UKH

Ok, you've just received a box of pennies. An admirer perhaps.

The pennies are wrapped in paper to form tubes of 50 pence. In the box is a total of 50 tubes. So £25 in total.

Now, you've been informed that one of the tubes if full of counterfeit coins. 

They are appear identical to the genuine coins, and the only way you can tell which coins are fake and which coins are genuine is by weighing them. The fake coins weigh 2.2 grams each, whilst the genuine coins weight 2.5 grams each.

Now, you could weigh each tube of coins, identify which one weighs 115 grams (paper weights 5 grams) and then identify which tube contains the fake coins.

This could take up to 50 uses of the weighing scales to identify the offending tube.

What method could be used to identify the fake coins using the weighing scales as few times as possible? Using your method, what would the maximum number of times that you would need to use the scales to guarantee that you find the offending tube?

The method that requires the least number of uses of the scales wins.

1

But, requires lots of prep.  So hardly worth it in real life.

4 max. 1 min.

E2A just realised my method uses balances rather than scales, to find the heavier of 2 items

1

Split the collection into two 12s and a single. If they balance, the single is the fakes. If they don’t balance the fakes are in the lighter 12. Split that side into 2 sixes. Repeat. Split lighter side into one each side of balance. If they balance, it’s the third one, if they don’t its the lighter one. 

5.

As an engineer, I would do  6  to double check the result.

I can do it in zero uses of the scales without about 2% accuracy.

Or 100% accuracy using 1 use of the scales and lot of perparation.

You don’t state if the scales are a mechanical balance between two sets of stuff, or a device that gives the weight of one set of stuff.  If the later, you don’t specify the precision of the readout.

Without those being clarified there is no answer, as there is ambiguity.

If you google “digital balance” you will find lots of devices that weigh one set of stuff, it’s common parlance in labs for example, so the use of the word balance doesn’t imply a “two sets of stuff” device.

zero.

They are rolls of coins, so you can create a balance by balancing one roll of coins on top of another one but at 90 degrees and precisely centred. Now take two more rolls of coins and balance them on opposite ends of the one you first balanced. Repeat 24 more times and you'll find that it balances nicely every time except one, the fake role has then been revealed. (you'd need to swap out the ones that make up the balance to do the final 25th time).

Obviously you'd want to have pretty steady hands.

he didn't say you couldn't construct a simple balance which you would check by balancing more than one pair of tubes before using it for the task

Also, another method using the scale zero times is just to pluck one out at random, declare them fake, and be lucky in getting the right one.

Not a very satisfying answer but it fulfils the requirements of the problem.

One weighing. I would place the tubes alongside each other in a row on one side of the balance, then find out what weight counterbalances them. Then I would use moments to calculate the position of the centre of gravity of the row of tubes and hence the position of the anomalous tube.

Very true, but if you allow methods making no use of the scales you just pick them all up in turn and see which one feels different.  A 15% weight difference is trivially discernible by hand, no need to build a simple balance or complex tiered balances out of the objects being weighed.

Take the bucket off the scales and put it in water. Measure the displacement to figure out the lighter roll.

0

I'm going to count any comparison of the rolls weight, by any method, as one use of the scales.

That's just a fancy way of weighing them, so they count I'm afraid.

That only works if the objects float. If their density is larger than 1g/cm3 they will sink and if they are the same dimensions (the OP specifies that mass is their only distinguishing difference), the water displacement will be identical for fake and real rolls.

This is probably true, but there are rules to follow:

"They are appear identical to the genuine coins, and the only way you can tell which coins are fake and which coins are genuine is by weighing them."

The type of scales I had in mind are the ones which give you a number. But if you want to solve the problem using balance scales I think that's fine.

They are precise enough to solve the problem.

Ok, I’ll play.

1 weighing.

Take 1 coin from roll 1, 2 from roll 2, 3 from roll 3 and so on up to all 50 from roll 50. 
 

Weigh this pile.

If I have done the sums right this should weigh 3187.5g however it will be under, the number of multiples of 0.3g which it is under gives the identity of the dodgy roll.

One, I think.

Mix the coins up such that you have one coin from roll one, two coins from roll two, three coins from roll three etc. Weigh that lot and see how much under weight it is. Subtract that weight from what it should have been if it had been with no counterfeit coins to get the total discrepancy. Divide the total discrepancy by 0.3 grams and that should give you the roll number? Or something like that. 
 

Edit: damn, beaten to it. 

I would be very interested to know what the downvoter doesn't like about this method. I've not tested it myself, so maybe it's a genuine concern about the precision required.

The fake coins have a mass of  2.2 grams each, whilst the genuine coins have a mass of 2.5 grams each.

Zero. I'd unwrap them all then shove them all in the machine down the bank which will happily spit out all the fakes.

One weighing. One coin from the first tube, two from the second, three from the third........  fifty from the fiftieth.

Edit: I see I was beaten to it.

One weighing. One coin from the first tube, two from the second, three from the third........  fifty from the fiftieth.

Edit: I see I was beaten to it.

We can all do that Robert!

The problem with this approach is you have mixed some of the fake coins into the pile you weighed. You’re going to have to separate them again.. unless your scales are big enough to have 50 separate piles of coins on!

I could just weigh them all individually to find the fake ones 🤔.

The only alternative I can think of without unwrapping the coins is to split the 50 wraps in half each time, so weigh 25, if it's got the fake then halve those 25 and weigh a pile of 13, again, if you find you have weighed the fake, split that and weigh 7, and so on.  You could do it in 6 weighings that way.

Nah, I liked your approach.. just go and buy some really big scales. But then will they be precise enough…? Oh no, this way lies interferometry! Be afraid. Be very afraid.

I first became aware of this bit of number theory during the Covid pandemic, when PCR tests were a scarse resource in some countries. 

Theoretically, by mixing samples we could vastly reduce the number of tests needed.

I don't think it was ever used in practice, but here's some more info 

https://math.stackexchange.com/questions/1254490/the-blood-test-riddle-numb...

Could we keep the coins in the original rolls, carefully removing the right amount from each one? So roll one has just one coin in it and there’s a neat pile on the floor with the 49 removed coins etc. The maths needs adjusting for the 50 x 5g weight of paper?

There are different ways you could solve this. Simply counting them on to the scales carefully in order would do it or marking the coins with a pencil or as you say different piles neatly stacked.

Of course the method is more complex and time consuming than taking a coin from each roll and weighing it.

Variant of the quantum bogosort.  Pick one roll at random and put it on the scales, then close all timelines where you didn’t pick the fake.

Fire them all ‘just’ past a black hole and measure the Kerr ring downs.. but we’re back to the interferometry.

Stick each tube on a ideally balanced wheel, equally distanced round the circumference. The wheel is mounted vertically and the scales read out in real time, at a high enough resolution that you can identify with a high speed camera which point corresponds to the minimum. Actually, there might be some phase shift (while since I did any balancing), so obviously it would be easier / better done with a tacho and balancer...

Technically one measurement, just a protracted one.

6 measurements was the least I could get to with actual measurements and not splitting tubes. 

It's a fun bit of theory but I can't see how that could be any use in practice. The trick only works if you know you have exactly one positive sample in the batch. You have been given a load of extra information there, that you wouldn't have in a real life experiment. In the example given you have 100 possible outcomes, which has an entropy of less than 7 bits. If you have 100 samples of which anything from 0 to 100 could be positive, then you have 2^100 possible outcomes, an entropy of 100 bits.

The ideally balanced wheel will always tend to deliver the light stack to the top position, no scale needed or more to the point, the wheel is the scale/balance. In the real world with bearing stiction you'd want to do a series of spin-down cycles in each direction, marking the top tube each time then taking the average.

If you're allowed extra kit then a stiff ruler and a couple of dowels in addition to the scale will find the light stack without unwrapping in one measurement.

Place the stacks on top of the rule, all touching so evenly spaced, perfectly centred 25 either side of the mid (balance) point of the bare rule. Place a dowel under the mid-line of the rule as a fulcrum. With the rule level, place the second dowel edge under the heavy end between the rule and the scale aligned with the outer edge of the last stack. Measure the moment, calculate the CoG then determine the position of the light stack.

If you're doing comparisons of stack weights as one step you can do it in 5 steps without opening the stacks and with the possibility of finding the light stack early:

1. 25 vs 25, discard the heavy stack
2. 12 vs 12 (1 left out), discard the heavy stack, id the odd one out if stacks are equal
3. 6 vs 6 discard the heavy stack
4. 3 vs 3 discard the heavy stack
5. 1 vs 1 (1 left out) same logic as step 2

By shuffling one known good stack back in at step 3 and step 4 you introduce more opportunities for a chance early ID of the odd one out without introducing more measurements overall.

I can't do any better than that so 5 steps max, <5 if you repeat the experiment and take the average result. I'll leave it for the reader to work that long run average out

jk

Better to split them into 3 instead of 2 each time, if the weighed pair are equal, the odd one is in the third group, so:

1. compare 17 & 17 (16 left over)

2. compare 6 & 6 (5 or 4 left over)

3. compare 2 & 2 (2, 1 or 0 left over)

4. If necessary compare 1 & 1

Weird, I thought about 3 way splits but somehow cocked up the logic!

jk

a fleet of identical small boats.  put one tube in each boat and see which sits the lowest in the water

Assuming the scales have an LED somewhere you could line up all the rolls of coins and observe the photons passing each one, the fake roll will have a smaller gravitational lensing effect.

you could build a high frequency oscillator with a physically large tuning inductance of sufficient diameter to admit a tube of coins.  This is connected to an accurate frequency meter.

It is fairly likely that the fake tube will cause an anomalous change in frequency as compared with the tubes of genuine coins if care is taken to ensure that insertion is performed consistently.

The odd tube with the heavy coins probably has a slightly higher thermal mass than the others too - same surface area though. 

Put them all in the fridge overnight, take them out and line them up on the bench in a nice warm room.  Use a thermal camera to check they're all the same temperature initially, and then pick out the one that's cooler than the others when they've all warmed up a bit.