Can you solve this minimum information puzzle?

[FunkyMonkee]

In a room are several sacks of gold pieces, as many as you like. Each sack contains several of these gold pieces, again, as many as you like. However; one sack is full of artificial gold pieces and they weigh differently. Let's say the gold pieces weigh a pound @. The artificial pieces weigh whatever you like. Now, you have a penny scale. You put the penny in and it spits out a card that tells you how much the gold weighs but, you only have one penny! How do you find out when sack has the fake gold in it?

[NathanDavis]

well... just keep it to minimal data to figure it out

- as many sacks as you want? make that two sacks...

- as many pieces of gold as you want? make that... one sack with two pieces, the other sack with three pieces of gold

- the artificial pieces of gold weigh as much as you want? make that... 1.5 pounds for artificial ones, the real ones are 1 pound

and then you weigh both sacks together and get your result... if the first sack with two pieces was the one with real gold, the total weight should be around 6.5 pounds... but if the second sack with the three pieces was the one with the real gold... then the total weight should be about 6 pounds

[FunkyMonkee]

But, you have to start out with an equal amount of coins in each bag.
Roughly, you've got it.

[NathanDavis]

you weren't specific and clear enough about this, but okay... I'll get back to it with the new rule

[FunkyMonkee]

That's the way it was told to me.

[NathanDavis]

okay, same principle then... each sack has whatever same number of gold pieces, but now I'll take one piece from a sack, and two pieces from the other sack, weight the three together... and still make the difference out of the result

the results would be either... 4 pounds, or 3.5 pounds

[FunkyMonkee]

That's the answer I'm looking for.

[GuyAnswersGirls123]

There would be only be one sack of gold in the room since I decide how many there are. So it has to contain the fake gold. I don't even have to use my penny.

[GuyAnswersGirls123]

By "how many you like", that means I get to decide, no? I do feel like I made this easier than intended. If it means it's a variable, that's a bit more complex.

Yet there's a simple solution there. Simply take a different number of gold pieces from each bag. For example, if there are 3 bags (A, B, C), take 1 from A, 2 from B, 3 from C.

Then take the total weight of the coins. If the weight is 5 (2+3), A has the fake coin. If the weight is 4 (1+3), then B has the fake coin. If the weight is 3 (1+2), then C has the fake coin.

[GuyAnswersGirls123]

Assuming the fake coin weighs 0 for simplicity. Yet we can also do it with any weight, provided it's not 1, by comparing the sum.

[FunkyMonkee]

As many as you like in all the bags. If one bag has 8, they all have 8.
As for the solution, you're on the right track.

[FunkyMonkee]

Oh! I see. Yeah, that works but, where are you gonna get things that weigh nothing?

[GuyAnswersGirls123]

I just used like X=0 to simplify it. If it's the fake piece is 2 pounds as another example (X=2), then a total weight of 7 means A is the fake bag (2*1+2+3), 8 means B is the fake bag (1+2*2+3), 9 means C is the fake bag (1+2+2*3).

Basically if the total weight is (X*A)+B+C, then A is the fake, if it's A+(X*B)+C, then B is the fake, and if it's A+B+(X*C), then C is the fake. I can generalize it further with a formula but I'm a bit out of this morning.

[FunkyMonkee]

No need. I'd say you've got it but slightly different than what I had in mind.