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

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Problem

Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true.

I. Bill is the oldest.
II. Amy is not the oldest.
III. Celine is not the youngest.

Rank the friends from the 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: All three friends have different ages; Statement I: Bill is the oldest; Statement II: Amy is not the oldest; Statement III: Celine is not the youngest; Exactly one of I, II, III is true; the other two are false; Answer choices give five possible orderings from oldest to youngest

Unknowns: The ordering of Amy, Bill, Celine from oldest to youngest

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #2 Make a Systematic List

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.

Execute — Answer: E

#3 Eliminate Possibilities K.MD.A.2 Step 1

Case 1: assume Statement I ("Bill is the oldest") is the true one.
Then II and III are both false.
The negation of II ("Amy is not the oldest") is "Amy IS the oldest." But Statement I already says Bill is the oldest.
Two different people cannot both be the oldest, so this case is impossible.

$$\text{I true} \Rightarrow \text{II false} \Rightarrow \text{Amy oldest} \;\text{and}\; \text{Bill oldest} \;\text{— contradiction}$$

💡 Comparing ages and seeing that two people cannot share the top spot is the Kindergarten "who is older" comparison.

#3 Eliminate Possibilities K.MD.A.2 Step 2

Case 2: assume Statement II ("Amy is not the oldest") is the true one.
Then I and III are both false.
From II true: Amy is not oldest.
From I false: Bill is not oldest.
So the oldest must be Celine.
But III false means "Celine IS the youngest." Celine cannot be both oldest and youngest, so this case is also impossible.

$$\text{Amy not oldest} + \text{Bill not oldest} \Rightarrow \text{Celine oldest}; \;\text{III false} \Rightarrow \text{Celine youngest} \;\text{— contradiction}$$

💡 Again pure pairwise comparison — once two people are knocked out of "oldest," the third must be oldest.

#2 Make a Systematic List K.MD.A.2 Step 3

Case 3: assume Statement III ("Celine is not the youngest") is the true one.
Then I and II are both false.
From I false: Bill is not the oldest.
From II false: the negation "Amy IS the oldest" holds.
From III true: Celine is not the youngest.
List what we now know.

$$\text{Amy oldest}, \;\text{Bill not oldest}, \;\text{Celine not youngest}$$

💡 Lining up the three facts side by side is the systematic-list move — write them down before drawing the conclusion.

#2 Make a Systematic List K.MD.A.2 Step 4

Finish the ranking.
Amy is the oldest, so the other two slots (middle and youngest) belong to Bill and Celine.
Since Celine is not the youngest, Bill must be the youngest, which puts Celine in the middle.
Oldest to youngest: Amy, Celine, Bill.

$$\text{Amy} > \text{Celine} > \text{Bill} \;\Rightarrow\; \textbf{(E)}$$

💡 Putting three people in age order from a few "older than" facts is exactly the Kindergarten compare-and-order skill.

[1] #3 K.MD.A.2 Case 1: assume Statement I ("Bill is the oldest") is the true one. Then II and I

[2] #3 K.MD.A.2 Case 2: assume Statement II ("Amy is not the oldest") is the true one. Then I an

[3] #2 K.MD.A.2 Case 3: assume Statement III ("Celine is not the youngest") is the true one. The

[4] #2 K.MD.A.2 Finish the ranking. Amy is the oldest, so the other two slots (middle and younge

Review

Reasonableness: 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.

Alternative: Tool #3 directly on the five answer choices: for each ordering, count how many of I, II, III come out true. (A) Bill, Amy, Celine — I true, II true, III false: two true, fails. (B) Amy, Bill, Celine — I false, II false, III false: zero true, fails. (C) Celine, Amy, Bill — I false, II true, III true: two true, fails. (D) Celine, Bill, Amy — I false, II true, III true: two true, fails. (E) Amy, Celine, Bill — I false, II false, III true: exactly one true. So (E) is the unique answer.

CCSS standards used (min grade 0)

K.MD.A.2 Directly compare two objects with a measurable attribute in common (Comparing ages pairwise ("oldest," "youngest," "not the oldest") and ordering the three friends from oldest to youngest once the constraints pin each one down.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 0)

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