The fake coin
7 coins look identical. One is fake and slightly lighter than the rest.
What is the fewest weighings that are certain to find the fake?
A balance answers three ways, not two
- 7 identical-looking coins, one lighter than the rest, and a balance that takes any number of coins in each pan.
- The obvious plan is to halve: put half in each pan, keep the lighter half, halve again. That works, in 3 weighings — 7 → 4 → 2 → 1.
- But a balance has three answers: tip left, tip right, or level — and level means the fake is in neither pan.
- So split into three roughly equal groups and weigh two of them. If it tips, the fake is in the lighter pan; if it sits level, the fake is in the group you did not weigh. Each weighing cuts the suspects to a third, not a half.
- Keep going and the count falls fast: 7 → 3 → 1. That is 2 weighings, 1 fewer than halving. No fewer is possible: 1 weighing can only sort out 3 coins, and there are 7.
How it works.
- 01
7 identical-looking coins, one lighter than the rest, and a balance that takes any number of coins in each pan.
- 02
The obvious plan is to halve: put half in each pan, keep the lighter half, halve again. That works, in 3 weighings — 7 → 4 → 2 → 1.
- 03
But a balance has three answers: tip left, tip right, or level — and level means the fake is in neither pan.
- 04
So split into three roughly equal groups and weigh two of them. If it tips, the fake is in the lighter pan; if it sits level, the fake is in the group you did not weigh. Each weighing cuts the suspects to a third, not a half.
- 05
Keep going and the count falls fast: 7 → 3 → 1. That is 2 weighings, 1 fewer than halving. No fewer is possible: 1 weighing can only sort out 3 coins, and there are 7.