In the investigation of the 12 coins, some prisoners are given a chance to earn freedom: they must overcome a difficult logic proof. They are placed in front of Twelve coinsa marker and one balance to two dishes. They know that coins all have it stress weightexcept one that is false And for this it has a different weight from the others, but they don’t know if it is lighter or heavier. And here’s the challenge: identify the false coin in sun 3 weighs. Those who make mistakes will be postponed to prison. Whoever succeeds will receive freedom and a booty in gold coins.
How will prisoners do to identify the counterfeit coin? Let’s see it together.
The solution to the tidy of the 12 coins
If the prisoners had all the weighs they want available, the simplest solution would be to weigh one coin at a time and compare them with each other: the lighter or heavier coin will be immediately identified. The prisoners, however, have available Only three weighs! What they will have to do, then, is split coins in subgroups And compare these subgroups with each other, gradually narrowing the field of suspicions. Let’s see how.
Let’s start by dividing the 12 coins in Three groups of 4. We then write a number on each coin (1 to 12) and compare two groups with each other. For example, we compare the 1-2-3-4 coins with 5-6-7-8 coins leaving the coins 9-10-12 out.
At this point, two things can happen: the two groups have it same weight or or the two groups have a different weight. Let’s start from the first case.
First case: the two dishes balance
If the 1-2-3-4 coins weigh as much as the 5-6-7-8 coins means that the 8 weighted coins are authentic and identical because they have the same weight, so the false It must necessarily be found between four remained out, 9-10-11-12. At this point, We take 3 Among the 4 remained outside-for example, 9-10-11-and comparison with three of the eight authentic coins, for example 1-2-3. If the two groups they balancethen the false coin must be the only one that has never climbed the scale, that is, the coin left by the third group, in our case the 12. In this way, with only two weighs, we found the counterfeit coin.
If instead the two dishes weighing in the second do not balancewe know that among the 6 coins on the plate there is a false currency. We therefore still have a weighing and we know that the false currency must be among the 3 extras in the second weighted, that is the 9, 10 and 11. This is because the 1-2-3-4 coins weighed equal to the 5-6-7-8 coins and therefore we are sure they are true. Before removing the coins indicted by the scale, let’s see if the three coins 9-10-11 weigh more or less of the three real coins 1-2-3 and we mark this information Writing a “+” If the 9, 10 and 11 weigh more or a “-“ if they weigh less.
At this point, we take two of the “suspicious” coins, for example the 9 and 10, and we weigh them, comparing them with each other. If the 9 and 10 weigh the same, then the false coin will be 11. If, on the other hand, they have a different weight, we know that one of the two will be the false one, but which one? To understand it, we look at the signs we had done on the coins. If we had marked “+”, it means that the counterfeit coin is heavier than the others and therefore the false coin is the one that weighs most between 9 and 10. If we had marked “-“, instead, the counterfeit coin is lighter than the others and therefore the false coin is the one that weighs less between 9 and 10.
In this way in just three weighs we have Find the false coin. But how do we do if the two groups are not balanced at the first weighted? Let’s see the second case.
Second case: the two dishes do not balance
If the coins 1-2-3-4 have a different weight From the 5-6-7-8 coins, the counterfeit coin will surely be among these 8. To understand what it is, let’s start by writing a “+” on the coins of heaviest group between the two, for example 1-2-3-4, and a “-“ on the lighter groupfor example 5-6-7-8. On the coins that remained out from the weighing (9-10-11-12) We write a “V” To remind us that those coins I am definitely real.
Now we need a strategy to find the false coin in only two weighs. To do this, we have to mix the group. There are several ways to do it, but the concept you follow is similar, let’s see one. Let’s take three coins with the sign “-“ above, for example the 5-6-7, and the we replace to three of those with the SEgno “+”thus creating group 1 (+) 5 (-) 6 (-) 7 (-). Then, let’s take the number 8 currency, the last remained with the “-” above, and we add three other real coins, creating group 8 (-) 9 (V) 10 (V) 11 (V).
At this point, we weigh the two new groups and see what happens. There are three possibilities:
- the group that previously weighed the most continue to weigh more;
- The group that previously weighed more now weighs less;
- the groups have it same weight.
If Group 1 (+) 5 (-) 6 (-) 7 (-) is still the heaviermeans that either the coin number 1 it’s that counterfeit And it is more heavy of the others, or it is the number 8 And it is more light of the others. To find out which is, we still make a weighing by comparing the 1 with a normal currency: if they are the same, then the counterfeit is 8; If they weigh differently, then the counterfeit is the 1. In any case, in three weighs we have solved.
If, on the other hand, group 1 (+) 5 (-) 6 (-) 7 (-) has now become lighterthis means that One of the three coins I moved is the counterfeit one And it is lighter than the others. So, the false coin is 5, 6 or 7. We then make the last weighing by comparing the coin 5 with the 6. If they are the same, the 7 is the false coin. If they weigh differently, the lightest one of the two is false.
If the two groups now have it same weightthen the 1-5-6-8-10-11 coins are all true and the false one must be among those that I removed from group 1-2-3-4. That is, the false will be the 2, 3 or 4, and it will be heavier than the others. Therefore, comparison coin 2 with 3: if they are the same, the 4 is the false currency. If they weigh differently, the heaviest one of the two is false.
In any case, with only three weighswe managed to Find the counterfeit coin.
