It is rather popular logical problem. There are twenty coins identical in appearance, one of which false, and scales with cups without weights. It is known that the false coin weighs less than real. It is necessary to find a false coin for three weighings.
It is required to you
- - twenty coins;
- - cup scales.
Instruction
1. Divide coins into three parts: in two there will be about seven coins, and in remained – six. On cup scales put two equal small groups. If scales are counterbalanced, so in two small groups on seven coins all coins real, and false is among the remained six coins. If balance was broken, miss the following item of the decision.
2. Take a small group from six coins, divide it into three parts. Put 2 coins in each cup of scales, leave 2 more. This second weighing. If scales are counterbalanced, so the false coin remained among two, lying on a table. If balance was broken, so a false coin among those two coins which were easier. Thus, you found couple of coins, one of which false, and it is easy to find it the third weighing, having just put in each cup of scales on one coin.
3. Take that part from seven coins which was easier. Let's remind that you do it in case at the first weighing the balance on scales was broken. Divide coins into three parts: in two there will be about three coins and in remained – one. Put three coins in each cup of scales. This second weighing. If balance of scales is not broken, so the remained coin also is false. The task is solved even quicker thanks to luck! If one cup of scales was easier, carry out the last weighing.
4. Put in each cup of scales on one coin from that part which was easier. The third coin it will be necessary to lie on a table. If scales are counterbalanced, so the remained coin also is false. If one of cups of scales is easier, then the false coin is in it.