国王查税：一次称重找出偷工减料的袋子

king demands a tax

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

国王向全国 10 个地区各征税 1000 枚金币。年末每个地区的税吏各交来一袋金币。

线人告诉国王：其中一个税吏在作弊，该袋金币每枚都比应有重量轻 $10\%$ ，但线人不知道是哪一袋。

国王知道每枚金币应重 1 盎司。如何只用一次称重装置就识别出作弊者？

A king demands a tax of 1,000 gold sovereigns from each of 10 regions of his nation. The tax collectors for each region bring him the requested bag of gold coins at year end. An informant tells the king that one tax collector is cheating and giving coins that are consistently $10\%$ lighter than they should be, but he does not know which collector is cheating. The king knows that each coin should weigh exactly one ounce. How can the king identify the cheat by using a weighing device exactly once?

解析

从第 1 袋取 1 枚、第 2 袋取 2 枚、…、第 10 袋取 10 枚，把这 55 枚一起称重。

若全为真币，重量应为 $55$ 盎司。

假设第 $k$ 袋为假币，每枚少 $10\%$ ，即少 $0.1$ 盎司。你从该袋取了 $k$ 枚，因此总少重 $0.1k$ 盎司。

称得重量为 $55-0.1k$ ，由缺重值 $0.1k$ 可直接反推出 $k$ ，从而确定作弊的税吏。