AMC 8 · 2003 · #17

Grade 1 logic

Archetype Constraint Graph Coloring └ Sub-archetype Shared-Attribute Partition

logical-deductionif-then-reasoningset-partitionsystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: logical-deductionsystematic-enumeration

📏 Medium solution 💡 3 insights

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Problem

Six children — Benjamin, Jim, Nadeen, Austin, Tevyn, Sue — come from two families of three siblings each. Each child is recorded with eye color (blue or brown) and hair color (black or blond). Children in the same family share at least one of those two traits. Given Jim has brown eyes and blond hair, which two of the others are Jim's siblings?

Pick an answer.

(A)

Nadeen and Austin

(B)

Benjamin and Sue

(C)

Benjamin and Austin

(D)

Nadeen and Tevyn

(E)

Austin and Sue

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Use Matrix Logic

This is a "who belongs with whom?" puzzle with two attributes per child — exactly the trigger for Tool #4 (Matrix Logic). Build a small table (child by trait), highlight Jim's traits (Brown, Blond), and the candidates for Jim's family must share at least one highlighted trait with Jim. That immediately narrows the pool. Tool #3 (Eliminate Possibilities) then handles the answer choices: any choice containing a child who shares no trait with Jim is out, and the remaining choice must also have its two named children share a trait with each other so all three siblings are pairwise compatible.

1STEP 1

Read the data as a table: Jim is brown-eyed and blond, so a sibling shares brown eyes or blond hair — Nadeen, Austin, and Sue match.

Child&Eyes&Hair&Shares with Jim? ; Benjamin&Blue&Black&No ; Nadeen&Brown&Black&Yes (eyes) ; Austin&Blue&Blond&Yes (hair) ; Tevyn&Blue&Black&No ; Sue&Blue&Blond&Yes (hair)

2STEP 2

Benjamin and Tevyn share nothing with Jim, so cross off (B), (C), (D) — that leaves (A) and (E).

Eliminate: (B), (C), (D) → remaining: (A), (E)

3STEP 3

Siblings must also match each other: Nadeen shares nothing with Austin, but Austin and Sue share both traits, so (E) stands.

(A) Nadeen + Austin: Brown/Black vs Blue/Blond→ no shared trait ✗ (E) Austin + Sue: Blue/Blond vs Blue/Blond→ both share (eyes and hair) ✓

Answer

Austin and Sue

Verify the family split. Jim's family is {Jim, Austin, Sue}: Jim–Austin share Blond hair, Jim–Sue share Blond hair, Austin–Sue share both Blue eyes and Blond hair. The other family is {Benjamin, Nadeen, Tevyn}: Benjamin–Tevyn share Blue eyes and Black hair, Benjamin–Nadeen share Black hair, Nadeen–Tevyn share Black hair. Every pair in both families shares at least one trait, so the split is consistent. A handy shortcut also confirms it: the hair-color column has exactly three Blonds (Jim, Austin, Sue) and three Blacks (Benjamin, Nadeen, Tevyn) — splitting by hair color alone already produces two valid families of three, and that matches (E).

💡Key takeaway

Two traits per child, one shared trait per sibling pair — this AMC 8 problem is a Kindergarten "sort by category" task in disguise, and answer (E) Austin and Sue falls out the moment you split by hair color.

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