# Story problem?

There are nine identical-looking coins. One of the coins is counterfeit and weighs less than the other coins. The only scale available is a balance scale, on which you can weigh any number of coins against each other. Using the scale only twice, figure out a way to find the counterfeit coin.

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- There are nine identical- looking coins. One of the coins is counterfiet and weighs less than the other coins. The only scale available is a balance scale, on which you can weigh any number of coins against each other. Using the scale only twice, figure out a way to...

divide into three piles of three

weight any 2 piles. If they are the same, the counterfeit is on the third, or it is in the pile that weight less.

Now weight two coin from the pile and if they are the same, the counterfeit is the remaining one or it is the one that weighted less

1) Divide the coins into three groups of three coins each.

2) Weigh two groups against each other. If they weigh the same then you know the counterfeit coin is in the third group. If they DO NOT weigh the same then the counterfeit coin is in the group that is lighter. Either way, you have determined which group the counterfeit coin is in.

3) Take the group of three coins that contains the counterfeit coin (which coin it is, though, is still unknown). Weigh two of them. If the two coins are the same weight then you know the counterfeit coin is the one that was NOT weighed. If the coins are not the same weight then the counterfeit coin is the lighter coin.

Put 3 coins on each side of the scale.. if the left side is lighter it is one of these 3, if the right side is lighter it is one of those 3, if the sides are equally balanced it is one of the 3 youâ€™ve not measured.

So now youâ€™re down to a one in three shot.

Put one of your 3 suspect coins on each side of the scales.

If the left side coin is lighter, this coin is counterfeit, if the right side coin is lighter, its counterfeit… if theyâ€™re the same weight the final coin is counterfeit.

Piece of cake….and people already beat me to the solution. The trick is to have three groups of three and then pick any two and weigh them.