Boxes and Coins: Newsletter Challenge (06/12/07)

I have a method that requires either previous knowledge of the good coin's mass, or two weighings. Put on the scales 1 coin from box 1, 2 coins from box 2 ....etc up to 10 coins from box 10 (55 coins total).

Let 'E' be mass we would expect if all coins were good. 'A' is actual mass of these 55 coins.

E-A = x.(5/100) ; x = number of box of bad coins.

If you don't already know the weight of a good coin then you need to check the weight of any individual coin. If 55 times the individual coin's weight is greater than the coins you previously weighed, then you have weighed a good coin. Otherwise you have weighed a counterfeit coin, so add 5 grams to get the weight of a good coin.

Seems to me that using the procedure described in post #1 ....

you don't need to know the weight of the coins. You just take the total weight of the 55 coins and divided by 54 coins and the remaining number will tell which box is counterfeit. if the number is 0.95 it's from box 1, if 0.90 it's from box 2 .... if it's 0.50 the counterfeit is in box 10.

This is based on a spring scale and not a leveling balance.

Dwight

I don't see how you could do this with less than two weighings, even if a spring balance were accurate enough. Please explain what you mean in more detail, taking (just as an example) coins with a correct weight of 1.0732 grammes.

Sorry ...... I was a bit too trigger happy with an equation ...... however, I believe there is an equation there with one weighing on an accurate spring scale ...... somewhere ...... I'll let you know if I stumble into it ......

Dwight

ok .... I think I got something ....

if you use the method of Post #1 and work thru the following 10 equations then the counterfeit box is revealed by equation where the number x that is closest to a whole number (one that doesn't go on and on in a decimal) ..... W is the total weight of the 55 coins.

if box Number 1 is counterfeit then equation x = (W+(1)(0.05))/55 is closest to a whole number.

if box Number 2 is counterfeit then equation x = (W+(2)(0.05))/55 is closest to a whole number.

if box Number 3 is counterfeit then equation x = (W+(3)(0.05))/55 is closest to a whole number.

if box Number 4 is counterfeit then equation x = (W+(4)(0.05))/55 is closest to a whole number.

if box Number 5 is counterfeit then equation x = (W+(5)(0.05))/55 is closest to a whole number.

if box Number 6 is counterfeit then equation x = (W+(6)(0.05))/55 is closest to a whole number.

if box Number 7 is counterfeit then equation x = (W+(7)(0.05))/55 is closest to a whole number.

if box Number 8 is counterfeit then equation x = (W+(8)(0.05))/55 is closest to a whole number.

if box Number 9 is counterfeit then equation x = (W+(9)(0.05))/55 is closest to a whole number.

if box Number 10 is counterfeit then equation x = (W+(10)(0.05))/55 is closest to a whole number.

It ain't pretty, but throw it on a spreadsheet and it works.....

Dwight

by the way ...... I just noticed that the number "X" for the counterfeit box is the only solution to the 10 equations of post 21 that is unique.

All other non-counterfeit boxes will yeild the same answer to the equation x=(W+(box #)(0.05))/55.

for example "x" for box #4 is x = (W+(4)(0.05))/55 with "W" being the total weight of the 55 coins of Post #1 ......

There, I'm done ....

Dwight

I don't understand. How do you put "W" on a spreadsheet?

Fyz

Hi Fyz,

You put x (individual coin weight) as the "changable variable" and W = (1)(x)+(2)(x)+(3)(x)+(4)(x)+...(10)(x)

Then you randomly chose a counterfeit box location and subtract 0.05 from the coin of that box (making it the counterfeit coin).

An example:

Lets say you chose box 2 as the counterfeit. The equation becomes W = (1)(x) +(2)(x-0.05)+(3)(x)+(4)(x)+...(10)(x).

Then chose any x (say 1.0732 grams) and you solve for W.

Then you "back solve" all the 10 equations of Post 21 and the unique solution is the counterfeit box.

This works for any value of x you chose and for any counterfeit box location.

Dwight

Hello again Fzx,

I forgot to say that with the solutions of post 21 and 32 you don't need to know the value of the coin weight "x" because by "back solving" the 10 equations, from the measured value of "W" the counterfeit box will yeild the only correct value for "x" (the real coin weights).

In effect you get the counterfeit box location and the weight of the individual coin from the same single measurement of Post #1.

Dwight

This is a curiously circular dialogue.

I really don't understand - unless you mean that you have 10 different readings for W, which is really not necessary (and isn't a single weighing by my reckoning).

Fyz

Hi Fyz,

No just just need one weight of the 55 coins of post #1 to get the counterfeit box location and the weight of the good coins.

Please ignore the spreadsheet references and take the measured weight W.

Put this value of W into each of the 10 equations of post 21 and solve for 10 values of x.

The counterfeit box is the only value for x that is not an on-going decimal and this value of x is the weight of the good coin.

Dwight

Suppose the total weight of the 55 coins is 121.55 grammes, which box contains the counterfeits?

Fyz

Hi Fyz,

Is this a trick question with the coin being a continuing decimal?

if the weight of the 55 coins is 121.54 then the counterfeit box is #9 with the non-counterfeit coins weight at 2.218.

if the weight of the 55 coins is 121.56 then the counterfeit box is #2 with the non-counterfeit coins weight at 2.212.

if the weight of the 55 coins is 121.55 it could be boxes 1 with coin weight 2.2109090909... or box 2 with coin at 2.21181818181.... or box 3 with coin weight at 2.2127272727... or box 4 with coin weight at 2.213636363... or box 5 with coin weight at 2.2154545454... or box 6 with coin weight at 2.21545454... or box 7 with coin weight at 2.216363636... or box 8 with coin weight at 2.2172727... or box 9 with coin weight at 2.218181818.... or box 10 with coin weight at 2.21909090....

you tell me...

by the way, that is a nice problem you developed.

Respectfully,

Dwight

No, it's not a trick question - it merely points out that coins should not be expected to have rational metric weights. If they have rational dimensions, then the volume will be a rational multiple of pi - but the weight will depend on the density of the alloy that is used. There is no reason that a coin should have a rational metric weight.

As there are now two solutions that don't require this, I had not realised that this is what you were assuming.

Fyz

Hi Fyz,

not sure if I convinced you that in most instances you can find 1) the counterfeit location, 2) the weight of the good coins and (by default) 3) the weight of the counterfeit coins based on the results of the 10 equations of post 21 with the one measurement of 55 coins described in post #1.

Dwight

No, you have no convinced me. Even with real coins that have nominally "round number" masses it would not work*. Take 55 coins of nominal mass +/- 0.2%, and assume the masses are similar because they are all from the same batch (0.2% is much better than US treasury standards). Say the nominal mass 2.5 gramme (that is the US cent - quite a light coin). The variability in the calculated result would be +/- 2.5*55/11*0.002 = +/- 0.025-gm. That actually bridges the differences between the coin weights - so in this case an absolute weighing could not work even if you knew the nominal mass of the coin.

Worse than that, you would need an accuracy in the nominal mass of the coins of +/- 0.01% in order to distinguish the beginnings of the recurring decimal from the general mass variation - and that is with a light coin whose nominal mass has no more than two digits resolution - the dime (for example) has a nominal mass of 2.268-gm.

Regards

Fyz

*And theoretical coins do not have to have round-number masses in grammes.

I should have added - most modern coins are made with rational weights in grammes to simplify the use of counting machines. On the other hand, very few of these are worth counterfeiting. An example of a collectible coin that has been widely counterfeited is the two rand gold coin. Its nominal weight is 248.76871779408 grammes, but I do not believe the weights to be uniform to that level of accuracy.

Regards

Fyz

248.76871779408

Can you explain Fyz. I am , as usual , confused.

Hi Kris

Dwight was assuming that the coins had masses that were precise decimal numbers that terminated after a few places of decimals. As he had 55 coins, if they were all good, dividing the mass by 11 would result in a reasonably short decimal outcome. The error due to the 55*N shortfall would result in a recurring fraction. Dwight therefore adds back all the possible errors in turn before dividing by 11. He will only get a decimal outcome that terminates if he has replaced the actual error.

The problem with this is the assumption that the coins have masses that terminate after a number of decimals that is short compared with their accuracy. I chose a coin that has been much forged as a counterexample.

However, this method would fail for almost any base-metal coin - as they are generally made with an accuracy of only about 1% (so you would have no chance of identifying even the start of a recurring sequence...)

Regards

Fyz

Hi Fyz,

Thank you for post 121.

I had thought that the accuracy of the spring scale in measuring both the individual coin and group of coins would result in a number that did not repeat and therefore could allow the identification of the tainted box. Perhaps the equations of post 21 (using post 1 to identify the 55 coins) would work for most instances....

However, I believe your discussion to be clear, complete, consise and correct.

I tip my hat to you.

Dwight

Thanks for the gracious acknowledgement

Fyz

Hi Fyz,

It's this bit that throws me ;

An example of a collectible coin that has been widely counterfeited is the two rand gold coin. Its nominal weight is 248.76871779408 grammes

Can you elaborate on how this fugure is arrive at ? I'm sure I'm missing something obvious in how it's derived . A two rand bit weighs how much ?? (cf my 'confused' link)

Thanks,

Kris

Hi Kris

It's wrong of course - my mistake using the first figure I found (it gave an incorrect mass, and stated it was in Troy ounces that I converted to grammes). Apologies for causing confusion.

I have now checked, and so far as I can establish:
The early one and two Krugerrand coins would have weighed one and two Troy ounces respectively, or 31.1034768 grammes and x2 that respectively. The nominal mass of a standard issue gold two rand coin is 7.9881 grammes. However, 1-rand coins with a mass of 1 Troy ounce have recently been made for collectors, and this is precisely the sort of coin that it would be attractive to forgers.

I still think the basic point was correct - although the data I used was *$%^!

Given your question - you probably know better than I...

Regards

Fyz

Here is the link info for 2 Rand coin;

.917 Gold Gold Content: 0.2354 oz Edge: Reeded Weight: 7.9881 grams Size: 22.05 mm Finish: Proof

No worries , I think I get your intended meaning Fyz !

Thanks for indulging me,

Kris

The one and two rand coins never ever weighed 1 oz.

The one and two rand coins are half soverign and sovereign equivalent coins.

You are thinking of Krugerrands, not the one and two rand coins, these indeed did hold 1 troy oz of gold but weigh more than that because they are .917 gold, so actual weight is 33.9305 grams.

sorry post in wrong place.

This makes no sense jdretired.

I usually dont. get confused?

Confusion is an equitable state of mind.

It is here. Glad I didn't step on that partivular boat.

Board of Directors suing Board of Directors. How wondrously superfluous.

From What I know it was individual investors who got ******d by Equitable. The directors continued to operate while they knew it was collapsing (and of course escaped safely themselves).

That aside , It's wonderfully accurate that the collective noun for Diectors ia a Board. The origin may not have been seen as a joke ,and is probably rooted quite literally in the usage as 'board of wood' , but a Board is often full of planks.

Yes, as well as splinters and termites.

"Confusion is an equitable state of mind."

Yes it is. Now excuse me while I play with my mental blocks!

This problem is starting to weigh on my mind! I hope I don't become unbalanced. If I do, I might try to scale the walls!

Sorry for getting so off topic, but if someone opens the door, the punster in me comes out.

Rather than doing all the calculations, simply stack all the boxes on the scale, get the toeal weight, remove one box at a time, and jot down the decrease in weight, and viola you can find what you seek. Since you put all the boxes on the scale at once, technically you are only performing one weigh-in because you would still have to remove the boxes from the scale, and removing them one by one is the least stressful way.

I know many will want to argue with this technique, but the instructions did not stipulate how the task was to be performed except the one weigh-in.

Not "W". He said to throw "it" on the spreadsheet.

Hmm, this is at least a solution that obeys to "one weighing on the scale". But what you calculate is every time the weight of one coin in case the hypothesis was true.

So by looking for the result closest to a whole number to validate the hypothesis, you assumes that the weight of one coin is a whole number. That would in reality by most rarely the case. So the procedure would not work.

But I also have no better solution.

barber

Hi Guest,

The coin does not have to be a whole number for this to work..... it just has to be reasonably close to a whole number.

Please see post #80 for an example of this.

Dwight

But we are not told that both the counterfeit and good coins are identical

in weight within their categories.

The snake said 'identical looking coins`.

The 10 boxes must be placed on the scale

the weight per box can now be determined

(Weight + 5 gram) / 10

The boxes can now be removed 1 by one.

when the weight of the balance equals the number of remaining boxes * the standard weight you have just removed the fake box.

I trust that this method would pass as one weigh.

I agree with this, I put my comment before reading any results and came up with the same answer. kutos

See Hendrik's later solution - post #20. That doesn't require taking ten readings - that is surely stretching the idea of a single weighing beyond breaking point.

I thought of weighing into this debate (crude pun).

I agree that 1 representative sample from each box would be sufficient. I agree that consecutive measurements would reveal the fake coin and hence the fake box.

This leaves the term 'weighing'. If a sample is measured and removed from the scale this would constitute a 'weighed' item. If it remains on the scale it is still being 'weighed'.

If I intend to weigh 10 coins (1coin per box) then placing the coins on the scale 1 at a time would still constitute a single 'weighing'. Noting the rate of change is secondary to the weighing and would reveal the fake coin ergo box.

Therefore the art is in taking a single representative sample from each box (marked accordingly etc) and placing it on the scale.

Otherwise it takes more than one attempt and is outside the design.

Alternatively, take all the coins and spend them as you only have a relatively small chance of being caught

Cliff

That is just playing with semantics - and avoids looking for physical insights. It's much more interesting to at least try to solve the problem using a single numerical reading from the weighing - which must be what the questioner intended.

limited to one weighing on the scale

I sense a difference of opinion about meaning will occur.

As any misunderstanding spawns serious differences!

And therwith spak this clerk, this Absolon,
"Spek, sweete bryd, I noot nat where thou art."
This Nicholas anon leet fle a fart,
As greet as it had been a thonder-dent,
That with the strook he was almoost yblent;
And he was redy with his iren hoot,
And Nicholas amydde the ers he smoot,...

Hey , I thought nobody would see my sly little joke !

Assuming that the questioner means the difference between the total weight of the coins in the box, that the counterfeit coins are identical, and we are allowed a differential weighing:

Place one coin form box 1, 2 coins from box 2, 3 coins from box 3 etc, up to 9 coins from box 9 on one side of the scales, and 45 coins from box 10 on the other side. The difference in weights will be 50-mg if the counterfeits are in box 1, 100-mg if they are in box 2, etc up to box 9, and 2.25gm (in the opposite direction) if the counterfeits are in box 10.

Fyz

P.S. Be careful how you place the coins, so you can isolate the bad ones afterwards

P.P.S. You only need to place a maximum of 15 coins on each side of the scales really. The above was my first conceptual thought.

Hard as I try , I can not argue with your excellent method Fyz. <applause>

I have frivolous arguments about the question that I will save for later .

I agree about the question. How about someone stating the 15-per-side - or is that just too obvious?

ps - didn't read your pps () . <Kelvinating>.

Take the coins out of the boxes and place them on the scale as shown below then add required weights to balance.

Side A Side B

Box 1 95 5

Box 2 85 15

Box 3 75 25

Box 4 65 35

Box 5 55 45

Box 6 45 55

Box 7 35 65

Box 8 25 75

Box 9 15 85

Box 10 95 5

The box of bad coins can then be determined.

For example:

If Side A is .5 grams lighter than side B Box 5 has the bad coins.

If Side A is 4.5 grams lighter than side B Box 9 has the bad coins

I didn't see physicists post until after I posted mine. I think I like his way better although they will both work.

Agreed - there is an large variety that satisfies the principle. Your method has an elegant symmetry.

In the end, however, I would prefer to use just 15 coins on each side - mainly because it's more practical to keep the stacks separate if the maximum height of a stack is five.

The type of scale may be a factor in the method used. We are told only that the scale is very accurate.

With a balance scale, Physicist's or Troy's method would work. But with the other type, as you add each box to the scale, note the increase for each box. When you add the box of counterfeit coins, the increase will be 5 grams less than previous increases.

Now here's another twist:

The counterfeit coins could be made of a material that is less dense than the real ones. In that case, you can ignore the scale at first. Remove one coin from each box, and perform a chemical analysis on them. If the chemical tests indicate which box has the counterfeits, use the scale for confirmation.
The difference in chemical compostion is entirely possible, because the only measurable difference in the coins is the weight. That means you cannot circumvent the process by measuring diameters or thincknesses of samples from each box.

I'm not proposing a final solution right now, just introducing a couple of factors that might have been overlooked.

I am twisted. The counterfeit coins contain a void.

Not overlooked - just not (in my view) answering the question. One of these solutions involves multiple weighings* (that could be significant if you are using one of those automated chemical balances, where you can't load or unload while the balance is active). The other damages the genuine coins as well as the fakes, and wouldn't work if the counterfeiter had plated his with metal from the originals.

(Given technical freedom, volumetric check followed by X-ray diffraction is probably acceptable. Similarly, I could drop that coins in a viscous fluid and time the fall. I'm certain that people could come up with several more methods. But these are not in my view answers to the question as posed)

Fyz

*Agreed that the principle would work on a 1970's pay-per-weigh machine - but that would not be sensitive enough to detect a 5-gramme differential. SFIK, modern pay-per-weigh machines are proof against this trick.

I have not seen pay-per-weigh scales in a while. I racall* being able to step on with a friend. You could take a reading then , and again if your friend stepped of carefully.

*I merely observed this of course.

Statute of limitations for minor crimes - up to the 1970s, most machines had a sort of lock that meant the weight measure would not come downwards until near zero, but you could progressively load them, and take the weights on an increasing basis. Maybe your friend was lighter than the difference between the lock and the combined weight.

Fyz

To heck with the law - I ripped somebodies scales off. Ha-ha.

I don't recall where or when this was , but it must have been c1970. My smaller mate got off first. It was a finely judged move. We had no interest in weight - just doing something that seemed naughty. Now I think about it , I haven't grown up.

Assume a balancing scale

on the left place

1 coin from box 1 , 2 coins from box 2 , 3 coins from box 3 , 4 coins from box 4 and 5 coins from box 5

on the right place

1 coin from box 6, 2 coins from box 7 , 3 coins from box 8 , 4 coins from box 9 and 5 coins from box 10

Now balance by adding weights

get the number of fake coins by dividing by 5/100 grams.

The box number can then be determined by the number of coins and the side the weights had to be placed.

example 4 coins on the right ==> box 9.

I did not read the other posts carefully before replying. I will change the method to be different and also not use weights.

put 1 coin each from boxes 1 to 5 on the left

and 1 coin each from boxes 6 to 10 on the right

note the heavy side.

now start adding coins in pairs on the opposite side of the original ie box 1 on the right and box 6 on the left, box 2 on the right and 7 on the left. etc

The moment the scale is in balance the coin last placed on the heavier side is fake.

OK - The owner of the scale will kill me dead.

I suppose modern scale work with readouts and not weights.

Nice work. At first glance I would have dismissed this as multiple weighings, but I'm sure it's the kind of logic the questioner had in mind.

The real question may end up being: How do you justify calling this "one weighing on the scale" ?

Davo

Hi Davo

I am puzzled - I would have thought that Hendrik's earlier reply (number 20), or the version of it that uses a digital read-out would be what the questioner meant. The main reason is that this is a single differential weighing. The other part of the reason is that he gives you sufficient information to do it in this way, and also states "very accurate" rather than "very sensitive".

If you are going to do the job by adding coins, testing the balance as you add is equivalent to this, and uses only a maximum of five coins per side. I can see that it doesn't go through the process of weighing - but is "going through the process" with uncalibrated weights (the coins) really any different?

Fyz (in pedantic mode?)

No

Excellent

Fyz

This is trial and error and also multiple weighings.

Post #20 is a single weighing and calculation - unless you count the traditional chemical balance method of weighing using calibrated weights as multiple weighings.

If it was a modern balance that used a load cell to sense the difference, would you allow that?

Fyz

Following the method of posting # 20: to reduce the sensitivity of the scale needed as much as possible, why not multiply all the coin numbers by twenty, and then there will be a difference of at least one gram between the two sides.

Perry

Hard to keep the large stacks of coins identified, for when we want to return the forgeries to the correct box?

Fyz

The difference per coin = 5/100

divide the boxes into pairs = 5.

remove the same number of coins from each pair.

each pair to have a different number of coins to the other pairs.

separate the boxes into two piles where each has a different number of coins in each box.

if n = number of coins in any box

and w = the wieght difference of the two groups of 5.

then the box with the lighter coins = (5-w)/(5/100) = n.

the box on the lighter side with n number coins is the counterfeit.

Did you spot the mistake?

n= w/(5/100)

Idealy the 5 boxes should contain 100,80,60,40 and 20 coins. Which gives an even scale.

5/.05 = the box containing 100 coins

4/.05 = the box containing 80 coins etc.

I forgot to make it clear about the weighing, put five boxes on each side of a balance scale (ten in total ) so that any two boxes on the same side do not have the same number of coins, but has a matching box on the other side with same number of coins. And with one weigh get the difference w. this divided by 5/100 is the number of coins in the conterfeit box on the light side. A one sided scale can only work if the weight of a good coin is known?

PUT both bowes on the scale then remove 1

subtract for your answer

I was wondering how a loose bolt could have been found inside a nuc plant - But I am certain now that a big nut left it there.

But 1_nut is completely Innocent.

Not sure if this was said, but I don't feel like reading through all the posts:

I would suggest sticking all ten boxes on the scale at once and take the average mass of the ten boxes. Then removing each box, measuring the mass of the remaining boxes then averaging it for how many box are left on the scale.

As the real coin boxes are removed the average value will decrease. Once the counterfeit box is removed the average will increase past the first average measured. I did it in a spreadsheet and it worked.

This complies with all the requirements above. Only uses the scale once, and says that the scale is very accurate so I don't see why this isn't feasible.

It works - but it is rather stretching the meaning of "one weighing on the scale". If you avoid the "what can I get away with semantically" route, you might find a more interesting solution.

The anonymous debating with the anonymous is a complete waste of space.

Surely that depends on the content? Personally, I don't care who wrote what - only whether it makes sense. (Except that knowing the originator sometimes makes me read more carefully...)

Fyz

You're absolutely right.

Fyz.

That would have been even funnier if you'd used your ID. But please don't spoof me again - I just can't handle the automatic upgrade to "Guru", so I'm posting my trivial comments without logging in.

Fyz

"Member titles are a fun way to track user participation on the site. The more posts you submit to CR4, the higher you'll rise" - CR4 FAQ

The title means nothing of any consequence . Why worry ? I have not heard of anybody being troubled by others seeking expert advice because they are shown as 'Guru' . Even I would not ask my own expert advice.

The absence of a log-in name makes life difficult for others following a thread. You could always add some form of disclaimer to your quote line. CR4 also has procedures if you feel that anybody is 'stalking' you etc. Perhaps you could even start a thread about the issues that concern you (?)

Kris

Yes Kris, hopefully the guests will find CR4 useful and enjoy it enough to sign up so we can get acquainted.

As far as the solution goes I am at a total loss with only 1 weighing. But, I am curious to see the final outcome next week. All of the thoughts presented are very interesting.

(aside) I don't think there is an on line calculator that will help.

Rich

Did you read Hendrick's post #20? The clue is not just that you know the value of the difference, but also that you know the sign (the counterfeits are lighter). Therefore, If the "scale" is a differential scale (such as a chemical balance) the value of the difference tells you not only how many fakes you have, but also which side the fakes are on.

(And, with stacks of coins that are at most five coins high, it should be practical to keep them tidy, so you can return the fakes to their original box. (But I doubt that there is a solution with a single-sided scale).

Fyz

I rarely bother to read postings by Guest any longer.

I'm starting to agree Ace. Even if somebody put's their name at the end , there is no way of knowing who really posted something.

Agreed.

I think this is an easy one if all of the coins are the same denomination and country. --- Place all ten boxes on the scale. Take the total or combined weight of all of the cubes and add 5 grams. Divide by ten to get the weight of a single box and then begin removing the boxes. When a box reduces the regisered weight by 5 grams less than expected, you have the counterfiet.

Again, multiple weighings.

Absolutely, except that you don't have to add anything. When 1 box is 5 grams less than the others, then you have the box of counterfit coins. Simple. Poundkatt

Take n number of balls from box n, n being from 1 to 10, call it set 1 and it has 55 balls.

Make set 2 with 55 balls from set 1.

Get the difference of the two sets, if the difference is 270, it is box no. 1.

If not, difference/5 is the number of the box.

-Abhishek

Take n number of balls from box n, n being from 1 to 10, call it set 1 and it has 55 balls.

Make set 2 with 55 balls from box 1.

Get the difference of the two sets, if the difference is 270, it is box no. 1.

If not, difference/5 is the number of the box.

-Abhishek

What does

"You are limited to one weighing on the scale!"

mean ?

On the weighing scale, put 5 boxes on both the pans.

Obviously one would weigh lower; that pan contains the counterfeit coins (Pan-1).

Remove one box from the pan that weighs more (Pan-2);

Removing one box from Pan-1; if the weight equals now, the box in your hand has the counterfeit coins; else replace the box, take another box from pan-1

Regards

Chockalingam

The type of scale to be used is a flat radial scale suspended by one point that is very accurate the center point of the ring. The "Plane" of the radial plate in side view must be set perfectly true. You then place all boxes on the ring and leave to settle. One side will rise very slightly. This will very accurately show the lightest box.

KennyT

Nice one. It fits the terms of the question and works fine.

Hi Kris..now't to do with me...

The private mesage system seems to be down.

Not my sort of Q...I'd have dissapeared up my own arse by now.(as us cats are prone to trying!)

I shall be an amused bystander (or bysitter...for those pedants out there) on this one.

Cheers...now play nicely

Del