1250.13 – False Coins no. 3

You have nine coins, supposedly made of silver. However, one of the coins is counterfeit and does not weigh the same as the other eight. The difference is slight enough that the coins all feel the same. You do not know which coin is the fake coin, and you do not know whether the fake coin is lighter or heavier than the others. You only know that its weight is different. What are you going to do?

Ah ha! Nearby is a two-pan scale. You can place as many coins as you want at a time on either pan. Can you figure out which coin is counterfeit, and whether it is lighter or heavier than the rest using the scale exactly three times?

Solution

This is a complicated problem with a flowchart-style solution. The first thing to do is label the coins: $a, b, c, . . .,$ and divide them arbitrarily into three groups: $abc$ being the first group, $def$ the second, and $ghi$ the third.

First, we weigh $abc$ against $def$ . There are three possibilities:

\begin{aligned} \text{Case 1: } abc &=& def, \\ \text{Case 2: } abc &<& def, \\ \text{Case 3: } abc &>& def. \end{aligned}

No matter which Case applies, the next step is to weigh $abc$ against $ghi$ . This also leads to three possible results:

\begin{aligned} \text{Case A: } abc &=& ghi, \\ \text{Case B: } abc &<& ghi, \\ \text{Case C: } abc &>& ghi. \end{aligned}

We will examine the nine possibilities (combining one of the cases 1, 2, 3 with one of the cases A, B, C). In each case, it is possible to figure out which of the three groups contains the odd coin and whether it is light or heavy. Then, making good use of the last weighing, we can isolate the counterfeit coin.

Consider, for example, what happens if Case 1 is true. Then the false coin must be in $ghi$ , but we don’t know whether it is light or heavy. The second weighing will tell us. Here's how.

First, if Case 1 is true, then Case A is impossible, as this would mean that all coins weigh the same, which we know is not true.

Next, if Case 1 and Case B are both true, then the false coin is in $ghi$ and is heavy. We know that it is heavy because we know that $abc$ and $def$ weigh the same, so, if $ghi$ is heavier than $abc$ , it must also be heavier than $def$ .

Thirdly, if Case 1 and Case C are both true, then we know that the false coin is in $ghi$ and is light.

Similar thinking applies to Case 2. If Case 2 and Case A are both true, then the false coin is in $def$ and is heavy. If Case 2 and Case B are both true, then the false coin is in group $abc$ and is light. Thirdly, it is impossible that Case 2 and Case C are both true, as this would mean that $ghi < abc < def$ , which implies multiple odd coins, which we know is false.

In summary, it is possible to know (after the first two weighings) which of the groups of three contains the odd coin and whether that coin is light or heavy. Once that is accomplished (and the reader/teacher/student should supply the analysis for the last three of the nine cases), the six coins we know to be normal can be set aside. They are no longer needed.

Finally, we weigh any two of the three remaining coins against each other. If the two coins weigh the same, then the third coin is odd and, we already know whether it is heavy or light. This is our answer. If the two coins do not weigh the same, we know that the third coin is normal, and the odd coin is one of the two that we weighed. If we know that the odd coin is light, then the lighter coin is odd. If we know that the odd coin is heavy, then the heavier coin is odd.

Ta-daa!