Thread: Coin sort.

OK if it's puzzle time...

You have 10 piles of coins each pile appears to have 10 silver dollars in it, however 1 pile contains only counterfeit coins. You are given the weight of a real silver Dollar and the weight of a counterfeit dollar plus a pair of accurate digital scales.

You may weigh the coins in any order or quantity you choose, the object is to find the counterfeit pile in as few weighing operations as possible.

What is the minimum number of weighings?

I can get it down to 3 assuming 'bad luck'. Or, I should say, I can have your answer in at most 3 weighings (as few as 2 with my method and some 'luck').

EDIT: I should add, that I don't need to know the weight of the good coins or bad coins - only that they are not equal. (end edit)

Correct?

I like this one because we can guess and not give away the method.

Cheers,
william

Originally Posted by william

I can get it down to 3 assuming 'bad luck'. Or, I should say, I can have your answer in at most 3 weighings (as few as 2 with my method and some 'luck').

EDIT: I should add, that I don't need to know the weight of the good coins or bad coins - only that they are not equal. (end edit)

Correct?

I like this one because we can guess and not give away the method.

Cheers,
william

All I can say without spoiling anybody else's fun is:

Any advance on 3/2?

Well...
At the prodding of billco,
I recall my previous post and claim it can be done with 1 weighing, every time (no luck required), provided we know the weight of the coins and have 2 scales.

So... only 1 weighing is required, and I emphasize, without the intervention of luck.

I won't spoil the fun for the rest and post the solution. But I will PM my solution to billco.

Once again, great post billco!

Cheers,
william

so if we place the coins on the two scales simalteneously, it is considered as one weighing?

I don't mind how many sets of scales are used so long as EVERY TIME a reading is taken it counts as one, so if you have 2 sets of scales and take a reading from each it is 2 weighings. (I did say in the original post "a set of scales")



If you get bored with that, try this one

I have a cube with side length 3, I wish to saw it into 27 * 1 unit cubes.
Now each time you make a saw cut you are allowed to re-arange the pieces in any way. The question is What is the minimum number of saw cuts needed in order to achieve the aim. Everybody knows you can do it in six, but can you do it in less??

Are you serious? THat one is easy

Answer in white (Highlight to spoil): SIX CUTS.

Originally Posted by Absane

Are you serious? THat one is easy

Answer in white (Highlight to spoil): SIX CUTS.

So if it's easy you can tell us the answer (by PM)...

He's given the answer - in white font. Just draw your cursor over the post, while holding down the left mouse key. It will appear.

I'm not sure that answer [whether or not correct] fits the question, either a lower number than six (with an explanation) or a short sentence on why six cannot be beat.

Well the cube is a 3x3x3, correct?

You'll notice that there is a cube in the very middle. It has 6 sides and you want to cut this 3x3x3 cube in into 27 different peices with the fewest cuts possible. However, you cannot cut the middle cube with less than 6 cuts.

The minimum number of weighings in the original question is: 1.

Sure. It depends on luckily picking the counterfeit stack to weigh. But it satisfies the semantics of the riddle.

No luck involved on this one, - there is a minimum number of 'weighings' which will ALWAYS find the answer. For example 9 weighings would always find it BUT there is a much more elegant solution with fewer moves.

I interpreted "a set of scales" as two scales....

Hmmm....

Instead of a "set of scales," is it more appropriate to think of a balance? Like the symbol for justice?

Like so


_A________B_
_____/\______

edit: Damn! I can't get my picture to post correct!

Originally Posted by william

Instead of a "set of scales," is it more appropriate to think of a balance? Like the symbol for justice?

Like so


_A________B_
_____/\______

edit: Damn! I can't get my picture to post correct!

How about

"A mechanical contrivance conceived to indicate the weight of an object or objects."

Type in 'pair of scales or 'set of scales' to your search engine.

Just think of the scale your drug dealer uses. You want to buy 1 kilo from your dealer, he puts a pile on and the digital display says 1.005 kg, so he cuts a tiny bit off.

Originally Posted by shmoe

Just think of the scale your drug dealer uses. You want to buy 1 kilo from your dealer, he puts a pile on and the digital display says 1.005 kg, so he cuts a tiny bit off.

And when you get home you find it's only 785.63gm

Originally Posted by billco

And when you get home you find it's only 785.63gm

But you're not terribly upset, given that you paid with counterfeit coins anyways.

Originally Posted by shmoe

Originally Posted by billco

And when you get home you find it's only 785.63gm

But you're not terribly upset, given that you paid with counterfeit coins anyways.

I'll have you know I lavished great care on those coins!

Answer below in white Highlight the field to see it.


THe answer is ONE weighing (at which you need to weigh 45 coins).

Take,
1 from the first pile
2 from the second
3 from the third .......
9 from the ninth.
(no need to weigh anything from the tenth.

This will give you 45 coins. compare the measurement against the weight of 45 real coins. and record the difference. Since you know the weights of both a real and a counterfeit coin you can tell how many of the 45 coins are the wrong weight. (if no difference then pile 10 is in error) The error weight /diff weight will tell you which pile is wrong.

E.G. If each real coin weighs 10 grams, each counterfeit coin weighs 11 grams and the 45 coins weigh 453 grams

then there are 3 extra grams therefore pile 3 is counterfeit!