The fake coin, step by step
Nine coins, one of them slightly light, and a balance scale.
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. Nothing has been weighed yet.
- The obvious plan is to halve: put half in each pan, keep the lighter half, halve again. That really does work, and it gets there in 3 weighings — 7 → 4 → 2 → 1.
- But look at what a balance actually tells you. It can tip left, tip right, or sit LEVEL — three answers. Halving uses two of them and throws the third away, and the third is the informative one: 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 at all. Either way one 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 — and no fewer is possible, because each weighing has only three outcomes, so 1 of them could only ever separate 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. Nothing has been weighed yet.
- 02
The obvious plan is to halve: put half in each pan, keep the lighter half, halve again. That really does work, and it gets there in 3 weighings — 7 → 4 → 2 → 1.
- 03
But look at what a balance actually tells you. It can tip left, tip right, or sit LEVEL — three answers. Halving uses two of them and throws the third away, and the third is the informative one: 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 at all. Either way one 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 — and no fewer is possible, because each weighing has only three outcomes, so 1 of them could only ever separate 3 coins and there are 7.