AMC 10 · 2003 · #12

Grade 4 arithmetic
divisibility-ruleslogical-deductionfactors caseworksystematic-enumeration ↑ Prerequisites: divisibility-rules
📏 Long solution 💡 3 insights
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Problem
Stack all nine cards in a single pile so red and blue alternate, and every red card divides evenly into each blue card touching it. Then add up the three cards sitting in the middle of the pile.

Pick an answer.

(A)
8
(B)
9
(C)
10
(D)
11
(E)
12
How to solve
Strategy Eliminate Possibilities

There are too many card orders to try one by one, so the smart move is to let the divisibility rule delete options until only one stack survives. Tool #2 (Make a Systematic List) first records exactly which red cards divide each blue card, turning a vague rule into a small table. Tool #14 (Extreme Principle) then spots the fussiest cards: the two reds that divide only a single blue have nowhere flexible to go and get pinned to the ends. From there Tool #3 (Eliminate Possibilities) chains each forced choice into the next until every slot is filled. Tool #1 (Draw a Diagram) keeps the nine slots in front of us so we can see the middle three.

1STEP 1

List which red card divides each blue

Alternating nine cards puts reds at both ends; tabulate which reds divide each blue: 3:{1,3}, 4:{1,2,4}, 5:{1,5}, 6:{1,2,3}.

3:{1,3}, 4:{1,2,4}, 5:{1,5}, 6:{1,2,3}
2STEP 2

Pin the fussiest reds to the ends

Reds 4 and 5 fit beside only one blue each, so they are forced to the two ends.

5 ∣ 5 only, 4 ∣ 4 only → reds 4,5 at the ends
3STEP 3

Chain the forced neighbors inward

Chaining the forced neighbours inward fixes the whole pile as 5,5,1,3,3,6,2,4,4.

5, 5, 1, 3, 3, 6, 2, 4, 4
4STEP 4

Add the three middle cards

The middle three are 3, 3 and 6, adding to 12, choice (E).

3+3+6=12 → (E)
Answer
12
Read the full stack 5,5,1,3,3,6,2,4,4 and check every touching pair: 5 ∣ 5, 1 ∣ 5, 1 ∣ 3, 3 ∣ 3, 3 ∣ 6, 2 ∣ 6, 2 ∣ 4, 4 ∣ 4 — all eight divisions come out even, and the colors alternate the whole way. Because every step was forced, this is the only stack that works, so the middle sum 12 is not one option among many but the answer, matching choice (E).
💡Key takeaway

Find the pieces that fit in only one place, lock those down first, and each forced choice will point you to the next until the whole puzzle solves itself.

  • List which red card divides each blue
  • Pin the fussiest reds to the ends
  • Chain the forced neighbors inward
  • Add the three middle cards