Competition · AMC preparation · step 4 of 4
AMC 8 · 2004 · #13
Grade 0 countingPick an answer.
AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Test statement I
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.
Comparing ages and seeing that two people cannot share the top spot is the Kindergarten "who is older" comparison.
K.MD.A.2Eliminate PossibilitiesTest statement II
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.
Again pure pairwise comparison — once two people are knocked out of "oldest," the third must be oldest.
If Statement II is taken as the single true statement, the case runs into a contradiction and must be thrown out.
▸ Why?
Assigning Statement II true and Statements I and III false forces Celine to be the oldest by one deduction and the youngest by another, and a single friend cannot sit at both ends of the age order.
▸ Why?
Statement II being true says Amy is not the oldest and Statement I being false says Bill is not the oldest, so the oldest spot — which one of the three friends must hold — is left to Celine, the only friend not ruled out.
▸ Why?
Statement III being false turns 'Celine is not the youngest' into 'Celine is the youngest,' but the oldest is older than each of the other friends while the youngest is younger than each of them, so chaining those comparisons makes Celine older than herself.
Test statement III
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.
Lining up the three facts side by side is the systematic-list move — write them down before drawing the conclusion.
K.MD.A.2Make A Systematic ListFinish the ranking
Amy is oldest; Celine is not youngest, so Bill is youngest and Celine is in the middle. Oldest to youngest lands on option (E).
Putting three people in age order from a few "older than" facts is exactly the Kindergarten compare-and-order skill.
K.MD.A.2Make A Systematic ListThree cases, two contradictions, one winner — this AMC 8 problem rides on the Kindergarten skill of comparing who is older.
- Test statement I
- Test statement II
- Test statement III
- Finish the ranking
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