You have 10 stacks of silver dollars, with 10 coins in each stack. The coins appear identical, but you know that all the coins in one stack are counterfeit. You know the weight of a genuine coin, and you know that a counterfeit coin weighs 1 gram less than this. How many weighings must you do to find the counterfeit stack?

One.

Yep, one. One stack on one pan, the other stack on the other.

That would work for two stacks, but there are ten of them.

Oh bugger, I didn’t see that. He’s not edited the post has he?

Damn. Oh well, that’s that.

Next time we’ll really, really fool you all.

Surely that one weigh would tell you which group of five had the counterfeits but not which roll it was?

You would need to subdivide the five into a three and a two and add one from the good group to the two; the lightest three has the counterfeits. Then remove one from the lightest three and add a good roll to that one and compare with the other pair. Then split the lightest pair into two singles and compare. You don’t need to know the weight and the answer is four.

As I said earlier, one:

Label the stacks “1”, “2”, “3” etc.

Take one coin from 1, two coins from 2, three coins from 3 etc and weigh them.

The counterfeit stack is indicated by how much the weight is short of what that number of of genuine coins should weigh.

But we don’t know that the counterfeit coins are restricted to the all-counterfeit stack. There could be another stack that’s five of each, for instance. Then a coin you draw from that stack could be genuine or fake, with equal probability. We don’t even know that there’s a single genuine coin in any stack.

So it reduces to “Here’s a hopelessly incomplete account of a problem you may have met in your childhood; which of you can remember the answer from those days?”

And what is the weight of a genuine stack?

Sadly, having weighed the 55 coins and determined that the 8th stack was the one with all the counterfeit coins, you drop and mix up all the 55 weighed coins.

What is the minimum and maximum number of weighings required to separate out the 8 counterfeit coins?

Are there different answers depending on whether one uses a strategy to minimise the minimum or minimise the maximum number of weighings.

Having solved that, and if not yet worn out, how does the number of weighings change between using a (one-dish) calibrated spring type of scale and a (two-dish) uncalibrated balance type of scale?

You will all be pleased to know that these are thought to be new puzzles, so we almost certainly won’t find the answers readily available on the WWW.

Best regards

Clearly, Chuckles has disappeared and left me holding the baby. Haven’t a blinking clue how to answer you.

Wot FrankH sed.

As per the question –

You have 10 stacks of silver dollars, with 10 coins in each stack.

The coins appear identical, but you know that all the coins in one stack are counterfeit.

You know the weight of a genuine coin, and you know that a counterfeit coin weighs 1 gram less than this.

So, ALL the coins in ONE stack are counterfeit. You do not need an upmarket chemical balance type scale, as we are not comparing masses of the stacks against one another.

A simple fairly accurate spring balance or similar accurate to a half or quarter of a gram will do.

As FrankH notes, take 1 coin from stack 1, 2 from stack 2 etc etc. and weigh them.

We will be weighing 1+2+3…+9+10 = 55 coins.

If they were all good coins they would weigh 55 X weight of good coin grams. However, in this case, it’s wight will be less than that by a few grams. The actual number being the number of the stack that contains the counterfeits.

Nigels puzzle is left as an exercise for the student ðŸ™‚

You’re assuming that the other stacks consist entirely of genuine coins. That is nowhere stated in the question. So your solution isn’t valid.

“The actual number being the number of the stack that contains the counterfeits.”

I repeat: we don’t know that there is a single “stack that contains the counterfeits”. All we know is that there is one stack that is all counterfeits.

“The coins appear identical, but you know that all the coins in one stack are counterfeit. ”

The givens are that one of the 10 stacks is counterfeit, but which one, is the question?

While we may indulge in endless levels of sophistry, we rapidly reach a point at which we decide that the stacks of coins are a figment of our imagination..:)

It’s called logic not sophistry. Just you point me to where it says that the counterfeits are confined to the all-counterfeit tower. You can’t; it doesn’t.

Tiptoes away.

I did too. ðŸ™‚

He said all the coins in one stack were counterfeit. You now have ten objects (stacks in this case) one of which is lighter. Forget about coins. So if five objects weigh less than the other five (determined by one weighing of five against five.) The lightest five contains the object in question. You can’t weigh two against five as two will be lighter whether or not the bad stack is in the pair. So you add one genuine stack from the heavy group to two of the lightest five and compare. (Second weighing.) Repeat then do a one against one. Four weighings. Wibble.