2004 AMC 8 第 13 题
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所有题目均经美国数学协会（MAA）官方合法授权使用。
13.
艾米、比尔和赛琳是年龄不同的朋友。以下陈述中恰好只有一个是真的。
。比尔年龄最大。
。艾米不是年龄最大的。
。赛琳不是年龄最小的。
请按从年龄最大到最小排列这些朋友。
Amy, Bill and Celine are friends with different ages. Exactly one of the following statements is true.
Bill is the oldest.
Amy is not the oldest.
Celine is not the youngest.
Rank the friends from the oldest to youngest.
比尔、艾米、赛琳
Bill, Amy, Celine
艾米、比尔、赛琳
Amy, Bill, Celine
赛琳、艾米、比尔
Celine, Amy, Bill
赛琳、比尔、艾米
Celine, Bill, Amy
艾米、赛琳、比尔
Amy, Celine, Bill
小提示：
如果比尔最大，检查会有多少个陈述为真
If Bill were oldest, check how many statements would be true.
大提示：
排除谁最大后，使用恰好只有一个陈述为真的条件
After ruling out the oldest person, use the fact that exactly one statement is true.
解答：
如果比尔最大，那么陈述 和 都为真，所以比尔不是最大的。
如果赛琳最大，那么陈述 和 都为真，所以赛琳也不是最大的。
因此艾米最大。陈述 和 为假，所以陈述 必须是唯一为真的陈述。因此赛琳不是最小，比尔最小。
顺序是艾米、赛琳、比尔。
所以正确答案是 E。
If Bill were oldest, then statements and would both be true, so Bill is not oldest.
If Celine were oldest, then statements and would both be true, so Celine is not oldest.
Therefore Amy is oldest. Statements and are false, so statement must be the single true statement. Thus Celine is not youngest, leaving Bill youngest.
The order is Amy, Celine, Bill.
Thus, E is the correct answer.
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