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

During a very long career in engineering I have worked for most of the major OEM's utilsing math's physic's,politics,economics etc and combined the clients requirements into sound and approved engineering designs, and been paid fairly well for it, whilst also supporting shareholder needs for good profit.

There is such a thing as "design proposal", ' concept buy off'? Project forecasts and regular cost and feasibilty reviews. Who locks anybody in a room for an undefined period without regular reviews of direction, progress, timing and fiscal control?

KennyT

You haven't listed good communication skills as one of your many attributes.

When you were doing all that great stuff, did you feel the need to constantly observe to your collegues all the great stuff that you'd done in the past, and assure them that the current task would be added to that list in time.

I take your point and declare that yes I have been blessed with many brilliant colleagues who have supported my design proposals, but only after we have all beaten the idea up to see if it held any water. I think it's good that no one is more critical of an engineer than another engineer. My biggest gain from it all is to see an end product that has been the result of good discussion, teamwork and careful planning and control. I love Engineering, maybe seen as sad by some people, but true.

KennyT

As well as who ?

I also had to google the answer. Couldn't stand it anymore

Clever, very clever.

I'm gunna google KennyT

I tried that as well to see who popped up, and found an MD, an Anesthetist and a music maker. The closest one that fits me I guess is the anesthetist. As I can bore people to sleep.

My real name does give several genuine google results, that I'm fairly proud of though.

KennyT

How can there be so many answers that work?

If the weight of a good coin is needed, The weight of a good coin is irrelevant and not needed!

1 - Mark the boxes from 1 to 10

2 - Take from box #1 ... 1 coin

Take from box #2 ... 2 coins

Take from box #3 ... 3 coins.........And so on

3 - Put the selected coins on the scale and notice the reading

4 - If there is a decrease in the reading of the scale by 0.05 gram from a round figure then the false coins are in box #1

If there is a decrease in the reading of the scale by 0.1 gram from a round figure then the false coins are in box #2

If there is a decrease in the reading of the scale by 0.15 gram from a round figure then the false coins are in box #3

And so on.....

Get an audio spectrum analyzer, drop each coin on to the scale without looking at the scale reading, the bogus coin will have a slightly different audio signature. Separate that coin from the rest, look at the scale reading and say " I've found the bogey with only 1 scale weighing!"

Suspend a pulley from the ceiling. From one side hang 1 coin from Box 1, 2 coins from Box 2, 3 coins from Box 3, 4 coins from Box 4, 5 coins from Box 5. On the other side hang 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. Take the side that moves toward the ground (heavier, all real coins) and place it on the scale. Adjust the height of the scale so that both sides of the pulley are at equal height (to negate differences in rope length).

Now weigh once. The weight reading should be the difference in weights of the two sides. Use this value value to determine how many coins on the other side are fake. Use this to figure out which box of coins is fake.

Of course this involves using a pulley, but definitely only uses one weighing no matter how you look at it.

What do you guys think?

~ Yogi

A remarkably smooth/lossless pulley... Why not save the trouble and use a digital balance...

Using a digital balance would be weighing, wouldn't it? Every solution so far seems like it involves incrementally adding/subtracting weights/coins from one of the sides and checking the scale each time. That seems like a weighing each time.

I like the circular balance and the ferris wheel ideas, but I guess it remains to be seen whether those are considered scales.

~ Yogi

No, I was proposing a two-sided balance, and using the same basic principle that you have used with the pulley (or that was used for #20 but some people objected to because there were several balances, though I would call that a single weighing). Early precision balances used a pointer to show a reasonably accurate differential weight up to perhaps 100 mg, depending on type (the pointer meant you could select the final weights more quickly, I think).

They seem to be hard to find now, but there was a period when you could buy a two-sided balance with a digital equivalent of the pointer. The digital differential measurement had ranges of at least 0.3-gm, and you used traditional weights for the precision weighing. I think that would be just what we need...

These days, it seems the term "balance" has been usurped by anyone purporting to offer a precise weighing instrument. The better ones use a reference mass ("weight") to calibrate them against variations in gravity - but they are not balances except in name.

N.B. My wife uses balancing scales and weights for routine kitchen measurements - but scientists generally use single sided scales that they only call balances. Is there an irony here, or what?

Fyz

While I applaud those that are mathmatically inclined, I try to avoid it at all costs. It seems to me that the easiest way to find the bogus box of coins and still only have one weighing would be to put all ten boxes on the scale and note the weight. Remove one box and subtract the current weight from the starting weight (yea, I know it's math but it's within my meager abilities). Keep removing and subtracting. Nine of the results should be the same and one should be 5 grams less. That's the box to send to the feds.

Drive fast, take chances.

I think that would be 10 separate weighings.

It saves a lot of arguing to play solo spoiler.

Afterwards , you can just relax and laugh.

I'll just talk to myself here.

Scales do not always have exact graduations.

Absolutely. you see , it's all a question of interpretation.

I already know that stupid !

Hang on a bit , I was only trying to help.

Oh , OK. Yes , I see. Quantitative versus qualitative.

That's right.

Now look , don't you patronize me !

Lets start again.

Why ? Who said so.

Dunno.

Well shall we just scrap anyway.

Err , if you thi...

<thwack>

A difference of 5 gram per 100 coins would be an acceptable for every day use coins. .

The coins in question would therefore be of gold or some precious metal.

Gold coins are minted in 1 ounce, ½ ounce etc.

The weight of each coin can therefore be established without a separate weighing.

The total weight divided by the number of coins will supply the range.

Method

Place the 10 boxes on the scale.

Take 5 coins from box 1, 10 coins from box 2, 15 coins from box 3 . . . . . . 45 coins from box 9 and 50 coins from box 10,

You have taken away 275 coins and are left with 725.

The coins should now be weighed and divided by 725 to establish the approx weight per coin.

For calculation purposes I am assuming a value of just under 1 ounce per coin

The correct weight of the 725 coins can now be calculated. = 725 ounces

The difference can now be calculated = 725 ounces - weight = xxx

The number of fake coins can now be calculated

The box number can then be established.

As you can see I don't have a idea of the conversion factor from ounces to gram.

The weight of the boxes are not considered.

Fortune favours the brave, and optimism is a wonderful characteristic, but...

Taking differences of 5 coins leaves you needing to resolve 0.25 grammes, or slightly less than 100th of an ounce. The accuracy needs to be in the order of a part in 150,000 - which I fear may be rather too accurate a mass even for a gold coin. That is without considering the mass of the box or the possibility of standard ounces, troy ounces or metric weights...

By the way, the "Royal Mint" estimate that 1% of pound coins in circulation in the UK are forgeries - so it's not "just" gold coins...

Regards

Fyz

The problem tells you to use a very accurate scale, and they do make graduated balances such as a triple beam. Take an even total number of coins from the 10 boxes, knowing how many came from each. (Ex 1 from box 1, 3 from box 2, 4 from box 3, 5 from box 4, 6 from box 5, 7 from box 6, 8 from box 7, 9 from box 8, 10 from box 9 and 11 from box 10.) This totals 64 coins. Arrange them int 2 piles of 32 (11, 10, 8 & 3) (1, 4, 5, 6, 7, & 9).

Now place both piles onto the balance and read off the weight difference, or slide the weights until both sides balance off. Divide this weight by .05g and this will give you how many coins weigh .05g less and thus tell you which pile and which box it came from. (Ex if it took .45 grams to equalize the sides, the pile of 9 coins, from box 8 was light)

Scott

We all agree that we can only have one weighing of the 10 boxes.

following the system as I outlined in #59, the first weighing is only of two boxes, next weighing is of four boxes and so on. By the time you load all ten boxes on the scale five each side you will have the answer already with just one weighing of the ten boxes.

Frank

Read it again Frank (sorry about the misquote) - the question (which is what we are describing loosely) says "one weighing" total, not "one weighing of the ten boxes".

If there was no difference between counterfeit coins and genuine coins, would the counterfeits be genuine ?

What would happen if the guy in the mint pressed the go button on the coin making machine a few times more than he was supposed to, and pocketed the products. Would they be counterfeit or genuine ?

If the quality control on the mint's coin making machine was a bit off [during a normal run], would the products be counterfeit or genuine.

If the people who made the coin making machine made themselves a similar machine and operated it, what would they be making ?

How come coins are couterfeit and not 'pirated' ?

If a white collar code cutting programer hacked into his own bank account and upped the balance, would his balance be counterfeit ? How would you weigh it on a suitably accurate scale ?

Oh Dear!

Double dratt, I've got sucked into this.

On the left scale pan.

Put 1 coin from box 1, 2 from box 2, 3 from box 3, 4 from box 4, 5 from box 5.

On the right scale pan.

Put 6 coin from box 6, 7 from box 7, 8 from box 8, 9 from box 9, 10 from box 10.

Read off the difference...then post it to FYZ so he can do the maths!

I think you meant 1 from box 6, 2 from box 7?

But it's too late anyway - Hendrik with post #20

Fyz

I think you meant 1 from box 6, 2 from box 7?

C'mon FYZ!

If that's what I meant I'd have said it!

In that case, my attempts at maths would have provided no illumination (on which subject, see the footnote to my most recent post under Bermuda triangle...)

Fyz

_______________________
I believe in "Miracles" - but only as a brand name.

Why don't you split the difference.

One last shot....

Left scale pan 99 coins from box 1.

Right pan 7 from 2, 8 from 3, 9 from 4, 10 from 5, 11 from 6, 12 from 7, 13 from 8, 14 from 9, 15 from 10.

If my arithmetic is right this gives 99 on each side...which answers any queries on 'balance'...

At this point I give the results to FYZ and ask him if this is any better?

My 'inbox' is on fire and smoking. Please, when is the answer?

I believe "The Answer" was posted at #20. If the official answer is significantly different, that will just be a disappointment.

You could always unsubscribe until next Tuesday.

Fyz

The official answer should include an apology for not telling us the actual weoght of a genuine coin.

Not if it is unnecessary to solve the problem - as in using one conventional weighing process with a good old-fashioned chemical balance. KennyT's CD balance also does the job...

Fyz

Hi Fyz,

Just realised that the post a couple of days ago in which I wrote "Kenny's method is very elegant and should be considered The Answer" (or something close) never got posted...

My post #220 refered only to the expected official solution with accurate scales - and I think that's what needs clarifying. I've assumed that what is described in the questions is a measuring device with a pan, a spring* and a readout. In this case, you do need the genuine weight, as there is nothing relative (genuine to counterfeit) in your measurement.

I've just read post #20 (missed it in the crossfire) and seen that this uses a balance (2 pans and a pivot) and calibrated weights. Since a balance measurement is relative, this method doesn't need the weight of a genuine coin.

* in principle - to be accurate it might be using a piezoelectric-derived current to provide the measurement

Hi Rose (should that be "English", or do you prefer ER - and if the latter should I insert a number in Roman numerals as on pillar boxes... ?)

Having spent some years with working piezo-electricity, the idea of a DC piezo-electric derived current was rather startling* - that's pure pedantry from me, but maybe a spur for your curiosity.

Fyz

*Piezo-resistive would be fine...

ER ?

Forget post boxes. The more urgent need here is for an Emergency Room. At least a half sensible ER link for those who read my drivel. Those with a humorectomy will not read me and thus miss out on a link. Those who like such progeams will look and not understand. I am utterly twisted.

No ER, a wet-bar. (it'll do more good) I'm absolut-ely parched after this commodious repartee.

LOL - ER is fine*.

Current/resistance...it had been a long day and I was being less than pedantic () in my approach...I had a picture in my head, and as often happens, used not quite the right words to describe it!

And put those spurs away - you'll have someone's eye out!

(and I was right about the official answer - there should be an apology there for the omitted info!)

*I didn't spot the obvious pillar box link when I came up with the name!

Non-pedantic? ER? You've just lost your status as a Chartered Pedant, and will have to re-qualify; the fee for a resit is £1000, to be posted to Rural Institution of Chartered Pedants, c/o Fyz, ...

Given my allergies, I'm not likely to be wearing spurs when sat on my high horse, so your eyes are safe.

N.B. To my mind, knowing the nominal weight of a coin would have made the problem rather trivial, especially given that KennyT came up with one working solution and Hendrik's #20 with another (though the wording didn't make it clear whether either would be allowable). The proposed solution also assumes that the difference in the total weight of the boxed coins is entirely due to the coins.

Regards

Fyz

Agreed. I think KennyT's solution is delightful and Hendrick's elegant.

I said all along that Kenny's idea was brilliant when nobody else seemed to think so.

Three cheers for Kenny and Hendrick (each of course).

Brad

Remove seven coins from each of the piles on the right, and 63 coins from the left, and you are similar to (but better than) my first submission... (why add unnecessary coins - it makes the piles even harder to keep separate?).

But I still prefer to use only fifteen coins on each side - perhaps I'm just old and feeble.

Fyz

I 've read all the post and finally I followed a road little different. The process I found is:

STEP 1- The test-group selection. Extract a group of coins from each box. You can choose from 0 to 100 coins from each box. The only condition is that the number of coins extracted must be different for each box. For being clear in the explanation imagine we select 1 coin from box-1, 2 coins from box-2,…….10 coins from box-10. Total 55 coins.

STEP 2- The weight an the scale. It doesn't matter what type of scale you use but for being different imagine we have one with a digital display like the one you find in any shop , supermarket, etc. We are going to use the scale in a normal way….we put the test-group of coin on it, only once, and read the weight. That 's all with the scale.

In respect to the expression "Using a very accurate scale" assume , only for now, that the scale does not have any error. The display shows the exact weight of the coin.

Don't get tangled with scale precision , errors, etc. More about this at the end of the process.

STEP 3- Time to elaborate the information. Take paper and pencil or better open an EXCEL sheet in your pc. Type in some cell the weight we got from the scale and begin to complete a table that will have 10 rows and 5 columns as follow:

Col.1 . Rows 1 to 10- enter a text that identify the box….("box-1, box-2…..etc").

Col 2. Rows 1 to 10- Enter the number of coins we extracted from each box and that is the conformation of the test-group……In our case the values are 1, 2, 3……..10.

Col 3. Rows 1 to 10 . We know that our test-group have some fake-coins. We know that only one box has fake-coins. We know how many coins were extracted from each box. Now we want to calculate what ,the test-group's weight, would be if ALL the coins (55 coins) were NO-FAKE. As we don ´t know which box deliver the fake-coins to the test-group we have to evaluate 10 cases…..considering that the fake coins could come from the 10 different box…..but as we know how many coins we extracted from each box (see col 2) then starting from the weight we got from the scale , we add 0.05 gr for each coin we extracted from each box and we get 10 different values for the weights of the "test-group-without-fake". Note that all the values are different because we ruled we choose different numbers of coins from each box. Note too that all that values are higher than the one we got in the scale because in the test-group are fake coins.

Col 4. Rows 1 to 10- We calculate what ,the test-group's weight, would be if ALL the coins (55 coins) were FAKE …..having the data of col.3 and knowing that for each 100 fake coins we loose 5 gr, is easy……we have to subtract to each one of the 10 values of "all no fake coins" of col.3 the difference of weight of 55 coins.

Col 5. Rows 1 to 10 . With the values of col 3 and col 4 we have the intervals of variation between 0 fake and 100% fake for each box that could deliver the fakes coin.

Then now we are going to recalculate the test-group's weight with that intervals and the real number of coins each box deliver to our test-group . Then we have to consider that box -1 could have deliver only 1 fake coin to our test-group and box -2 , 2 coins ….etc. The calculation is lineal because for each box we have the range of variation from 0 to 55 fake coins…..

One of the values that we obtain in this column MUST match with the value we obtained from the scale. This is that way because what we get in col.5 are ALL the possible values that our test-group can weight. There are 10 different values depending on which box could deliver the fake coins and one of them must match with the box that deliver the fake coins to our real test group.

The box that deliver the fakes coin is the one indicated in the row (col 1) where that match is found

Now is clear that this algorithm depends on the quality of the scale because from that value we go forward and backward with the algorithm and expect to match that value with the obtained from the calculation. Then my interpretation of the phrase "Using a very accurate scale" is "don´t worried with the precision…..solve the problem ".

Finally I did an Excel sheet where I could test the algorithm …If someone wants to check-it just send me a mail and will reply with the Excel.

I assume this list doesn't have the option of sending files attached.

Federico

foehninger@tenaris.com

foehninger@yahoo.com.ar

I would ask you to send the spreadsheet, except that I don't divulge my e-mail. So I'll just describe the issue as I see it:

The only information you have is
Measured_Weight = 55*Good_Weight - N*0.05gm
where N is the number of fakes

Whatever the actual weight is, and whatever N you choose, I can calculate a value for the weight of a good coin that satisfies this equation.

That means that you are making an unstated assumption, such as that the coin weight is expressible using a limited number of decimal places (measured in grammes, ounces, or Troy ounces??), or maybe you are even further restricting things by using some specific weight for a good coin.

Which is it?

Note that the first assumption is only valid for modern coins, and even then will only work for very valuable ones, because the tolerance on the mass of many coins is in the percent region (clearly the accuracy of the scale becomes irrelevant under these conditions).

Fyz

Fyz:

You are completely wright....forget my idea.........

But now I come with a different idea…..I will use a scale with 2 plates…..

I will use all the coins of the 10 box arranged as follow:

(sorry for the presentation of the following list but I couldn't gave it a better format. The editor seems no to accept multiples spaces)

source / left plate / right plate / dif / Grs. to be balanced

box1 / 11 /89 /78 /3.9 in right plate

box 2 /17 /83 /66 /3.3 in right plate

box 3/ 29 /71 /42 /2.1 in right plate

box 4/ 41 /59 /18 /0.9 in right plate

box 5/ 47 /53 /6 /0.3 in right plate

box 6/ 53 /47 /-6 /0.3 in left plate

box 7 /59 /41 /-18 /0.9 in left plate

box 8 /71 /29 /-42 /2.1 in left plate

box 9 /83 /17 /-66 /3.3 in left plate

box 10/ 89 /11 /-78 /3.9 in left plate

total /500 /500

I ask the operator of the scale to balance the plates and inform me how match weight he added and in what plate. ( only one reading)

The colm "grs. to be balance" shows that are only 5 possible values to balance the plates. The values are not periodicals....are exacts. The combination plate / value of weight added to equilibrate indicate without any doubt which box has the fakes coins.....

It doesn't matter the individual coins's weight (fake/ no fake) . I am weighting the difference of weight between them…..of course I'm assuming the difference of weight is equally distributed

Could it be?????

Regards,

Federico

Hi Frederico

Assuming that the good coins are all the same as each other, and the bad coins are the same as each other, and the difference is in the coins (rather than the boxes) that should work. I think those are the assumptions the questioner made - but we will have to await the published answer to be certain.
The principle is the same as number 20, but there are advantages and disadvantages. The advantage is that you have used more coins and evened out the spacing in error mass, each of which could give the opportunity to make the process more tolerant of minor variations in mass. The disadvantage is that you've included unnecessary coins on both sides, which more than makes up for the improvements. With this number of coins, you also need to be rather careful to avoid mixing up the coins.

As "the solution" will be published shortly, I think the ideal numbers for averaging errors (if you can avoid mixing up the coins) are something like:

1 . 11 on left
2 . 33 on left
3 . 55 on left
4 . 77 on left
5 . 99 on left
6 . 11 on right
7 . 33 on right
8 . 55 on right
9 . 77 on right
10. 99 on right

That gives the maximum possible difference between adjacent weights, and the maximum practical averaging of coin weights. Personally, I'd really have to work at keeping all the piles separate.

(Otherwise, I'd go for #20 on the grounds of economy of effort)

Regards

Fyz

Firstly I question anyone only allowing the use of the scales only once ....this is pure stupidity if you are trying to find the fakes.. Secondly if you read the question correctly it is not the coins that weigh the difference it is the BOX !!!!

Now for the answer simply whip out a high speed centifuge and drop all coins in the ... the lighter fakes should be closer to the center and the real ones towards the outside then confirm this with the scales...

Yes a silly answer for an even more silly question !!!

somebody needs a hug

The scales would have to go in the centrifuge as well.

Your killing me, Kris.

Then you can go in the centrifuge as well. I like to keep the juicy bits , and it's less messy the sieving.

I only pop in and out of this discussion when I can, and have held back a little to some of the comments so far. However I see that there are still a couple of doubters ref my solution to the problem. Namely regarding accuracy.

The photo I submitted of the concept of a radial balance, was balanced first without any loading. Using the central indicator underneath and a fixed and plumb point on a piece of paper. Plus an old T square with a long needle attached to stand in for a vernier height gauge. To gauge the rim constant throughout 360deg's. I then added a 0.1grm piece of Blu-Tack and it went way off balance. I then prepared very constant weight items and tried to shape them into little boxes, and positioned them using a plastic sheet template as a spacer gauge. Working from a center point marked on each 'box'. As you can see in the photo. It took a while with Blu-Tack and a digital balance. I then checked again for rim constancy and lower central indicator movement.

When I was happy that under constant loading all things were equal, I substituted one 'box' for a lighter 'box' of 0.2grm. It went of balance quite noticeable as you can see. It didn't plummet into ground as one 'Guest' said it would. It took about 1.5 minutes to steady with the air con off and the door closed.

You may mock my use of Blu-Tack but it does allow for quite fine additions and subtractions of weight rolled into little balls until corect. Then it can be moulded to shape without it changing mass. Actual size control is difficult but the mass is very easy to maintain. All materials used can be found in most offices. So try it.

When built to more exacting standards and a incorporating a few more design changes, and contained within it's own see through box this instrument is a very precise and accurate scale. The radial rings simply drawn around the fixed plumb point can also be graduated to very fine tolerances of deflection.

Whilst I'm inclined to agree that it is no substitute for the ease of use of a modern digital scale, for everyday use. My simple design does in my opinion cover the questions requirement of 1 weighing and identifying the lightest box.

KennyT

Nice one KennyT , but I bet somebody will argue .

argue argue

Hello KennyT

I'm the 'guest' who questioned the aspect of what would stop the imbalance in weight from travelling as far as the mechanism would allow. It was a genuine question.

I can see from your photo that what you've done is create a pivot point which is above your balancing plane [ie top of the central blu tak]. What this does is create a system whereby the effective lever lengths are varied depending on the angle of the plane.

I can now acknowledge that yes a small imbalance will result in a small movement, a large imbalance will result in a large one.

As I say, it was a genuine question, I'm glad that there was a genuine answer.

Hi Guest

I appreciate your words. Thank you.

KennyT

If x= the weight of 100 good coins then Y = x-5 grams. If 10 coins were taken from box 1 and 20 from box 2 and 30 from box 3 and so on... than the weight(of good coins) would equal 5.5(x). However one box is counterfeit and hence the weight would equal .5 grams times the #of the box the coins were taken from(n) + the weight(w) or 5.5x =.5n + w

In my last post, I said I would wait on the official answer, but now I don't care about it as much as I did then. I am still curious as to what it is, but not enough to use Kris' spoiler. I do recommend Kris' advice to stop and enjoy a laugh.

Last night, I worked out a solution that does not require knowing the weight of a good coin, nor the use of a spreadsheet. It uses only the information given, two simple equations, and a technique to achieve only one weighing. However, I have not confirmed the validity of my math or my method, and probably will not be able to do so until Monday.

Now, in keeping with my own practice, I insert my spoon into the pot for one more stir by making the following comment: Just because the problem as given says we are restricted to one weighing, it does not mean we have to weigh anything. If using a balancing device, we can isolate the box of counterfeit coins without weighing anything, just look for the imbalance. The problem does say to use the scale, so the chemical analysis metioned in my first post is out of the question.

To remind everyone, I said in my last post that we can say that one weighing means placing an object, or objects, on the scale and taking one reading. I'll have more in my next post.

Weekend Thinking! Use the old method of fluid displacement to establish the lightest SG coins. I don't think this is weighing as such, in the true sense of "Weighing" but it will ID the lightest coins. The weighing statement may be the red herring. Last minute argument needed here I think.

KennyT

Non "Weekend Thinking" Ignore the above post.

KennyT

Not necessarily.

If you're thinking of dropping the coins into a graduated cylinder with water in it, and checking the displacement, that won't work, because all coins will displace the same volume.

However, if you place a bouyant body into the cylinder (such as a toy boat or a piece of cork), and add the coins one at a time to the body, then the coins will cause a change in displacement based on the change in weight. The counterfeit coin will cause the smallest change.

You may have to weigh the sample coins altogether and do some math to determine if the bouyant body will support the coins.

"The counterfeit coin will cause the smallest change."

Isn't this just the same as doing several weighings and taking the lightest one?

Actually, you don't have to use the scale at all. As the coins (one from each box) add weight to the bouyant body, it displaces more water and causes the water level to rise. All you have to do then is check the change in water level against the markings (graduations) on the cylinder.

Or did you think that a graduated cylinder had a diploma? (Just kidding!)

Sorry if I didn't make that clear.

To my mind, a graduated bath is an Archimedian weighing scale with a floating scale-pan.

Cynic

Hi 3Doug

I had a bouyant round tray in mind. The fluid contained in a slightly larger glass tube filled with water, with graduations on the side. But felt that I was arguing against my point of one weighing. Still it's worth the debate I think.

KennyT

We put all the boxes on the scales, then calculate - [total weight + 5]/10 = what the box should weigh. We then remove the boxes one at a time until we see the scale show we have removed the box 5g less than the established weight.

Got the answer after lot of brain scratching. Water has gone bit down deeper! Here is the answer.

First take out the coins in each box to arrange the boxes in following fashion.

Box 1 shall have only 10 coins,

Box 2 shall have only 20 coins,

Box 3 shall have only 30 coins,

.

.

.

Box 9 shall have only 90 coins,

Box 10 shall have all 100 coins.

Considering all coins are genuine, and weight of the box with 100 coins is X, the weight of above boxes will be as follows:

Box 1 - 0.1X; Box 2 - 0.2X: Box 3 - 0.3X;......... Box 9 - 0.9X; Box 10 - X

Hence total weight of all these boxes (if all coins were genuine) would be 5.5 X

If coins in Box 1 are counterfeit, then weight if that box will be less by 0.5 grams, Similarly if it is for Box 2 it is less by 1 grams, Box 3 - 1.5 grams ....... so on.

Now measure weight of all boxes together, the weight and result table is depicted below.

Weight Counterfeit Box

5.5X - 0.5 Box 1

5.5X - 1 Box 2

5.5X - 1.5 Box 3

.

.

.

5.5X - 4.5 Box 9

5.5X - 5 Box 10

Thus we can identify the box with bad coins in only one weigh.

Cheers!

Surely that only applies if you already know what X is and the boxes are weightless.

Cynic

I spoke to the original designer of the South African Kruger Rand and he gave me the following info.

The ounce of fine gold = 31.103 grams

The coin is made from 22 carat gold and the weight of the coin is between 33.93 and 34 grams. the 1/2 = 16.965 - ? , 1/4 = 8.482 - 8.502, 1/10 = 3.3?? - 3.4??

my pen went dry and I cannot really read the dents.

The weight of the coins are described as very accurate.

He has an official gold scale. At least capable of 3 decimal places.

Counterfeits are made of gold plated brass, but the weights deviate more than 5 g/100. The fake coins are therefore made of different grades of gold.

The question limits one to one weighing of the coins that will itself not exclude any other method or procedure.

Method

weigh 5 coins from box 1, 10 from 2, 15 from 3, . . . . 50 from box 10,

Now place the same coins in water and measure the combined volume. (weigh it if you want)

Calculate the specific weight and compare against the standard SW of 22 carat gold.

Eureka - the rest is easy. I did not even had to run naked in the street.

Look at the comment by Mark 3rd down on this. I think the Rand was an inspired choice of coinage (for me anyway).

One unjustifiable assumption* = N wrong answers, where N is indeterminate, but probably greater than one.

Cynic

*Who told you it was a Krugerrand?

It must be. It is the best.

Let me offer some options.

The answer is not exactly right. The principle is correct, however the weights given are not. The differences would be 0.05; 0.10 grams etc. the total weight difference of 100 coins is 5 grams not the individual coins.

Could you explain a little more.....in my opinion not only the numbers are incorrect......How would I calculate which is the difference (0.05, 0.10,etc) if I don´t know the total weight ( the 1000 coins weight) ????? For calculating a diff I need 2 values..... the only weight I have is for 945 coins all togethers......

I understood that the only data we had was the difference of 5 grams not the real weights of the fakes/not fake 1000 coins......

Am I in an error?????

Federico

You also know the weight of 1000 genuine coins is the weight of all 10 boxes together + 5grams

Yes that sentence is true but the weight of the genuine coins can not be derived starting with the weight of the 945 coins that are used to have a weight.......You only know that this group of 945 coins contain from 90 to 99 fake coins.....then is not possible to calculate the value of the 1000 genuine coins ....I accept that you can calculate 10 possible values but as the solution says that all the coin are weight together I don't know how to determine which of the 10 values is the correct.

Federico

I agree with you that the posted solution requires information you don't have - but not just a single piece of information either - in order to calculate an expected weight you need something equivalent to the weight of the individual coins - and the total of coins plus boxes. This also requires a ridiculously high precision for knowing the total weight in the first place - even if you have the second piece of information it would be simpler to weigh the coins you have removed rather than the boxes.

It's really very disappointing.

Fyz

You are right we would still need to know the weight of a good coin.

There is a problem with the posted "answer". The question DOES NOT STATE that the coins weigh different. It is not clear from the question whether the difference in weight is due to the boxes or due to the coins. The posted answer should state the assumption, or, the question should state that all boxes have equal mass.

Harry N.

If the weight of one empty box was different than another empty box then there would be more than one measurable difference as it is stated that the one box of coins are counterfeit and 5g/100=the weight of a counterfeit coin.

Harry: If the fake was 1 of the boxes , not the coins, the solution would be wrong too because it weights the 10 boxes. The solution varies the quantity of coins not the boxes

Federico

How do we obtain a calculated weight with just the information given?

Troy:

Assuming the numbers expressed in the solution should be hundredths of grams instead of grams, the "calculated weight" should be the corresponding to 945 genuine coins.

The problem is because the way the sample have been selected it contains from 90 to 99 fakes coins and as it has been weighted as a hole it can not be determined how many fakes coins really are. Then, at least me; can't determinate the weight of the 945 genuine coins...

Federico

Take one coin from each box, keeping track of which coin is from which box.

Go down the hallway to the vending machines, whichever coin screws up the machine is counterfeit.

You are wrong! "the box of bad coins weighs 5 grams less". Assuming the boxes (empty) all weigh the same, the weight difference of a single coin is 5 grams / 100 coins = .05. Therefore if the difference is .05 grams it is box #1

Less than what?

"Assuming the boxes (empty) all weigh the same," How do you justify this assumption?

Unless I am mis-reading this, your answer should be "If the reading differs from the calculated weight by .05 grams, then box #1 is counterfeit; if the difference is .10 grams, then the second box is counterfeit, etc." because the question states that "The only measurable difference is that the box of bad coins weighs 5 grams less than the others." Since the box is 5 grams less, this would mean each bad coin is .05 grams less than the good coins.

It seems to me that the answer given for this challenge would be true only if EACH counterfeit coin weighed 5 grams less than a legitimate coin. However, the challenge states "the box of bad coins weighs 5 grams less than the others." Therefore, the weight difference of each counterfeit coin would be .05 grams less than an authentic coin. You would need a very accurate scale, indeed, to determine which box held the counterfeits.

Hmmm... It looks like Guest was right in post #1 and the rest of us have spent a week being frustrated by a poorly worded question, again.

I agree with you Davo, and as a result of the poorly worded question I'm awarding first prize to kennyT [even though I did spend the week trying to dismantle his solution]

Well done KennyT, in the absence of an official solution that works, your solution wins because it works !

Thank's 'Guest'

I liken the answer to a statement from a good accountant, "what figure did you have in mind".

Onwards and upwards,

KennyT

The difference being that a good accountant will create the figures in a way that appears reasonable - and be able to justify them (at least sufficiently to stay clear of jail).

I thought the team at Global Spec were supposed to have checked each of these challenges out...

Fyz

Worse than poorly worded - "the answer" requires the total weight (including boxes) to be known to a very high (and unnecessary) degree of precision, and also that we know the nominal and highly repeatable weights of the coins. (And even if you had the data, it would be much more straightforward to weigh the coins you took out than the residue in the boxes)

So I would divide the prize for first presentation of a "good" answer (one that can work without making unreasonable assumptions) equally between Hendrik's #20 (only equal coin weights would have been needed for it to have been the standard interpretation any time before about 1950), and KennyT's. The reason for sharing the award is that KennyT's arrangement requires the scales to be built specially.

Fyz

Solution has holes in it - 1. But if box #1 is bad (and thus the box weighs 5g less) then 1 coin difference will be 50mg NOT 5 grams! 2. Only the difference in weight between the boxes is given, not what a valid box weighs. [RWR]

Dear Sirs, Your answer is OK. But what is "the calculated weight"? You can not calculate it because the weigt of the box is unknown value in accordance with the terms of this task...

Wonderful! The official answer is, in my opinion, absolutely the WORST one given to this challenge. What "calculated weight"?

I'm still looking in the shops for a radial set of scales!

Errrrr.... I only occasionally read the challenge and most of the time, the questions are ambiguous and the answers plain wrong. This weeks was a particularly bad example. Who is this guy/girl and is he an engineer??????