let us say it's x) and if we take the product of $ 1*x $ it should be colored with red (by condition "c"),but as we assumed it was blue,so it's contradiction.By condition "b", "5" is colored with blue ($1 + 4 $).And now we have to consider these four variants : 1. "2" and "3" are colored with red ; 2. "2" and "3" are colored with blue ; 3. "2" is colored with red and 3 is colored with blue and 4. "2" is colored with blue and "3" is colored with red.First of,let us consider first one : this fails easily,because by condition "b", $1+2$ should be colored with blue ; 2nd : if both are blue,then $2+4$ is blue,$3+4$ is also blue and $2*4$ is red (this is one solution) ; 3rd : $2*3$ is colored with red,$2+5$ is colored with blue and "8" is allowed to be colored with both colors (this is 2nd and 3rd solution) ; 4th : this fails with condition b ($1 + 3$ is colored with red).So the solutions are : I. 1(blue),2(blue),3(blue),4(red),5(blue),6(blue),7(blue),8(red) II.1(blue),2(red),3(blue),4(red),5(blue),6(red),7(blue),8(red) III. 1(blue),2(red),3(blue),4(red),5(blue),6(red),7(blue),8(blue).
AK001
05.08.2021 23:03
Blitzkrieg97 wrote: it's easy solution by considering variants :
" 1 " is colored with blue,because if it was red,by given "a" condition we have 1 blue number ( let us say it's x) and if we take the product of $ 1*x $ it should be colored with red (by condition "c"),but as we assumed it was blue,so it's contradiction.By condition "b", "5" is colored with blue ($1 + 4 $).And now we have to consider these four variants : 1. "2" and "3" are colored with red ; 2. "2" and "3" are colored with blue ; 3. "2" is colored with red and 3 is colored with blue and 4. "2" is colored with blue and "3" is colored with red.First of,let us consider first one : this fails easily,because by condition "b", $1+2$ should be colored with blue ; 2nd : if both are blue,then $2+4$ is blue,$3+4$ is also blue and $2*4$ is red (this is one solution) ; 3rd : $2*3$ is colored with red,$2+5$ is colored with blue and "8" is allowed to be colored with both colors (this is 2nd and 3rd solution) ; 4th : this fails with condition b ($1 + 3$ is colored with red).So the solutions are : I. 1(blue),2(blue),3(blue),4(red),5(blue),6(blue),7(blue),8(red) II.1(blue),2(red),3(blue),4(red),5(blue),6(red),7(blue),8(red) III. 1(blue),2(red),3(blue),4(red),5(blue),6(red),7(blue),8(blue).
Can this be a solution BRBRBRBR? B = blue, R = red
AK001
05.08.2021 23:04
Blitzkrieg97 wrote: it's easy solution by considering variants :
" 1 " is colored with blue,because if it was red,by given "a" condition we have 1 blue number ( let us say it's x) and if we take the product of $ 1*x $ it should be colored with red (by condition "c"),but as we assumed it was blue,so it's contradiction.By condition "b", "5" is colored with blue ($1 + 4 $).And now we have to consider these four variants : 1. "2" and "3" are colored with red ; 2. "2" and "3" are colored with blue ; 3. "2" is colored with red and 3 is colored with blue and 4. "2" is colored with blue and "3" is colored with red.First of,let us consider first one : this fails easily,because by condition "b", $1+2$ should be colored with blue ; 2nd : if both are blue,then $2+4$ is blue,$3+4$ is also blue and $2*4$ is red (this is one solution) ; 3rd : $2*3$ is colored with red,$2+5$ is colored with blue and "8" is allowed to be colored with both colors (this is 2nd and 3rd solution) ; 4th : this fails with condition b ($1 + 3$ is colored with red).So the solutions are : I. 1(blue),2(blue),3(blue),4(red),5(blue),6(blue),7(blue),8(red) II.1(blue),2(red),3(blue),4(red),5(blue),6(red),7(blue),8(red) III. 1(blue),2(red),3(blue),4(red),5(blue),6(red),7(blue),8(blue).
Also another quick approach is to see 1 & 3 must have same color ( since 1 + 3 = 4)
AgentC
28.12.2023 16:43
AK001 wrote: Can this be a solution BRBRBRBR? B = blue, R = red Isn't it mentioned above?
Solrac-sam
24.11.2024 19:57
Blitzkrieg97 wrote: it's easy solution by considering variants :
" 1 " is colored with blue,because if it was red,by given "a" condition we have 1 blue number ( let us say it's x) and if we take the product of $ 1*x $ it should be colored with red (by condition "c"),but as we assumed it was blue,so it's contradiction.By condition "b", "5" is colored with blue ($1 + 4 $).And now we have to consider these four variants : 1. "2" and "3" are colored with red ; 2. "2" and "3" are colored with blue ; 3. "2" is colored with red and 3 is colored with blue and 4. "2" is colored with blue and "3" is colored with red.First of,let us consider first one : this fails easily,because by condition "b", $1+2$ should be colored with blue ; 2nd : if both are blue,then $2+4$ is blue,$3+4$ is also blue and $2*4$ is red (this is one solution) ; 3rd : $2*3$ is colored with red,$2+5$ is colored with blue and "8" is allowed to be colored with both colors (this is 2nd and 3rd solution) ; 4th : this fails with condition b ($1 + 3$ is colored with red).So the solutions are : I. 1(blue),2(blue),3(blue),4(red),5(blue),6(blue),7(blue),8(red) II.1(blue),2(red),3(blue),4(red),5(blue),6(red),7(blue),8(red) III. 1(blue),2(red),3(blue),4(red),5(blue),6(red),7(blue),8(blue).
Nótese que si 3 y 5 tienen el mismo color, en este caso azul, por la condición b, 8 debería ser rojo, la solución 3 queda descartada, también nótese que si 2 y 4 tienen el mismo color, por la condición c, 8 debería ser azul, así que la solución 2 queda descartada también