# Quarter Wit, Quarter Wisdom: Solving the Weighing and Balancing Puzzle

Let’s continue the discussion on puzzles that we began last week. Today we look at another kind of puzzle – weighing multiple objects using a two-pan balance while we are given a limited number of times to weight the objects against each other.

First of all, do we understand what a two-pan balance looks like?

Here is a picture.

As you can see, it has two pans that will be even if the weights in them are equal. If one pan has heavier objects in it, that pan will go down due to the weight. With this in mind, let’s try our first puzzle:

One of twelve coins is a bit lighter than the other 11 (which have the same weight). How would you identify this lighter coin if you could use a two-pan balance scale only 3 times? (You can only balance one set of coins against another, i.e. you have no weight measurements.)

There are various ways in which we can solve this.

We are given 12 coins, all of same weight, except one which is a bit lighter.

Let’s split the coins into two groups of 6 coins each and put them in the two pans. Since there is one lighter coin, one pan will be lighter than the other and will rise higher. So now we know that one of these 6 coins is the lighter coin.

Now split these 6 coins into another two groups of 3 coins each. Again, one pan will rise higher since it will have the lighter coin on it. Now we know that one of these three coins is the lighter coin.

Now what do we do? We have 3 coins and we cannot split them equally. What we can do is put one coin in each pan. What happens if the pans are not balanced? Then we know the pan that rises higher has the lighter coin on it (and thus, we have identified our coin). But what if both pans are balanced? The catch is that then the leftover coin is the lighter one! In any case, we would be able to identify the lighter coin using this strategy.

We hope you understand the logic here. Now let’s try another puzzle:

One of 9 coins is a bit lighter than the other 8. How would you identify this lighter coin if you could use a two-pan balance scale only 2 times?

Now we can use the balance only twice, and we are given an odd number of coins so we cannot split them evenly. Recall what we did in the first puzzle when we had an odd number of coins – we put one coin aside. What should we do here? Can we try putting 1 coin aside and splitting the rest of the 8 coins into two groups of 4 each? We can but once we have a set of 4 coins that contain the lighter coin, we will still need 2 more weighings to isolate the light coin, and we only have a total of 2 weighings to use.

Instead, we should split the 9 coins into 3 groups of 3 coins each. If we put one group aside and put the other two groups into the two pans of the scale, we will be able to identify the group which has the lighter coin. If one pan rises up, then that pan is holding the lighter coin; if the pans weight the same, then the group put aside has the lighter coin in it.

Now the question circles back to the strategy we used in the first puzzle. We have 3 coins, out of which one is lighter than the others, and we have only one chance left to weigh the coins. Just like in the first puzzle, we can put one coin aside and weigh the other two against each other – if one pan rises, it is holding the lighter coin, otherwise the coin put aside is the lighter coin! Thus, we were able to identify the lighter coin in just two weighings. Can you use the same method to answer the first puzzle now?

We will leave you with a final puzzle:

On a Christmas tree there were two blue, two red, and two white balls. All seemed the same, however in each color pair, one ball was heavier. All three lighter balls weighed the same, just like all three heavier balls weighed the same. Using a 2-pan balance scale only twice, identify the lighter balls.

Can you solve this problem using the strategies above? Let us know in the comments!