2003 AMC 12A Problem 12

Attempt Problem 12 of the 2003 AMC 12A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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
Answer: E
Concepts:divisibilitylogical deduction
Small hint:
Among the blue numbers 3-6,3\text{-}6, the red 44 divides only 44 and the red 55 divides only 55
Big hint:
Work inward from the forced ends R4,B4R4,B4 and B5,R5B5,R5
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,R4, B4,B4, R2,R2, B6,B6, R3,R3, B3,B3, R1,R1, B5,B5, R5.R5.
The middle three cards are 6,3,3,6,3,3, which sum to 12.12.
Thus, the correct answer is E.
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