Est. 1998 — 27 years of woodworking knowledge
Tuesday puzzle
An easy one:
I have 10 bags of pearls, one of which contains counterfeits. The counterfeits weigh 0.1 grams more than real ones, but are otherwise indistinguishable. Each bag has somewhere between 100 and 200 pearls in it, but I don't know how many, and they are not all the same. I have a scale that is as accurate as I need it to be. How can I find the bag with the fake pearls using the scale only one time?
Re: Tuesday puzzle
ALL of the pearls in the bag containing counterfeits are counterfeit, right? If not, I'll have to think about this some more.
Subtraction
If you put all the bags on the scale then remove one at a time by reading the remaining weight you will know which bag last removed weighs more than the others.
right, all are bogus
except
They do not have the same number in each bag, and you do not know how many are in any given one.
Re: right, all are bogus
Then if I'm right, the scale only needs to read to the nearest gram, although it would be easier (in implementation, not in concept) if it read in tenths of a gram.
;-)
If there is a tenth of anything remaining
on the scale you continue to remove the bags. When there is no tenth reading on the scale then the bag you just removed is the conterfeit bag. However if there are any multiple of 10 counterfeit pearls then you can't know because .1 x 10 is the same as a regular pearl. Still it's a 90% probability of being right and you don't have to count the pearls in the bag to verify.
not it
There is still some level of chance in your method. The way I am thinking has no element of chance.
Re: If there is a tenth of anything remaining
Besides the point Dan made, you would have to read the scale each time you removed a bag, which I think goes against the spirit of the requirement that you use the scale only once.
True, didn't think of that. (CRS syndrome ;-))
Answer
number the 10 bags from 1 to 10. Take one pearl from bag #1, 2 from #2, etc. When you get done, you will have 55 pearls. weigh them. if it weighs 55.3 for instance, the bogus pearls would be in bag #3. If it weighed 56, then the bogus pearls would be in bag #10.
Re: Answer
Doesn't this apply only if you know the weight of a real pearl, which wasn't a given in the original question ... or am I missing something?
Re: Answer
It actually would not matter. The total weight of the pearls would have to be an even multiple of 55, so any tenths would still be the bag the fakes were in, or if it came out even, then it would be bag 10.
It is just mod 55. Divide by 55 and the remainder times 10 is the bag it is in. That is assuming that the real ones are of integer weight, which I should have stated.
Re: Answer
correction, take the closest multiple of 55 and subtract it from the weight you got, then the remainder times 10 is the answer.
Re: Answer
Ed raises an excellent point, and one I totally missed (I assumed integer weight for the real ones). Without the integer assumption, you can't tell, e.g., whether a weight of 55.1 grams represents 54 real ones and 1 gram plus a counterfeit at 1.1 grams or 49 real ones at .9898 grams plus six counterfeits at 1.0898 grams.
If the weight of a real one is integer number of grams, then the answer is easier than you gave; just look at the digit to the right of the decimal. No need for subtracting a multiple of 55. And to save yourself a little work, number the bags 0-9, so you don't even need to open one of them.
Re: Answer
I missed the integer weight restriction when I posted the question. It is actually in the original that I have. I tutor a math lab for the local community school as a volunteer and I collect interesting puzzles that I sometimes give them. They end up learning some math or thinking skills without realizing that they are doing it ;-)
Nice give-back. Sounds rewarding.