Solve the Colombo Gold Coin Brain Teaser

[marlon]

I have a nice brain teaser coming from the Colombo-series.

Suppose you have three bags each filled with the same amount of gold-coins. In two bags, there are real gold coins and in the third sack, the coins are false. You can suppose that the weight of a false coin has another value (more heavy or lighter, you can chose) then that of a real gold coin. You have an apparatus to measure the weight of an object. Now can you determin which bag has the false coins by performing just ONE measurement. So you can read of the weight only once...

marlon

 

[Jimmy Snyder]

I know this one. But it is not stated correctly. To solve this puzzle, you need to know the weights of both real and false coins.

 

[marlon]

I know this one. But it is not stated correctly. To solve this puzzle, you need to know the weights of both real and false coins.

no, you do not...because you can assign any number that you want for the weight if you think that is necessary. I did not say you were not allowed to do that. you can do anything you want, as long as you perform just ONE measurement...

marlon

 

[ArielGenesis]

weight, do you mean weighing and get a number or using balance to know which one is lighter(i choose lighter is the fake one). if using balance, [colour="white"]take 2 bags and balance them, if they balance, the third bag is fake. if they are not balance, the lighter one is fake[/colour]

 

[DaveC426913]

You will need to know the normal weight of the real coins AND the weight of the fake coins.
A:
- Take one coin from bag 1, two coins from bag 2 and three coins from bag 3.
- Put all six coins on the weigh scale together.

- If the weight is off by the difference between 1 real and 1 fake coin, the fake coins are in bag 1.
- If the weight is off by the difference between 2 real and 2 fake coins, the fake coins are in bag 2.
- If the weight is off by the difference between 3 real and 3 fake coins, the fake coins are in bag 3.

 

[Jimmy Snyder]

Here is a solution to the problem if I know the weights of the real and the false coins. As yet, I have no solution for the problem as stated.
If I knew that the weight of a real coin was 1 oz. and the weight of a false coin was 1.1 oz. I could take one coin from bag 1, 2 coins from bag 2 and three coins from bag 3, and weight the 6 coins. If the weight was 6.1 oz, then bag 1 contains the false coins, if the weight is 6.2 oz., then bag 2 contains the false coins and if the weight is 6.3 oz. then bag 3 contains the false coins.

However, if I don't know both weights, then this is no solution to the problem. The reason is given below.

If I know that the real coins weigh 1 oz, but do not know the weight of the false coins, then for instance if the weight was 6.1 oz. I would know that there was .1 oz overage, but I wouldn't know whether I had 1 false coin weighing 1.1 oz, or two false coins weighing 1.05 oz. each, or 3 false coins weighing 1.0333... oz.

If I know that the false coins weigh 1.1 oz, but do not know the weight of the real coins, then for instance if the weight was 6.1 oz. I would not know if there were 3 real coins weighing .9333... oz. each, or 4 real coins weighing .975 oz. each, or 5 real coins weighing 1 oz each.

If I don't know the weights of the real or the false coins and the weight were 6.1 oz. I would have no clue how to interpret it.

 

[Jimmy Snyder]

Sorry Dave, I was composing while you were posting.

 

[marlon]

You will need to know the normal weight of the real coins AND the weight of the fake coins.
A:
- Take one coin from bag 1, two coins from bag 2 and three coins from bag 3.
- Put all six coins on the weigh scale together.

- If the weight is off by the difference between 1 real and 1 fake coin, the fake coins are in bag 1.
- If the weight is off by the difference between 2 real and 2 fake coins, the fake coins are in bag 2.
- If the weight is off by the difference between 3 real and 3 fake coins, the fake coins are in bag 3.

good and NO you do not need to be gicen specific weight values...why is that so hard to see ? Did you mention any numbers in your solution for the weight ...Pitty you already knew the solution from somewhere

marlon

 

[marlon]

Here is a solution to the problem if I know the weights of the real and the false coins. As yet, I have no solution for the problem as stated.

Again, i said that you do not need to be given specific values for the weight. I did not say it is invalid to assume a certain number, as long as you keep in mind that a false coin has a different weight (be it either bigger or smaller) then a real gold coin. The way this problem has been stated with the RELATIVE weights being given, is part of the difficulty of the problem...Besides, that is also what Colombo was struggling with...His wife actually solved the problem :rofl:

marlon

 

[arildno]

Off to see "Inspector Colombo" right now..

 

[BicycleTree]

Did you mention any numbers in your solution for the weight

Yes, he did. Specifically he mentioned the number "the normal weight of the real coins" and the number "the weight of the fake coins." The problem is unsolvable without knowing those two numbers.

 

[BicycleTree]

Say that you follow his solution and put on the scale 1 from bag 1, 2 from bag 2, and 3 from bag 3, and weigh them, and the weight is 10 ounces. Now then, given only this information, tell me which bag contains the counterfeit coins.

 

[TenaliRaman]

why are we looking at more complicated answers?
after all its just "three" bags?

Furthermore,
in the original problem statement

the weight of a false coin has another value (more heavy or lighter, you can chose)

I presume that this means we know whether if the false coins are lighter or heavier

In which case, just take one coin from each bag and weigh any two of them and u are done.

Now if marlon were to rephrase his original question to say that we don't know whether it is lighter or not then we may have to look at the complicated solutions.

-- AI

 

[DaveC426913]

good and NO you do not need to be gicen specific weight values...why is that so hard to see ? Did you mention any numbers in your solution for the weight

So you weigh the bag of six coins. It weighs 7 ounces. Please tell us which bag contains the fake coins.

Pitty you already knew the solution from somewhere marlon

what makes you think so?

 

[BicycleTree]

why are we looking at more complicated answers?
after all its just "three" bags?

Furthermore,
in the original problem statement


I presume that this means we know whether if the false coins are lighter or heavier

In which case, just take one coin from each bag and weigh any two of them and u are done.

Now if marlon were to rephrase his original question to say that we don't know whether it is lighter or not then we may have to look at the complicated solutions.

-- AI

Oh... duh. Many of us myself included were not thinking about this problem right. I see ArielGenesis also said something along these lines.

 

[marlon]

Off to see "Inspector Colombo" right now..

:rofl: :rofl: :rofl:

Yeah, tomorow evening...

marlon

 

[arildno]

Actually I did see it yesterday; the murderer was General Hollister.
An okayish Columbo episode.

 

[Jimmy Snyder]

No need to wait til tomorrow, here is the dialog from that show. Look at item #14 at this site.

http://www.columbo-site.freeuk.com/faq.htm

In the Columbo puzzle, he knew the weights of both the real and the false coins.

Marlon, I think what you mean is that the problem is solvable no matter what the weights are. This is true. But the problem is not solvable if you don't KNOW what the weights are. In the way that you worded the problem, the weights aren't known, so the puzzle is unsolvable.

Several people have asked you essentially the same question, and I ask it too. If you don't know the weights of the coins and you take the 1 coin from bag 1, 2 coins from bag 2 and 3 coins from bag 3, and you weigh the six coins and get 6.1 oz., then which bag contains the false coins and why?

Or do you have some other solution in mind?

 

[ArielGenesis]

let's wait for marlon...