2003 AMC 10A Problems/Problem 24

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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 the other 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 $3+3+6=12 \Rightarrow E$ .

2003 AMC 10A (Problems • 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

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2003 AMC 10A Problems/Problem 24

From AoPSWiki

Problem

Solution

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