2003 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2003 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12A solutions, or check the answer key.

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Concepts:divisibilitylogical deduction

Difficulty rating: 1500

12.

Sally has five red cards numbered 11 through 55 and four blue cards numbered 33 through 6.6. 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?

88

99

1010

1111

1212

Solution:

Since 44 divides only 44 and 55 divides only 55 among 3,4,5,6,3,4,5,6, the two ends must be R4,B4,,B5,R5.R4,B4,\dots,B5,R5.

Because 22 divides only 44 and 6,6, the next card is R2,B6,R2,B6, and since 33 divides only 33 and 6,6, the full stack is R4,B4,R2,B6,R3,B3,R1,B5,R5.R4,B4,R2,B6,R3,B3,R1,B5,R5.

The middle three cards are 6,3,3,6,3,3, which sum to 12.12.

Thus, the correct answer is E.

Problem 12 in Other Years