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It was something like this: There are twelve bags, each bag containing a random number of gold coins. Eleven bags contain genuine gold coins, each genuine coin weighing one-ounce apiece. One bag contains counterfeit coins, each counterfit coin weighing 9/10-ounces apiece. Using a scale to make only one measurement, how can you determine which bag is counterfeit?

You have as many bags as you please, (a minimum of two). You can have as many pieces of gold in the bags as you like and you can determine the weight of the coins as well, but the fake coins will weigh differently (more or less, it's up to you). You have a penny scale, meaning you can place something on a scale put a penny in and then receive a slip with the weight amount on it.

Answer:

For the answer i will use simple denominations of weight. There are three bags with five pieces of gold in each bag. Each real piece weighs one pound and each fake piece weighs 1 pound one ounce.

So you have bags A, B, and C and one is full of gold that is slightly heavier (fake).

You take one coin from the first bag, two coins from the second, and three from the third and you put all 6 coins on the scale and weigh them.

Now if all coins were real this should amount to 6 pounds, however, that isn't the case and you can determine the the fake bag by the overage weight.

So for instance if the reading comes back with 6 pounds 2 ounces then bag B is the one with the fakes. Since every fake piece is one extra ounce and you're two ounces over it must be the bag that you took 2 coins from which would be bag B.. If it was 6 pounds 3 ounces then it would be bag C and 6 lbs one oz would be bag A.

Alternatively, since you decide the amount of coins in each bag you could specifically say that you want one coin in the first bag two coins in the second, three in the third etc. and just weigh the bags together. Then you just follow the same train of thought to determine the fake bag.

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