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 re: better -- within the constraints of the puzzle by rod hines)

"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."

No, the numbers I gave were

0  1  2  4  7  13  24  44

indicating that zero (no) coins were to be used from bag 1, and the table included cases where bag 1 was one of the bags containing fake coins, as in

1  2  3       92

where the 1 fake coin from bag 2 and the 2 fake coins from bag 3 make the total come out to 3 grams less than the total of 95--that is, 92. This incorporates the fact that no coins were used from the first bag.

Your proposed 1   2  3   6  12  24  48, is what I would have described as 0  1   2  3   6  12  24  48.

However, in keeping with your bag numbering, where it's bags 1 through 7 that have 1   2  3   6  12  24  48 coins represented, respectively, I note that

If bags 1, 2 and 4 have the fake coins, the total will be 87 (i.e., 96 - 1 - 2 - 6). But if it were bags 3, 4 and 8(the left-out one) that had the fakes, the total would also be 87 (this time 96 - 3 - 6).

If bags 1, 2 and 5 have the fake coins, the total will be 81 (i.e., 96 - 1 - 2 - 12). But if it were bags 3, 5 and 8(the left-out one) that had the fakes, the total would also be 81 (this time 96 - 3 - 12).

If bags 1, 2 and 6 have the fake coins, the total will be 69 (i.e., 96 - 1 - 2 - 24). But if it were bags 3, 6 and 8(the left-out one) that had the fakes, the total would also be 69 (this time 96 - 3 - 24).

If bags 1, 2 and 7 have the fake coins, the total will be 45 (i.e., 96 - 1 - 2 - 48). But if it were bags 3, 7 and 8(the left-out one) that had the fakes, the total would also be 45 (this time 96 - 3 - 48).

Posted by Charlie on 2008-10-24 01:57:08