Counterfeit Coins
You have 11 piles of coins, with 10 coins in each pile. You know what a single coin should weigh. You also know that one of the piles is made up entirely of counterfeit coins. Each fake coin weighs one gram more than a real one. What is the minimum number of times you would have to weigh the coins to determine which pile has the counterfeits?
This is a slightly modified version of a problem found in Martin Gardner's book "My Best Mathematical and Logic Puzzles".
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1 solution
The pile containing the fake coins can be determined with only one weighing. Start by numbering each pile 0 through 10. Place 0 coins from "Pile 0" on the scale, 1 coin from "Pile 1" on the scale, 2 coins from "Pile 2", and so on until you have put each pile's corresponding number of coins on the scale. Next calculate what the weight on the scale SHOULD BE if none of the coins were fake. Subtract this from the actual weight on the scale. This number represents the pile containing the fake coins. For example, if the "extra" weight is 6 grams, "Pile 6" contains the counterfeits. If the "extra" weight is 0 grams, "Pile 0" contains the counterfeits.
It works this way because there is a different number of coins from each pile on the scale. For example, imagine you had two coins, one fake and one real. If you just weigh a single coin you determine whether it is fake or real, and you can use this information to determine the "identity" of the other one. This idea can be extended to as many piles of coins as you please, as long as you can put a different number of coins from each pile on the scale.