You have eight bags, each of them containing 48 coins.
Five of these bags contain only true coins, the rest of them contain fake coins. Fake coins weigh 1 gram less than the real coins.
You do not know what bags have fake coins and what bags have real coins. You do not know also, besides that it is an integer value, the weight of the real coins.
You can use a digital or analog reading scale with precision up to 1 gram.
Making only one weighing and using the minimum number of coins, how can you find the bags containing the fake coins?
(In reply to
better -- within the constraints of the puzzle by Charlie)
Charlie: Not to quibble with your solution, but it looks like you've only considered the situation where coins from all 3 fake bags actually get weighed. What if the bag from which 0 coins were drawn was actually one of the fakes, so only coins from the other two get weighed.
I've hit upon another set of coin pulls for bags 1 to 7; that is
1 2 3 4 5 6 7
1 2 3 6 12 24 48 = 96 coins. If all were actually real, their weight would be 96x grams. But coins from 2 or 3 fake bags get weighed as well, so the scale would indicate 96x grams - 1 gram for each fake coin weighed.
I worked this out for all possible positions considering coins from either 2 or 3 fake bags getting drawn/weighed, similar to your table, to determine the number of grams less than 96x each would result in. Surprisingly, all the resulting numbers (as grams) were unique on each table. I was hoping the numbers were actually unique between the two tables as well, which they essentially are with the exception of consecutive 15/27/31 results on both. That is, Bag 3 + Bag, 5, 6 or 7 considering 2 bags only, and Bag 1 + Bag 2 +Bag 5, 6 or 7 with 3 bags.
Not sure if it's at all significant that Bags 5, 6 and 7 in both cases involve 12/24/48 coins, or whether any other part of this is helpful. It's just my first stab at it!
re: better -- within the constraints of the puzzle
