Simple Logic Problem

[HedgeHog]

This is a relatively easy problem, so maybe the geniuses can refrain from answering right away (PM me if you know the answer Ganch, Data, Justin, Donjuan, etc...). For everyone else:

You have 3 bags of coins with an unknown # of coins in each. 2 of the bags are fake (fool's gold), while 1 bag has real gold coins. The coins are identical except that the real gold coins weigh 1.1 oz per coin while the fake coins weigh 1.0 oz each. You have a very accurate scale, but are allowed just one weighing. How do you determine the bag of real gold coins for sure with just the one weighing?

[HedgeHog]

I should have started this in the Main Forum. Any chance it could get linked to there, too?

[HedgeHog]

Thanks for the move, Willie. Any ideas?

[Justin7]

This is a variation of "8 fake balls, 1 real ball, and you get 2 weighings"...

[HedgeHog]

TLD has PM'd me with the right answer. Keep trying.

[Ganchrow]

You have 3 bags of coins with an unknown # of coins in each. 2 of the bags are fake (fool's gold), while 1 bag has real gold coins. The coins are identical except that the real gold coins weigh 1.1 oz per coin while the fake coins weigh 1.0 oz each. You have a very accurate scale, but are allowed just one weighing. How do you determine the bag of real gold coins for sure with just the one weighing?

Let's not forget we can also make use of the [EXTRA][/EXTRA] tag.

Answer has been ROT13'd:

Jrvtu 1 pbva sebz ont O naq 2 pbvaf sebz ont P.

Vs gbgny jrvtug == guerr cbvag mreb bhaprf gura Ont N pbagnvaf gur erny pbvaf.
Vs gbgny jrvtug == guerr cbvag bar bhaprf gura Ont O pbagnvaf gur erny pbvaf.
Vs gbgny jrvtug == guerr cbvag gjb bhaprf gura Ont P pbagnvaf gur erny pbvaf.

[HedgeHog]

Pico has PM'd the right answer too. Also I think Ganch has taken a slightly different rout to the correct answer.

Extra credit for Ganch. He figured out how to do this by using just 2 bags instead of all 3--something I never thought of previously.

[Ganchrow]

Pico has PM'd the right answer too. Also I think Ganch has taken a slightly different rout to the correct answer.

Extra credit for Ganch. He figured out how to do this by using just 2 bags instead of all 3--something I never thought of previously.

Since you never specified that minimizing stress on the scale was a goal of the problem I'm afraid I can't accept any extra credit for my solution.

[Louisvillekid1]

allowed one weighing or only allowed to way one bag?

[bettilimbroke999]

allowed one weighing or only allowed to way one bag?

That's what I was thinking, if you can weigh more than one bag at once it's pretty easy, otherwise I haven't figured it out yet.

[bigugly]

I think you should be able to determine which bag it is by weighing 1 coin from one bag, 2 coins from another bag, at the same time. Right?

[janus]

given a common weight per bag (net of coins), the answer would be that the real gold total bag weight would be divisible by 1.0oz rather than 1.1oz.

[Ganchrow]

given a common weight per bag (net of coins), the answer would be that the real gold total bag weight would be divisible by 1.1oz rather than 1.0oz.

What if a particular bag weighed 110 oz?

[HedgeHog]

I think you should be able to determine which bag it is by weighing 1 coin from one bag, 2 coins from another bag, at the same time. Right?

You're on the right track. Finish your thought.

[janus]

What if a particular bag weighed 110 oz?

as long as the bag itself was standard/a common weight, it doesnt matter.

[HedgeHog]

as long as the bag itself was standard/a common weight, it doesnt matter.

You don't know how many coins are in the bag. If you weigh one bag and get 110 oz as Ganch suggests, then you could have 100 real coins or 110 fake ones--the weight would be the same (assume the weight of the bag itself is zero or subtracted from the equation). Only the coins and their weights matter.

[janus]

You don't know how many coins are in the bag. If you weigh one bag and get 110 oz as Ganch suggests, then you could have 100 real coins or 110 fake ones--the weight would be the same (assume the weight of the bag is zero or subtracted from the equation).

ah, i see.

[bigugly]

You're on the right track. Finish your thought.

Well either that is right or it isn't. I just thought I'd not thoroughly explain it all.

If the 3 coins add up to 3 oz, then you know the real gold is in the bag you did not draw from.

If the 3 coins add up to 3.1 oz, the real gold is in the bag you took one coin from.

If the 3 coins add up to 3.2 oz, the real gold is in the bag you took 2 coins from.

How's that?

[JNGDist]

woooo

It wasnt specified but can you open 2 of the bags ?

[HedgeHog]

Well either that is right or it isn't. I just thought I'd not thoroughly explain it all.If the 3 coins add up to 3 oz, then you know the real gold is in the bag you did not draw from.If the 3 coins add up to 3.1 oz, the real gold is in the bag you took one coin from.If the 3 coins add up to 3.2 oz, the real gold is in the bag you took 2 coins from.How's that?

Yes you got it. You took the same route that Ganch did, so I needed a little more explanation for it to sink in.

Another way to get the same answer is to take 1 coin from Bag 1, 2 coins from Bag 2, and 3 coins from Bag 3 --a total of six coins. If the scale reads 6.1 oz then the Gold was in Bag 1, 6.2 oz means Bag 2 had the gold and of course a 6.3 oz reading means Bag 3 had the gold coins (remember the gold coins are 1.1 oz instead of 1.0).Several people figured this out in one form or another. Sharp bunch here

[Data]

You have a very accurate scale

You need to specify the kind of scale, the correct solution depends on it.


The posted solution is good for the scale above but not for the scale below.

[HedgeHog]

Obviously I was thinking the first type. Put items on and get a digital reading. You're right, I should have been more specific.

[Ganchrow]

With a balance scale the solution is obviously trivial.

Measure one coin from bag A against one coin from bag B. The heavier side corresponds to the bag of real coins. If both sides are qeial the real coins are in bag C.

[Data]

With a balance scale the solution is obviously trivial.

Right, I understand why you are a bit upset. Still, I hope you appreciate the irony that an ancient instrument gives simpler and faster solution.

[Ganchrow]

Right, I understand why you are a bit upset. Still, I hope you appreciate the irony that an ancient instrument gives simpler and faster solution.

Actually, the real challenge would be figuring out a solution using one of these scales:

[swede96]

Ganchrow, that was priceless. I'm a sucker for fun with puns.

[Ganchrow]

Ganchrow, that was priceless. I'm a sucker for fun with puns.

As the late, great Douglas Adams once penned, "You can tune a guitar, but you can't tuna fish. Unless of course, you play bass."

What's long, brown, and sticky?

A stick.

What's the difference between a nun and a woman in a bathtub?

The nun has hope in her soul.

[Data]

one of these scales

Yes, scales. I am glad you noticed.

[operaman]

I was thinking the same type of scales data was originaly.

The balance can be referred to as either scale or scales I think.

In the intended version doesn't the given solution neglect the
possible 1-1-1 distribution of coins in bags? (assuming you are not allowed to divide the coins. )

The way I read it all three bags could contain one coin.

[Ganchrow]

In the intended version doesn't the given solution neglect the
possible 1-1-1 distribution of coins in bags? (assuming you are not allowed to divide the coins. )

The way I read it all three bags could contain one coin.

Good point.

I suppose it all depends on how you read the phrase "3 bags of coins". Is there a copy editor in the house?

[HedgeHog]

Didn't realize how ambiguous my puzzle is--valid points on the type of scale used and minimum # of coins in each bag. I need at least 3 coins in each bag for my solution (just 2 for the method Ganch and others described).

[Data]

In the intended version doesn't the given solution neglect the
possible 1-1-1 distribution of coins in bags?

I think the wording ("bags of coins) is sufficient to assume that there are at least two coins in each bag. A very good point nonetheless.

[Data]

This was intended to be a PM but that did not work out.

Man, you are just a bag of surprises! What the hell were you doing in this old thread? Let me guess, a routine utilizing an RND made you do this, no?

Happy New Year to you and Jenbird!

Best wishes from Mrs. Data and myself.

[bztips]

Admittedly off-topic -- I can't get any of Ganch's "Show Answer" buttons to work. Do I need to install something special in my browser? (running Firefox 3.6 on a brand-new Win64 machine). Thx.

[TexansFan]

Was the answer C?