Monster box puzzle

[paulmerton]

Just for some Friday fun, a quick topical puzzle...

You buy a monster box of 1oz silver Britannia coins off your dodgy mate down the pub. You have a quick look inside, and as expected, the box contains 20 tubes, each containing 25 coins (500 coins in total).

But after you leave the pub, your dodgy mate phones you because he's just remembered that he "accidentally" replaced the contents of one of the tubes with 25 fake Britannias! He assures you the remaining 19 tubes are genuine, and helpfully points out that the fake coins can easily be identified as they only weigh 0.95 troy ounces each. The fake coins are extremely convincing and cannot be identified by sight (they even have milkspots!); the weight difference is too slight to detect from handling; and the only testing equipment you have with you is a digital scale that can measure up to 10kg and is accurate to the nearest 0.1g. You do not know in advance how heavy an empty tube is.

So the question is... What is the smallest number of times you would have to use the scales to always guarantee finding the tube of fakes?

[Derv]

Once - finding said dodgy tube first pick 😬😋

[paulmerton]

Just now, Derv said:

Once - finding said dodgy tube first pick 😬😋

How would you know it's the dodgy tube first?

And what if you don't pick the dodgy tube first?

[Derv]

It was tongue in cheek. I haven't worked out the math. It would be 0.05 toz ×25 is what the tube would be less a legitimate tube. Finding it first pick is the conundrum

[LawrenceChard]

47 minutes ago, paulmerton said:

Just for some Friday fun, a quick topical puzzle...

You buy a monster box of 1oz silver Britannia coins off your dodgy mate down the pub. You have a quick look inside, and as expected, the box contains 20 tubes, each containing 25 coins (500 coins in total).

But after you leave the pub, your dodgy mate phones you because he's just remembered that he "accidentally" replaced the contents of one of the tubes with 25 fake Britannias! He assures you the remaining 19 tubes are genuine, and helpfully points out that the fake coins can easily be identified as they only weigh 0.95 troy ounces each. The fake coins are extremely convincing and cannot be identified by sight (they even have milkspots!); the weight difference is too slight to detect from handling; and the only testing equipment you have with you is a digital scale that can measure up to 10kg and is accurate to the nearest 0.1g. You do not know in advance how heavy an empty tube is.

So the question is... What is the smallest number of times you would have to use the scales to always guarantee finding the tube of fakes?

On a first attempt basis, I make it 5.

If instead of a digital scale, we had an old-fashioned balance, there may be a different answer.

😎

[paulmerton]

Just now, LawrenceChard said:

On a first attempt basis, I make it 5.

If instead of a digital scale, we had an old-fashioned balance, there may be a different answer.

😎

I'd be interested in your approach! (Using just a digital scale)

[ady]

Take one coin from tube one, two coins from tube two and so on. Should equal 1110 ounces. If tube one is fake it weighs 1109.5, if tube 2 is fake it weighs 1109 and so on....

[LawrenceChard]

6 minutes ago, paulmerton said:

I'd be interested in your approach! (Using just a digital scale)

1st Iteration: Weigh any 10; the rogue tube is in the lighter 10

2nd Iteration: Weigh the suspect 5; the rogue tube is in the lighter 5

3rd Iteration: Weigh 3 of the suspect 5; this will identify either 3 suspects, or 2 suspects.

4th Iteration (a): Weigh 2 of the suspect 3; if one is light, problem solved, if not put the third tube to one side.

4th Iteration (b): Weigh one of the remaining 2; if one is light, problem solved, if not the third tube you put to one side is the target.

It is almost always good practice to show your method.

😎

 

 

 

[LawrenceChard]

4 minutes ago, ady said:

Take one coin from tube one, two coins from tube two and so on. Should equal 1110 ounces. If tube one is fake it weighs 1109.5, if tube 2 is fake it weighs 1109 and so on....

Well done!

😎

[paulmerton]

9 minutes ago, LawrenceChard said:

1st Iteration: Weigh any 10; the rogue tube is in the lighter 10

As you don't know the weight of the empty tubes, this step alone would require two weighings, and could be optimised by weighing 7 and 7 rather than 10 and 10. If both 7s are the same weight, then you can deduce that the fake tube exists amongst the 6 unweighed tubes.

Also by that stage you will know what a tube weighs.

I think this kind of approach is probably going to be the fastest.

You can actually solve this problem with just one weighing, but it's far from quick! 😎

[paulmerton]

23 minutes ago, ady said:

Take one coin from tube one, two coins from tube two and so on. Should equal 1110 ounces. If tube one is fake it weighs 1109.5, if tube 2 is fake it weighs 1109 and so on....

Not quite sure about your maths there 😃 

You seem to have plucked 1110oz of silver out of a 500oz monster box and still have some spare!

[ady]

2 minutes ago, paulmerton said:

You seem to have plucked 1110oz of silver out of a 500oz monster box and still have some spare!

Yes your right should be 210 was getting ahead of myself, but the principal works for one weighing,  Glad you spotted the deliberate mistake😁

[paulmerton]

Just now, ady said:

Yes your right should be 210 was getting ahead of myself, but the principal works for one weighing,  Glad you spotted the deliberate mistake😁

You also seem to have assumed that the fake coins now weigh only 0.5oz. My mate ain't that dodgy! 🤠

[LawrenceChard]

7 minutes ago, paulmerton said:

Not quite sure about your maths there 😃 

You seem to have plucked 1110oz of silver out of a 500oz monster box and still have some spare!

But he has identified an optimal method in terms of the number of weighings, I think.

😎

[LawrenceChard]

12 minutes ago, paulmerton said:

Not quite sure about your maths there 😃 

You seem to have plucked 1110oz of silver out of a 500oz monster box and still have some spare!

I could find him a job!

😎

[ady]

15 minutes ago, paulmerton said:

You seem to have plucked 1110oz of silver out of a 500oz monster box and still have some spare!

You really need to see my monster boxes!

[daca]

1 hour ago, paulmerton said:

What is the smallest number of times you would have to use the scales to always guarantee finding the tube of fakes?

 

Zero  times.

You open all the tubes and eliminate the ones that have milk coins or scratches, the remaining tube with perfect coins is the one with fakes.

 

[paulmerton]

31 minutes ago, daca said:

 

Zero  times.

You open all the tubes and eliminate the ones that have milk coins or scratches, the remaining tube with perfect coins is the one with fakes.

 

But:

2 hours ago, paulmerton said:

The fake coins are extremely convincing and cannot be identified by sight (they even have milkspots!)

 

[James32]

Honestly after all this frustration and wrong answers...personally I'd just dump the whole monster box...better safe than sorry.

[Stu]

I could beat up your dodgy mate if it expedites things? 

[Bigmarc]

After I give the dodgy guy a good hiding and making him bring me the full tube of originals back I would be more concerned about what I do with the fakes. Is there some kind of amnesty bin I can put them in to prevent them from getting back in the system? 

I would use the scales once, weigh the whole box and take a tube out at a time until there is not a consistent subtraction. 

[paulmerton]

3 minutes ago, Bigmarc said:

I would use the scales once, weigh the whole box and take a tube out at a time until there is not a consistent subtraction. 

A slightly sneaky attempt at cheating there, but it wouldn't work anyway - the scales can only weigh a maximum of 10kg!

[LawrenceChard]

1 hour ago, paulmerton said:

As you don't know the weight of the empty tubes, this step alone would require two weighings, and could be optimised by weighing 7 and 7 rather than 10 and 10. If both 7s are the same weight, then you can deduce that the fake tube exists amongst the 6 unweighed tubes.

Also by that stage you will know what a tube weighs.

I think this kind of approach is probably going to be the fastest.

You can actually solve this problem with just one weighing, but it's far from quick! 😎

So, by my sub-optimal method, the answer would be 6, not 5.

@ady was the first and only one to get the correct answer, and method, but failed due to faulty arithmetic.

Thanks for setting us this problem.

Now, how about the correct gross weight of gold sovereigns?

 

😎