AMC 10 · 2003 · #12
Easy mode Grade 4Sally has five red cards, numbered 1,2,3,4,5, and four blue cards, numbered 3,4,5,6. She piles all nine cards into one stack so the colors take turns: red, blue, red, blue, and so on. She also makes sure that each red card divides evenly into every blue card touching it. Add up the numbers on the three cards in the middle of the stack. What is that sum?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Five red cards numbered $1,2,3,4,5$ and four blue cards numbered $3,4,5,6$; The colors must alternate down the stack; Each red card must divide evenly into every blue card next to it; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$
Unknowns: The order of the nine cards in the stack; The sum of the numbers on the three middle cards
Understand
Restated: 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.
Givens: Five red cards numbered $1,2,3,4,5$ and four blue cards numbered $3,4,5,6$; The colors must alternate down the stack; Each red card must divide evenly into every blue card next to it; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #2 Make a Systematic List, #14 Extreme Principle, #1 Draw a Diagram
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.
Execute — Answer: E
4.OA.B.4 Step 1 List which red card divides each blue
- Nine cards with alternating colors must run red, blue, red, blue, red, blue, red, blue, red, so reds take both ends and each blue is squeezed between two reds.
- For each blue, write down which red numbers divide it: blue $3$ is divided by reds $1,3$; blue $4$ by reds $1,2,4$; blue $5$ by reds $1,5$; blue $6$ by reds $1,2,3$.
- This table is the whole rulebook for what may sit next to what.
💡 A red can only sit beside a blue it divides, so listing those divisors up front tells you every legal neighbor.
4.OA.B.4 Step 2 Pin the fussiest reds to the ends
- Turn the table around and ask where each red is allowed.
- Red $5$ divides only blue $5$, and red $4$ divides only blue $4$; each of them fits beside just one blue.
- But a red placed in an interior slot touches two blues and would have to divide both.
- Reds $4$ and $5$ cannot do that, so they are forced to the two ends of the stack, each with its single matching blue beside it: red $5$ next to blue $5$, and red $4$ next to blue $4$.
💡 A card that fits only one spot cannot be squeezed in the middle where it would need to fit two, so it gets pushed to a corner.
4.OA.B.4 Step 3 Chain the forced neighbors inward
- Blue $5$ needs two red neighbors, both from $\{1,5\}$; red $5$ is one, so red $1$ must be its other neighbor.
- That builds one end: red $5$, blue $5$, red $1$.
- Only reds $2$ and $3$ are left for the two inner red slots, and only blues $3$ and $6$ are left for the two inner blue slots.
- Beside blue $4$ at the far end, its other neighbor must divide $4$; between reds $2$ and $3$ only red $2$ works, so red $2$ sits next to blue $4$ and red $3$ takes the center.
- The blue between reds $3$ and $2$ must be divisible by both $3$ and $2$, which is blue $6$, leaving blue $3$ between reds $1$ and $3$ (both divide $3$).
💡 Each time a blue has only one red left that divides it, that neighbor is forced, and one forced choice hands you the next.
2.NBT.B.5 Step 4 Add the three middle cards
- The finished stack of nine cards reads $5,5,1,3,3,6,2,4,4$ from top to bottom.
- The three cards in the very middle are the fourth, fifth, and sixth: blue $3$, red $3$, and blue $6$.
- Their sum is $3+3+6=12$.
- So the sum of the numbers on the middle three cards is $12$, which is choice (E).
💡 Once the order is locked, the answer is just adding the three numbers standing in the center.
4.OA.B.4 Nine cards with alternating colors must run red, blue, red, blue, red, blue, red 4.OA.B.4 Turn the table around and ask where each red is allowed. Red $5$ divides only bl 4.OA.B.4 Blue $5$ needs two red neighbors, both from $\{1,5\}$; red $5$ is one, so red $1 2.NBT.B.5 The finished stack of nine cards reads $5,5,1,3,3,6,2,4,4$ from top to bottom. T Review
Reasonableness: Read the full stack $5,5,1,3,3,6,2,4,4$ and check every touching pair: $5\mid5$, $1\mid5$, $1\mid3$, $3\mid3$, $3\mid6$, $2\mid6$, $2\mid4$, $4\mid4$ — 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).
Alternative: Instead of starting from the ends, start from blue $5$: its only possible neighbors are reds $1$ and $5$, so it is locked between them immediately. Then blue $4$ forces reds $1$ or $2$, blue $6$ needs a red from $\{1,2,3\}$ on each side, and pushing these constraints together rebuilds the same unique stack and the same middle sum.
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Deciding which red numbers divide each blue number, which fixes every legal neighbor and forces the whole arrangement.)2.NBT.B.5Fluently add and subtract within 100 (Adding the three center cards $3+3+6$ to get the final sum $12$.)
⭐ 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.
⭐ 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.
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