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Difference between revisions of "2003 AMC 12A Problems/Problem 12"
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{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #12]] and [[2003 AMC 10A Problems|2003 AMC 10A #24]]}}
 
== Problem ==
 
== Problem ==
 
Sally has five red cards numbered <math>1</math> through <math>5</math> and four blue cards numbered <math>3</math> through <math>6</math>. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?  
 
Sally has five red cards numbered <math>1</math> through <math>5</math> and four blue cards numbered <math>3</math> through <math>6</math>. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?  
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<math>R_2</math> doesn't evenly divide <math>B_3</math>, so <math>B_3</math> must be next to <math>R_1</math>, <math>B_6</math> must be next to <math>R_2</math>, and <math>R_3</math> must be in the middle.  
 
<math>R_2</math> doesn't evenly divide <math>B_3</math>, so <math>B_3</math> must be next to <math>R_1</math>, <math>B_6</math> must be next to <math>R_2</math>, and <math>R_3</math> must be in the middle.  
  
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This yeilds the following arrangement from top to bottom: <math>\{R_5,B_5,R_1,B_3,R_3,B_6,R_2,B_4,R_4\}</math>
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This yields the following arrangement from top to bottom: <math>\{R_5,B_5,R_1,B_3,R_3,B_6,R_2,B_4,R_4\}</math>
  
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Therefore, the sum of the numbers on the middle three cards is <math>3+3+6=12 \Rightarrow E</math>.  
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Therefore, the sum of the numbers on the middle three cards is <math>3+3+6=\boxed{\mathrm{(E)}\ 12}</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=Q30Dt8q7vm4&feature=youtu.be
  
 
== See Also ==
 
== See Also ==
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*[[2003 AMC 12A Problems]]
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{{AMC10 box|year=2003|ab=A|num-b=23|num-a=25}}
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*[[2003 AMC 12A/Problem 11|Previous Problem]]
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{{AMC12 box|year=2003|ab=A|num-b=11|num-a=13}}
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*[[2003 AMC 12A/Problem 13|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 17:04, 28 December 2020

The following problem is from both the 2003 AMC 12A #12 and 2003 AMC 10A #24, so both problems redirect to this page.

Problem

Sally has five red cards numbered $1$ through $5$ and four blue cards numbered $3$ through $6$. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?

$\mathrm{(A) \ } 8\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } 12$

Solution

Let $R_i$ and $B_j$ designate the red card numbered $i$ and the blue card numbered $j$, respectively.

$B_5$ is the only blue card that $R_5$ evenly divides, so $R_5$ must be at one end of the stack and $B_5$ must be the card next to it.

$R_1$ is the only other red card that evenly divides $B_5$, so $R_1$ must be the other card next to $B_5$.

$B_4$ is the only blue card that $R_4$ evenly divides, so $R_4$ must be at one end of the stack and $B_4$ must be the card next to it.

$R_2$ is the only other red card that evenly divides $B_4$, so $R_2$ must be the other card next to $B_4$.

$R_2$ doesn't evenly divide $B_3$, so $B_3$ must be next to $R_1$, $B_6$ must be next to $R_2$, and $R_3$ must be in the middle.

This yields the following arrangement from top to bottom: $\{R_5,B_5,R_1,B_3,R_3,B_6,R_2,B_4,R_4\}$

Therefore, the sum of the numbers on the middle three cards is $3+3+6=\boxed{\mathrm{(E)}\ 12}$.

Video Solution

https://www.youtube.com/watch?v=Q30Dt8q7vm4&feature=youtu.be

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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