AMC 10 · 2003 · #24
Grade 4 number-theoryPick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
List which red card divides each blue
Colors must alternate, so reds take both ends and each blue sits between two reds; list which reds divide each blue.
A red can only sit beside a blue it divides, so listing those divisors up front tells you every legal neighbor.
4.OA.B.4Make A Systematic ListPin the fussiest reds to the ends
Red 5 divides only blue 5 and red 4 only blue 4, but an interior red must divide two blues — so reds 4 and 5 are forced to the ends.
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.
A card that fits only one spot cannot be squeezed into the middle, so it gets pushed to an end.
▸ Why?
A red can only sit beside a blue it divides, so the divisor list fixes every legal neighbour.
▸ Why?
With every other position ruled out, the remaining one is forced.
Chain the forced neighbors inward
Blue 5 then forces red 1, blue 4 forces red 2, and red 3 lands in the center: 5,5,1,3,3,6,2,4,4.
Each time a blue has only one red left that divides it, that neighbor is forced, and one forced choice hands you the next.
4.OA.B.4Eliminate PossibilitiesAdd the three middle cards
The fourth, fifth, and sixth cards are blue 3, red 3, blue 6, so the middle sum is 12, choice (E).
Once the order is locked, the answer is just adding the three numbers standing in the center.
2.NBT.B.5Draw A DiagramFind 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