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Is it possible to find out the lightest coin from 8 different coins by weighing 3 times only on a balance without having a weight?
Wiki User
∙ 12y ago
If only one coin is lighter than the others (which are all the same weight), this is definitely possible. Let's say that all of the 'normal' coins are of weight X, and the lighter coin has a weight of Y.
Step 1:
We will weigh two completely randomly divided sets of four coins. This means that one of the sets, set 1, must contain the coins X,X,X and X. The other set, set 2, will contain the coins X,X,X and Y.
After weighing these two groups, you will know which of the sets contains Y, since this side of the scale will be lighter than the other side. We will discard the other group, since we know that the lightest coin cannot be in that group.
Step 2:
We now have a set of four coins that we know contains the coin Y (from the side of the scale we isolated in step 1). Next, we will weigh two randomly divided sets of two coins. This means that the two sets must be either X and X or X and Y.
After weighing these two groups, you will know which of the sets contains Y, since this side of the scale will be lighter than the other side. We will again discard the other group, since we know that the lightest coin cannot be in that group.
Step 3:
We are now left with just two coins that are possibilities for coin Y. Now all we have to do is weigh the two against each other, and it will be clear which of the two coins is lighter. We have now successfully isolated coin Y - the lightest coin of the eight possibilities.
Wiki User
∙ 12y ago
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