AMC12 2003 A

AMC12 2003 A · Q12

AMC12 2003 A · Q12. It mainly tests Logic puzzles, Divisibility & factors.

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?

Sally 有五张红色卡片，编号 $1$ 到 $5$，以及四张蓝色卡片，编号 $3$ 到 $6$。她将卡片堆叠，使得颜色交替，并且每张红色卡片上的数字都能整除与其相邻的每张蓝色卡片上的数字。中间三张卡片上的数字之和是多少？

(A) 8 8

(B) 9 9

(D) 11 11

(E) 12 12

Answer

Correct choice: (E)

正确答案：（E）

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 $3+3+6=\boxed{\mathrm{(E)}\ 12}$.

设 $R_i$ 表示编号为 $i$ 的红色卡片，$B_j$ 表示编号为 $j$ 的蓝色卡片。 $B_5$ 是唯一一张能被 $R_5$ 整除的蓝色卡片，所以 $R_5$ 必须在牌堆的一端，且 $B_5$ 必须紧挨着它。 $R_1$ 是唯一另一张能整除 $B_5$ 的红色卡片，所以 $R_1$ 必须是紧挨着 $B_5$ 的另一张卡片。 $B_4$ 是唯一一张能被 $R_4$ 整除的蓝色卡片，所以 $R_4$ 必须在牌堆的一端，且 $B_4$ 必须紧挨着它。 $R_2$ 是唯一另一张能整除 $B_4$ 的红色卡片，所以 $R_2$ 必须是紧挨着 $B_4$ 的另一张卡片。 $R_2$ 不能整除 $B_3$，所以 $B_3$ 必须挨着 $R_1$，$B_6$ 必须挨着 $R_2$，而 $R_3$ 必须在中间。于是从上到下的排列为：$\{R_5,B_5,R_1,B_3,R_3,B_6,R_2,B_4,R_4\}$ 因此，中间三张卡片上的数字之和为 $3+3+6=\boxed{\mathrm{(E)}\ 12}$。

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