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17 球问题

17 Balls

专题
Brainteaser / 脑筋急转弯
难度
L2

题目详情

有 18 颗豆子，其中 17 颗完全相同，1 颗更重。使用天平，至少需要称几次才能保证找出更重的那一颗？

There are 18 beans, 17 of which are identical and one is heavier. What is the minimum number of weighings using a balance scale to guarantee finding the heavier bean?

解析

天平一次称量只有 3 种结果：左边重、右边重、两边相等。因此称量 kk 次最多区分 3k3^k 种情况。要在 18 颗豆子中保证找出唯一更重的那个，必须满足

3k≥183^k \ge 18

最小的整数 kk 为 3，因此最少需要 3\boxed{3} 次称量。

Original Explanation

A balance scale has three outcomes:

  1. Left side heavier
  2. Right side heavier
  3. Both sides equal

Each weighing can give us at most 3 outcomes, so in general, if we have n items and k weighings:

3k≥n3^k \geq n

is the minimum requirement to guarantee identification of one odd item. This is a standard approach from "coin weighing problems."

We have n=18n=18 beans. We want the smallest integer kk such that:

3k≥183^k \geq 18

So the minimum number of weighings is 3\boxed{3}