AMC10 2003 A

AMC10 2003 A · Q24

AMC10 2003 A · Q24. 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 有五张红牌，编号 1 到 5，和四张蓝牌，编号 3 到 6。她将牌堆叠，使得颜色交替，并且每张红牌的数字整除相邻蓝牌的数字。中间三张牌的数字之和是多少？

(A) 8 8

(B) 9 9

(D) 11 11

(E) 12 12

Answer

Correct choice: (E)

正确答案：（E）

Solution

(E) Let R1, ..., R5 and B3, ..., B6 denote the numbers on the red and blue cards, respectively. Note that R4 and R5 divide evenly into only B4 and B5, respectively. Thus the stack must be R4, B4, ..., B5, R5, or the reverse. Since R2 divides evenly into only B4 and B6, we must have R4, B4, R2, B6, ..., B5, R5, or the reverse. Since R3 divides evenly into only B3 and B6, the stack must be R4, B4, R2, B6, R3, B3, R1, B5, R5, or the reverse. In either case, the sum of the middle three cards is 12.

（E）令 R1，…，R5 和 B3，…，B6 分别表示红色和蓝色卡片上的数字。注意到 R4 和 R5 分别只能整除 B4 和 B5。因此，这一叠必须是 R4，B4，…，B5，R5，或其逆序。由于 R2 只能整除 B4 和 B6，我们必须有 R4，B4，R2，B6，…，B5，R5，或其逆序。由于 R3 只能整除 B3 和 B6，这一叠必须是 R4，B4，R2，B6，R3，B3，R1，B5，R5，或其逆序。无论哪种情况，中间三张卡片的和都是 12。

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