Problem 11 · 2004 AMC 8
Medium
Logic & Word Problems
process-of-elimination
The numbers −2, 4, 6, 9 and 12 are rearranged according to these rules: The largest isn't first, but it is in one of the first three places. The smallest isn't last, but it is in one of the last three places. The median isn't first or last. What is the average of the first and last numbers?
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Answer: C — 6.5.
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Hint 1 of 2
The question only asks about the endpoints, so don't solve the full ordering — just figure out which numbers are banned from the ends. Each rule kicks one specific number off at least one end.
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Hint 2 of 2
The strategy is answer only what's asked via elimination: rather than placing all five numbers, rule out who can't be on the ends. Whoever's left must be on the ends — and you never needed the middle three.
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Approach: rule out the ends, ignore the middle
- Identify the three special values: largest 12, smallest −2, median 6. Now read the rules as bans on the endpoints: the largest 'isn't first' and lives in the first three (so not last either) → 12 is off both ends; the smallest 'isn't last' and lives in the last three (so not first) → −2 is off both ends; the median 'isn't first or last' → 6 is off both ends.
- Three of the five numbers are forbidden from both endpoints, so the two endpoints must be the survivors: 4 and 9.
- Average of the ends: (4 + 9) ÷ 2 = 6.5.
- The transferable lesson: when a puzzle asks for one feature, attack that feature directly. We never determined whether 4 or 9 is first — and we never had to, because their average is the same either way.
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