AMC 8 · 2004 · #13

Grade 0 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

logical-deductionif-then-reasoningcasework caseworkcontradiction-elementary ↑ Prerequisites: logical-deduction

📏 Short solution 💡 2 insights

Problem

Amy, Bill, and Celine all have different ages. Of the three statements — (I) Bill is the oldest, (II) Amy is not the oldest, (III) Celine is not the youngest — exactly one is true and the other two are false. Rank the three from oldest to youngest.

Pick an answer.

(A)

Bill, Amy, Celine

(B)

Amy, Bill, Celine

(C)

Celine, Amy, Bill

(D)

Celine, Bill, Amy

(E)

Amy, Celine, Bill

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

There are only three possible cases — Statement I is the true one, or II is, or III is. Tool #3 (Eliminate Possibilities) tests each case and throws out the ones that lead to a contradiction. Tool #2 (Systematic List) keeps the casework tidy: write down each case, work out what "oldest" and "youngest" have to be, and check for clashes. Whichever case survives is the answer. No algebra is needed — only careful comparing.

1STEP 1

Case 1: if I is true, II is false, so Amy is the oldest too — but I says Bill is. Two oldest people is impossible, so drop this case.

I true → II false → Amy oldest and Bill oldest — contradiction

2STEP 2

Case 2: if II is true and I is false, neither Amy nor Bill is oldest, so Celine is — but III false makes Celine youngest. Also impossible.

Amy not oldest + Bill not oldest → Celine oldest; III false → Celine youngest — contradiction

3STEP 3

Case 3: III true is the last option. I false → Bill not oldest; II false flips to Amy is the oldest; III true → Celine not youngest.

Amy oldest, Bill not oldest, Celine not youngest

4STEP 4

Amy is oldest; Celine is not youngest, so Bill is youngest and Celine is in the middle. Oldest to youngest lands on option (E).

Amy ＞ Celine ＞ Bill → (E)

Answer

Amy, Celine, Bill

Check the surviving ranking Amy ＞ Celine ＞ Bill against all three statements. (I) Bill is the oldest — Amy is, so I is FALSE. (II) Amy is not the oldest — Amy IS, so II is FALSE. (III) Celine is not the youngest — Bill is, so III is TRUE. Exactly one true and two false, matching the problem's condition. Cases 1 and 2 each produced a clear contradiction, so (E) is the only ordering that works.

💡Key takeaway

Three cases, two contradictions, one winner — this AMC 8 problem rides on the Kindergarten skill of comparing who is older.

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