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

Am I daft or do I read your solution wrong, the question seems to indicates that the whole box weighs 5 grams less and not each coin- surely then you put all 10 boxes on the scale and remove them 1 at a time until you get a 5 gram difference in reading

Regards

barry williams - registration in progress

This is a BAD answer to a BADLY posed question! The question said that the BOX of coins weighed 5 grams less, not that the COINS in the box weighed 5 grams less.

Also, what numbers are given that will give you the data to CALCULATE the weight proper weight as you stated in the question.

Whoever submitted this should be barred from participating.

Whoever submitted this should be barred from participating

I'm a wonder'n if the original questions was wrote goodly and the editorial staff "tweeked" it to make it exciting (300+ posts is pretty interesting)......

Even if you had the coin weighs and the total weight, you wouldn't want to weigh nearly everything to find a small difference.

Fyz

I don't think the answer is right. It doesn't say that each coin is 5 grams less, but the entire box. The weight difference per coin would be .05 grams if each coin was identical in weight.

The box weighs 5 grams less not each coin, the difference is .05 grams per coin

I've been waiting a week to see if my solution was right and that's what I get.

Booooooo!!!

That solution almost hurt.

I still think mine is right:

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 into 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

That should work - if you have a weighing balance with two pans rather than a scale. But the least unreliable version (most resistant to variations in coin weights) is to have the largest possible equal spacings in coin numbers so 11 33 55 77 99 on each side.

Fyz

What the heck is the "calculated weight" to which you are comparing the weight? Me thinks this will not work. Response #20 seems the most reasonalbe answer.

This doesn't really effect the method but the counterfeit box weighs 5 grams less which makes each coin weigh 0.05 grams less. Otherwise the box would weigh 0.5 kg less. Thats 0.011025 pounds or 1.1025 pounds less for us stubborn Americans.

I have not read every post, so please forgive me.

Two problems with this answer. First, the stated difference of 5 grams is per box, not coin, so the measured difference would be 0.05 grams per coin. Lastly, you still need to know the weight of one coin, which is not given. This would required a second weighing, voilating the constraints of the original challenge.

You might find posts 20 and 212 interesting.

I did. Thank you. Both give a solution to the challenge that fits the challenge, and both are better than the "answer"

From where do you get the "calculated weight" to which you compare the actual weight of all the boxes (minus the removed coins)?

THE ANSWER IS BOGUS.

QUESTION SAYS: "THE BOX OF BAD COINS WEIGHS 5 GRAMS LESS THAN THE OTHERS", NOT EACH BAD COIN WEIGHS LESS THAN THE OTHER COINS.

SO A WEIGH-IN OF 5 GRAMS DIFFERENCE RESULTING IN BOX 1 BEING BAD, DOESN'T MAKE ANY SENSE.

IF THE RESULT WAS 0.05 GRAMS, THEN I WOULD AGREE.

How can this be so?

Fact 1: you do not know the weight of the coins and therefore do not know the total weight

Fact 2 :The bad box of coins weighs 5 grams less than the others. Does this mean 1 bad box of 100 weighs 5 grams less than 9 boxes of 100 good coins or does it mean 1 bad box of 100 coins weighs less than 1 good box of 100 coins

Fact 3: 1 bad coin weighs 0.05 grams less than 1 good coin

Fact 4: the calculated weight is unknown

Just for laughs, I thought I'd mention that your "fact 2" is actually a question. ;)

Seriously, I think the questioner is "not well balanced"........:)

Also depends on the scale of the matter ...

.......or what scale one uses to determine how well one is balanced ;)....

...and I think this whole matter of the counterfeit coins is becoming centsless...

Maybe this will clarify the question and answer to someone!

Assume 10 stacks of 100 coins sitting on a platform scale with 10 numbers on the platform #1--------#10.

You are told that one of the 10 stacks weighs 5 grams less than any of the other 9 stacks.

Take 55 coins off the platform 1 from stack 1, 2 from stack 2, -------10 from stack 10.

You are left with 945 coins on the scale.

The weight indicated now on the scale will be one of ten possibilities.

For clarity of weight possibilities assume that each good coin weighs 1 gram.

Possibility#1 if #1 Stack is Bad= 944 good + 1Bad = 944.995g

Possibility#2 if #2 Stack is Bad= 943 good + 2Bad = 944.990g

Possibility#3 if #3 Stack is Bad= 942 good + 3Bad = 944.985g ,

4

5

6

7

8

9

Possibility #10 if #10 stack Bad = 935 good + 10 Bad= 994.5g

The Weight of a good coin is not given and is not necessary to find the answer to the question as asked

"Which stack is the light one"

Ok, let's say I have loaded the coins on the scale as you said. Then I take a single weight. It reads 425.237 grams. So, you tell me, which box is counterfeit?.........

All the boxes are real - but not many of the solutions

Fyz

This low weight gives a coin weight of less than .5 grams per 10 coins which is the light coin error weight or

approx .4999-0.995= -.4951g negative weight which is not possible unless they are helium balloons

Ok, if you need heavier coins, the scale reads 2000.732 grams. Now you tell me which box is counterfeit. I be checking back to see your answer (but I won't be holding my breath in the meantime;).

Hi Jessernew

The weight of each good coin is not given and is not necessary to obtain the answer to the Question " Which stack is the Light one?" See #343

Maybe this will clarify the question and the answer to SOMEONE!

Assume 10 stacks of 100 coins each are sitting on a platform scale in indicated positions #1 to #10.

You are told that one stack of coins weighs 5 grams less than each of the other 9 stacks

Take 55 coins off the scale 1 from #1, 2 from #2, etc.

You are left with 945 coins on the scale and the wieght shown on the scale will be one of ten possibilities

Assume for clarity of weights that each good coin weighs 1 gram

Possibility #1 =944 good + 1Bad = 944.995 grams

Possibility #2 =943 good + 2Bad = 944.990 grams

3

4

5

6

7

8

9

Possibility #10 = 935 good + 10Bad = 994.950 grams.

The weight of each good coin is not given and is not necessary to obtain the answer to the Question " Which stack is the Light one?"

Can anyone see a fault with this?

Don

I was trying to edit the "Preview Comment" screen and hit enter twice which sent my comment #320. Not realizing where it went I retyped it and sent it as #323

Using the method you described, you do need the weight of a good or bad coin. How else can you create the points to which you are comparing your measurement to?

The weight of a coin can be X

Possibility#1= 944X + 1X-0.995grams

Possibility#2= 943X + 2X-0.990grams

Possibility#3= 942X + 3X-0.985grams

etc

Possibility#10=935X + 10X-0.950grams

The actual coin weight is irrelevent and is not needed to answer the question "which stack is lighter"

I have chosen a value for X and a box number for the fake coins that gives a measured weight is 1732.643 grammes. You tell me which box I chose. (As I understand it, you have one equation and two variables)

Fyz

See Post#343

Dear Fyz

I do not have one equation and 2 variables. All that is necessary is the location of the light box. This can be found by using 1 good coin weighs 1 gram and post#323

Don

ps: login

Hi Don

I don't think even you believe this system works any more. Otherwise you would tell me the answer to #333, the important part being repeated below (mainly to satisfy your objection to the unsigned original)

I have chosen a value for the coin weight and a box number for the fake coins that gives a measured weight of 1732.643 grammes. You tell me which box I chose.

Fyz

P.S. I know I said the previous one would be the last - I weakened.

Dear Fyz

I cannot tell the Box Number of the fake coins, given only the weight of the measured sample or the measured remaining coins without 2 weighings.

The question does not give this information and does not ask for it.

The scale is used to confirm the position of the fake coin box by comparing the diferences in weight which are introduced by the selection of 1 to 10 fake coins selected in the sample or the 90 to 99 remaining in the box.

The question does not stipulate that a coin weight cannot be chosen so for simplicity of comparing weights I chose 1 good coin weighs 1 gram.

This makes it much easier to see the effects of the error introduction and which of the 10 possible error weights is the one which contains the fake coins and hence the box number is given by the calculation.

Sincerly

Don

Hi Don

I think you are now saying that, in order to solve the problem, you also need to know the mass of a good coin. As we aren't given this, the question was unanswerable in the way proposed in the "official solution". Had you said that and proceeded to state the conditions* under which the solution works, that would have been fine. But you actually state:
"The weight of each good coin ... is not necessary to obtain the answer to the Question 'Which stack is the Light one?' "
I can see no way of reading this that is consistent with the reality, which does not make that particular intervention too helpful.

You now say that the question does not say that the weight of the coin cannot be chosen for simplicity". That too is almost* fine, so long as you state that you will need to know the weight of the cons, and you are giving an example for a coin weight of one gramme.

*Actually, the proposed solution requires that you know more than the 50-mg difference and the weight of a good coin, because the weighing includes the boxes: therefore, if you want to use this method, you would also need to know the total weight of the boxes, which is quite unnecessary if you weigh the coins you have removed instead of the residues.

Regards

Fyz

_________________

Only lawyers should try to defend the indefensible - anon

Hi again

The question askes for a box# and this can be determined by making each box identifiable by a difference in weight for each box.

Taking a different number of coins from each box and keeping track of which box each coin came from will accomplish this.

I assume that all the boxs containing the coins are identical, I hoped to clarify this by placing Stacks of coins on the platform scale eliminating the boxes.

The weight of the coins can remain unknown because it is not asked for, or necessary to obtain the answer. So use X for the good coin weight you do not need to know what X is.

You will be able to identify the # of the fake stack of coins by the weight of the sample taken from it or the weight of the coins remaining on the scale.

There will be 900 coins weighing X grams each and 100 coins weighing X - 0.05 grams on the scale before the samples are taken and 945 coins on the scale weighing a choice of 10 possible weights - #1= 9X + X-.05 grams #2= 8X + 2X- 0.1 grams #3 = 7X + 3X-0.15grams - - - - etc.

The weight of the fake coins will be the choice that indicates the actual measured weight on the scale.

I hope that this will help you to understand how my thoughts are working or not working.

Sincerely

Don

Impossible to tell - as I don't understand what you are trying to say.

Fyz

This answer is incorrect!!!!! Let's see who can figure out why....

Later

The answer is flawed, as well as the way the problem is defined.

What is the calculated weight ? There is no indication as of the weight of a good (or fake) coin.

I hate say it (I don't really) That after 328 posts so far and something plus of 1 weeks discussion. My answer is realistic, and proven to work. I have identified the weight of the lightest box quickly and accurately. I know this will really get up the noses of some of the purist's, but that's life.

KennyT

Hey Kenny , I voted for ya ! Your idea was perfect. They really should add a hand-clap type of smilie.

I like your 'John McEnroe' footnote. Very true.

KennyT

Hi Kenny

The only serious reservations* I recollect about your method were: that (as described) it needed refinement to cope with unbalanced boxes; and that it was not clear whether it fell within the intended terms-of-reference of the questioner. In the light of the answer presented, that latter reservation can be discounted.

On the other hand, I suspect the question to have been borrowed from a rather old source, and somewhat corrupted. If you translate it to a sensible form based on that idea (old fashioned unionised bank that uses a two-pan balance for all weighings, small variation in the weight of good coins etc.), I think you will agree that an alternate answer developed from #20 would be the expected and adequate solution.

Fyz

Take credit where it is due - but share it on the same basis

I have been waiting for the true failing of the design to be identified but it hasn't so far. The downside of the radial scale is it can only identify "1 lightest box" or "1 heaviest box". If there are several light boxes intermingled with the heavy it will balance out and sit at equilibrium. Especialy if there are many boxes. So it is not perfect. However under the auspices of the question we were only looking for one box of coins. The lightest one.

It doesn't need refinement to cope with 9 common weights and one lighter box. As in the question. As the weights are all loaded with the scale at rest. Then it is raised as in the photo and fairly quickly it becomes apparent which is the lightest box. The question clearly did not give enough data as we know now for sure, for a purely mathematical answer to be calculated.

What did come out of the question for me was the good discussion between the same names, when it was still a question. The onslaught after the answer was given from non participants is fairly typical of the spectator world we now live in. I see the same names come up again in the next question with the same get up and go spirit of problem solving and having fun. Smashing.

kennyT

What a memory...

The refinement needed to your initially described arrangement was to allow you to weigh apparently symmetrical boxes where the coins were offset internally. Someone (apols for lack of acknowledgement - too many posts to search) quickly pointed out that suspending the boxes would sort that (hang it all).

Fyz

To find Genuine coin weight weigh all ten stacks of 100 coins add 5 grams then divide by 1000

Fake coin weight = genuine coin weight minus .05 grams

Calculated weight of 944 coins plus one fake in stack one is 944 times genuine coin weight plus .995 grams.

etc etc

You do not need to know the Genuine coin weight which is not asked for and would require two weighings so use a value of 1 gram = 1 genuine coin weight.

Now weigh the 10 stacks of 100 coins after removing the 55 coins and the given answer makes sense if you change the 5 grams to .05 grams and 10 grams to .1 grams etc.

Don

Don

I will definitely read this more carefully once you have demonstrated your system works by answering post #133 correctly.

Fyz

Post 133 states

I think that for the intent of this question as described we must assume that the weight of each genuine coin is exactly identical to all other genuine coins regardless if it is nnn.xxxxxxxxxxx--- grams. The weight of each fake coin is also identical to to all other fake coins which is .05 grams less than a genuine coin.

What is to demonstrate?

Try Post 323 using information in post 337- Good coin weight = 1 gram

If you are saying that your method only works if the mass of the correct coin is 1-gm, I can accept that it works under those conditions and under no other. As you do not know that the coin mass is 1-gm that is not a great deal of use - mine coins certainly did not weigh 1-gm each.

So, if your method works, show me by giving me the answer to my question. Otherwise, I shall simply assume it doesn't.

Fyz

If it works with a weight of 1 gram per good coin what else do you want?

The Question is

"Which is the lightest box" and does not ask how much it weighs.

So the weight of each good coin is not needed and using 1 gram is acceptable and works.

If your coins are more or less than 1Gram each you would have to use another weighing to find out what each of your Good coins weigh but why bother, The weight of a good coin is not needed.

Don

I want it to work with the actual currency I have used. I have no evidence so far that it can.

The sole reason for giving you any time is that you are insistent that your system works without needing to know the weight of a good coin. You are the person asserting that it works, so it is up to you to put your calculations where your mouth is. If you, fully knowing and understanding your system, can't or won't use it to give a specific answer to my question, how do you expect anyone else to go into the details of a method that at first sight looks totally unpromising.

This is my last word on your system unless and until you give me the box number that corresponds to my post number 333.

Regards

Fyz

Hello Extremist,

See post #335.

I am still waiting for you to demonstrate that you can tell me which box is the counterfeit based on the weight I read.

I agree that there are 10 possible readings and that they would differ by multiples of .05 grams. So what? You only get one reading. There is nothing to compare it too. Take the challenge, if your method works, tell us which box is counterfeit.

Again:

The one weighing does give you the # of the box with the light coins without knowing the actual coin weights!!

The one actual weighing will be the one of the ten possible wieghts calculated that is the same as the scale.

Don

In that case, why don't you either answer my question #348, or say how you calculate the ten possible weights when you don't know the weight of a good coin.

Fyz

Dear Fyz

I have a problem with your post# 348 you do not specify whether your chosen weight (1732.643 grams) is 1 coin 1 box of coins or 10 boxes of coins, nor do you say how a chosen box number for the bad coins may help to solve the challenge.

I do not use the weight of a good coin but rather use the differences in weight created by the removal of the 55 coin sample.

The actual wieght of 1 coin does not help to solve it because we are looking for a box location with an identifiable difference in weight.

Use X to indicate the total Measured weight of the 945 coins left on the scale after the 55 sample coins have been removed.

A weight difference between the 10 boxes is created by the removal of 1 to 10 fake coins from each box.

We know that each fake coin removed leaves a Location Marker of -.05 grams

And we know the number of coins removed from each box.

So this allows us to calculate weight of each box by the number of -.05 gram multiples it takes to equal the measured weight.

The weight differences are quite small compared to the total weight of the 945 coins remaining on the scale and would require a scale with .001 gram accuracy but these are available.

I will stay subscribed

Don

It is the final weighing made according to your chosen method.

The reason for trying to see what you make of a specific case followed according to your method is that I can't make heads nor tails of what you write - so I think our best hope of establishing common ground is via an example.

Specifically
1) I don't understand how you establish the difference without having an initial weighing or additional information

2) Even once you have this difference, I don't understand how you can estimate how much of the difference is due to the removal of good coins and how much is due to the removal of bad coins unless.

Fyz

I had enjoyed reading the contributions by all before "the answer" was given. There were some very creative ways to meet the challenge. As I said in post 301 and 311 (as a guest before I registered) I found the solutions in post 20 and 212 most satifying to the question and logical.

When "the answer" was so disappointing, it was almost comical. It was then I decided to register and offer a comical contribution as a way of expressing my dismay at "the answer". (see post 315 thru 319).

It then became apparent that there were still some who did not understand that the answer was faulty, still insisting that they could clarify it for "SOMEONE". After repeated request (see post#'s 321,335,333,348,341,346,342) for one contributor to give demonstrations that his method worked, there remains none.

What began as an entertaining enjoyable exchange has become irritating and time wasting. I must say that this question and "the answer" has generated a lot of traffic. Could it be that the refusal by one contributor to demonstrate that his method works is intended to prolong the discussion for advertizing purposes? I hate to think so. I rather suspect that someone is backed into a corner and is resisting admitting it. In any case, this vain of discussion has become too irritating to continue. So, for this question, its "Good bye".

Hi Jesseernew

Sorry for the irritation but there were so many that wanted proof giving me a coin weight or box weight etc. that I could not individually contact them all.

See post #50

I cannot tell the Box Number of the fake coins, given only the weight of the measured sample or the measured remaining coins without 2 weighings.

The question does not give this information and does not ask for it.

The scale is used to confirm the position of the fake coin box by comparing the diferences in weight which are introduced by the selection of 1 to 10 fake coins selected in the sample or the 90 to 99 remaining in the box.

The question does not stipulate that a coin weight cannot be chosen so for simplicity of comparing weights I chose 1 good coin weighs 1 gram.

This makes it much easier to see the effects of the error introduction and which of the 10 possible error weights is the one which contains the fake coins and hence the box number is given by the calculation.

Sincerly

Don

Ps: Don't Quit

Maybe En who submitted the question could comment on whether the question was tickled up a bit by the owners of the page, and could tell us what his original words were.

if we review sentence no. 2 from the question, it is obvious that the 5 gram difference is referring to the overall weight of the box - not the individual coins.... therefore, with the method you described, the difference would be in multiples of 50 milligrams (ie 5 grams/100 coins). however in this method, you still have to know the exact weight of of the genuine coins to determine the weight of 55 coins.

as a solution, i would suggest taking 10 coins from each box and record the scale reading every time 10 coins are placed... (or is this allowed? maybe not...)

otherwise, we can take 1 coin from box#1, then 2 from box#2 and so on until box no.9. from box #10, take 55 coins so there will be a total of 100 coins. since we have stablished the fact that a bad coin weighs 50 milligrams less than the original coin, telling which box is counterfeit would be peanuts...

It is interesting the more you discuss, the more you go round in circles and loose the original plot.

This idea may help and confirms an earlier posting.

Take a beam balance, that is a balance with 2 pans. When the 2 pans are empty, the needle pointer shows 0 on a graduated scale. That graduated scale is marked to show 0.05 grams divisions.

From box 1, take 1 coin, box 2 2 coins to box 5 5 coins and place them on the left hand side of the balance. From box 6 take 1 coin, box 7 2 coins to box 10 10 coins.

If the coins were all good there are 15 coins on the left and 15 coins on the right so both sides are in balance

If the graduated scale shows 0.05 grams to the left that means 1 coin is lighter on the left hand side. The right hand pan is heavier and the needle moves to the left. If it shows .1 gram, there are 2 coins light to 0.25 grams which is 5 coins light.

If the pointer is to the right of zero the same applies.

You know which coins came from which box and can therefore know which box contains the bad coins.

No formulae, no spreadsheets, a single weighing and obvious if you know the type of balance to use.

This would need the use of an analytical balance which are standard pieces of equipment and still readily available today.

As I mentioned, this was stated quite early

It's a good answer as far as it goes but the question actually states that the box containing the dud coins is five grams lighter. It does not state that all the boxes are identical in mass and the deficiency amounts to a 50mg shortfall on each coin. I agree it's not an unreasonable assumption to make but to be true to the question one could postulate a whole bunch of lead coins in a really light weight box....

In which case the only proposed solution would be KennyT's - assuming that you read that as conforming the the question.

Couldn't agree more Fyz, I've been a fan of the radial scale ever since it was first mentioned. I'm not entirely sure but I think I was the first to suggest dangling the boxes. I'm certainly not going back to confirm!

I stumbled on this problem in a TechSpec newsletter, and managed to view the page (I'm unlikely to 'register' as I have difficulties coping with computers and internet access -- just a humble old-fashioned mechanical engineer). Though I note that 'guest' entries are not welcome in certain quarters (Posts 40, 54, 87, ...) and the topic has been flogged to death and back again, let me add my tuppence worth to the discussion.

Since the problem is titled "Boxes and Coins" I imagined that the boxes would play some role in the problem, but apparently only stacks of coins are essential. My first understanding was that the boxes are 'sealed' or that individual coins cannot be removed at will. However the original problem-setter appears to have been thoroughly discredited by now.!

Despite some bickering on the semantics, posts 20 and 58 provided some elegant solutions. Indeed all that the 'radial' scale requires is for the disc to have ten little holes drilled around the periphery (equal radius, 36 degree spacing) small enough for a string to pass through but hold a knot. Using ten strings (of equal weight) we can suspend the ten boxes, and avoid the arguments regarding centre of gravity or box weight variations. We only need to ensure that adjacent weights don't touch when suspended -- they can even be alternately staggered up and down. A similar hole at the exact disc centre can accommodate a suspension string with the knot below. If the disc (uniform metal sheet) is thin enough to safely hold the weights without buckling or wrinkling, the slight downward deflection around the periphery should lower the cg of the hanging points sufficiently below the centre to provide stable equilibrium, and a readily noticeable tilt.

The 'weighing' is also possible if a uniform straightedge with accurately graduated linear markings is used as a 'scale'. Then if the boxes are suspended using strings at equal (or known) distances, five symmetrically on each side of the exact centre, then the position at which a known weight must be placed to achieve balance will yield the solution. Or else the weight at a known distance from the centre. This may be a more semantically acceptable than a radial scale, avoids multiple weighing, and also works with sealed boxes. I hope the related algebra isn't faulty, but it is similar to post 20. So this is not very original.

I'm not sure I can find this web page again, but I'll try. =TeeSquare= (Madras, India)

Hi TeeSquare ,

I don't think it's the case that 'Guests' aren't welcome. CR4 deliberately allows it (I think in the hope that they may join ). When a lot of people join a topic and are shown as 'Guest' it makes the thing hard to follow (even if they use a name at the end of their post ). There are lots of advantages to joining , so I hope you persue this and do so. Just so you are aware , the title Guru by my name means nothing except I post a lot - the reason being that there is so much good stuff to be found within CR4. Anybody who is interested will be welcomed here , it's not some kind of exclusive site. Hope you manage to navigate your way to joining .

Kris.

ps - have a good explore around the site and you will see why to join.

Hi Guest,

So far as I am concerned, anyone who considers his comments as you have is more than welcome. The advantage (to me) of your providing an identity is that I'm more likely to read what you say.

That is a very nice summary of the main stream. The only omission that I can see is the "improved" arrangement that can cope with greater (but still limited) variation in the weights of each type of coin. Then we would want the differences to be as large a proportion of the total as possible (perhaps 1, 3, 5, 7, 9 coins on each side). Then, if we know that there is a limited number of coins at either end of the range, there could be additional advantage to using the maximum possible number of coins that gives equal differences (11, 33, 55, 77, 99 on each side).

Please continue to post. Actually, however, registering is ridiculously easy (click on register and enter a name and a password).

Regards

Fyz

Your procedure would work, except each counterfit coin is not 5 grams less than each genuine coin. The whole box containing the counterfit coins is 5 grams less than each of the other boxes that contain the genuine coins.

Steve

I must be as thick as 2 short boxes of coins!

The question says the only difference is the box of bad coins weighs 5 grams less than the others. Only 1 weighing is aloud.

This has 2 meanings

a) the bad box + 5 grams = weight of 9 boxes

or

b) the bad box + 5 grams = any other individual box

A box + good coins = unstated weight.

Just using some numbers for clarity, what happens if

box 1 weighs 500 gram and 100 coins weigh 1500 grams =2000 grams (weight of coin =15 grams)

box 2 weighs 1250 grams and 100 coins weigh 750 gram= 2000 grams (weight of coin=7.5 grams)

box 3 weighs1500 grams and 100 coin weighs 500 grams =2000 grams (weight of coin=5 grams)

box 4 weighs 1625 grams and 100 coins weigh 375=2000 grams (weight of coin=3.75 grams)
grams

and so on

Take 1 coin (15 g)from box 1=15 grams

take 2 coins (2 * 7.5) from box 2 =15 gram

take 3 coins (3 * 5) from box 3 =15 gram

take 4 coins (4* 3.75) from box 4=15 grams

and so on with the weight of coin +1 =15 gram

The difference is 15 grams so which box is selected

box 1 has coin at 15 gram

box 4 has 4 coins at 15 gram

We do not have any facts because the first statement is ambiguous; we can make assumptions to arrive at any answer we wish

Dear Tony

Your meaning choice - b) is the one that I assume to be correct by the wording of the question. I also assume that all the BOXES are of equal weight. Also the 900 coins in 9 of the boxes are of equal weight, which leaves 1 box of 100 coins which each weigh .05 grams less than a genuine coin.

The question askes for a box# of the bad coins and this can be determined by making each box identifiable by a difference in weight for each box.

Taking a different number of coins from each box and keeping track of which box each removed sample (1 to 10) came from will accomplish this identification.

Sample #1 = 1 coin came from box 1

Sample #2 = 2 coins came from box 2

etc

Sample #10 = 10 coins came from box 10

I assume that all the BOXES containing the coins are identical, I hoped to clarify this by placing STACKS of coins on the platform scale eliminating the boxes.

The weight of the coins can remain unknown because it is not asked for, or necessary to obtain the answer. So use X for the good coin weight you do not need to know what X is.

You will be able to identify the Box# on the fake stack of coins by the weight of the sample taken from it or the weight of the coins remaining on the scale in the corresponding positions.

There will be 900 coins weighing X grams each,

and 100 coins weighing X - 0.05 grams each

on the scale before the samples are taken

And 945 coins on the scale weighing a choice of 10 possible weights -

#1= 9X + 1X-.05 grams

#2= 8X + 2X- 0.1 grams

#3 = 7X + 3X-0.15grams

#3 = 6X + 4X - etc

Thus each of the ten boxes will contain

10X (-.05grams)

to

10X (-.5 grams).

The weight of the fake coins will be the choice that indicates the ACTUAL scale reading.

I hope that this will help you to understand how my thoughts are working or not working.

Sincerely

Don

Hi Don

Your wording again implies that you don't need to know the weights of the coins - though when I give you an example to solve you state that it can't be done because you don't know the weight of the coins. Please try to make your text convey the reality...

Fyz

As I see it, there were two options for actually solving the question - based on how it was posed:

Design a piece of equipment that will do the job precisely as asked (KennyT's approach) - but this does demand that the specific use of the term "weighing" does not involve knowing a weight; or

Assume that the question is badly worded, but can be solved using equipment that is in some sense standard. In that case we would go for the solution that works for a rewriting of the question that is closest to the question as posed; in this case that uses an old-fashioned chemical balance - Hendrik's #20 presents the most economical version, though there are arrangements that are less sensitive to small variations in coin weights.

Then there are numerous attempts (capped by the "official" answer) that need additional information and much higher precisions of coin weights than either of the above. Personally, I don't regard any of those as "good" answers to the published question.
Then there are methods that are mere delusions...

Fyz

Nicely summarised Fyz. Whilst I like KennyT's solution best , I would have to admit that the question stated 'weigh' and not 'balance' . The statue outside the Old Bailey (OK , Justice and the Central Criminal court) does not appear to have a graduated weighing device , but the scales usually function.

As you say , it was (again) a poorly worded question. I asked STL for clarification on this precise issue re the 'Chocolate' question. The questions are worded with the intent of being presented ambiguously (that is my conclusion - STL gave a perfectly good description to the edit team , and look what emerged). I suppose it is preferable (to have plenty of room to debate ) than an answer that can be goggled like the current question.

How about that, another wrong answer.

Not only does it make a bad assumption, it also contains a simple error of arithmetic.

Bad assumption: that there is a "calculated weight" that is known.

Arithmetic error: since the box of 100 counterfeit coins is 5 grams lighter than the other boxes, each counterfeit coin (assuming uniform coins within a box) would be 0.05 grams lighter than an authentic coin, not 5 grams lighter.

Sheesh, whoever submits these questions really needs to go find the old puzzles book that they half-remembered them from, and quote the questions accurately. Part of the appeal of this kind of question is checking the assumptions, making sure they are clearly stated, and finding an elegant solution that meets them.

It would not be such a bad idea if Global Spec also checked the quality of the official solution before posting. A subsequent check against comments in the thread followed by clarification of the question and (where necessary) correction of the answer would also be appreciated...

Fyz (pronounced as in "bubbles", please)

That wouldn't hurt.

As it stands, it's kind of like doing a crossword puzzle only to find that you have to misspell 3 words to finish it.

In the end, I suppose it's up to each participant to decide whether the entertainment value is worth the frustration. Usually I find that it's not.

Fitz (pronounced as in "interference", thank you)

I thought as much. Sounds a bit tight - you've not been at the single malt?

Fyz

I like your idea of the publisher acknowledging what occurs in a thread. Blindly posting the 'answer' is a touch arrogant.

btw. Are you a Chimp ? Recent playful behaviour is noted.

Annoying (pronounced 'Kris')

In view of the quality of some of the answers, more than a touch, I fear.

BTW, this is supposedly a technical site. Cockney abbreviated references to possible sexual orientation are not always appreciated. (=Chim******)

Bubbles = long disappeared Cimpanzee. What confused part of london do you hail from ? Has Bow Bell damaged your hearing Fyz ?

My earing is fine, thanks. Born in Bow, but dragged up in Middlesex, which may account for your confusion - and my querying your use of playground euphemisms. Is Bowbell your transliteration of cheese? Otherwise, there are sevral as can be eard from 'ighgit 'ill, wear Dicky Wittington sat ishelf darn.

I was born in Lambeth Hospital behind the bedlam. First 10 years brought up in Kennington.

Using the assumption that the good coins in all boxes weigh the same see post 357

Almost the furthest arc of the county, then. My birthplace hasn't existed for nearly fifty years...

Should #357 read box ten 5 coins? Same as post #20?

Fyz

I just know you went to a poshe skule and now speak mockney. Same thing happened to Nigel Kennedy on fine day.'Kris' is a cunningly disguised form of 'Kes'.

I prefer to speak as is natural to me, and not adopt a false accent (perhaps this is pure hypocrisy, as I probably couldn't carry it off in any case). Consequently, like many British born people whose first language was not of these isles, the closest description of my mode of speech is "received pronunciation". I suspect that this could be apparent from my usual writing style. And I believe my grammar as normally spoke could even be acceptable to Lynne Truss. BTW, unlike 'arrow, hEton and such like, no-one went to my skool because it or they were posh. In any case, my parents could never have paid those sorts of fees... (No, not deprived - I'm not planning to start the old Pete-n-Dud routine).

Fyz (not a cunningly disguised form of anything)

___________________________

A knave for a Kestrel?

Relax there Fyz ,your English is impeccable . I try to speak here pretty much as I do in person , but as you will surely know things don't translate in writing. The Pete 'n' Dud routine has been worked several times on CR4 and so lacks appeal , though I had not foreseen it's possibility here. My posts get littered with obscure connotations in an attempt to counter the predominantly American cultural ones. Being a Brit you will understand also that people talk a different language just a few miles apart. I don't pursue all American cultural references , especially not cheesier ones. You're always welcome to post or message asking for clarification if something offends. It's about 2 am * here now , so you see I shall indeed hear the Milkman.

Kris (disguised by language limitation)

I don't do 9-5 , or anything else conventional really.

Relax ysen - I hadn't taken offence - it's usually blatantly obvious when I do.

P.S. Milkman?? Ours vanished some years ago when all the locals decided they couldn't put up with sour milk any longer (no need to start excavations, we just stopped using that dairy when the old milkman retired).

That sounds like Yorksher. Having lived (a long time ago) in Rutland I acquired a mix of accents as a kid. When I first read Maggie T had used 'frit' it sent all sorts of confused thoughts rushing round my head. It was amazing the way her language and manner of speech changed. Hearing the accents used by C4 teen-show presenters makes me squirm in a similar way. I like the news coverage , but have to run for it afterwards in case I hear the street-cred horrors.

I don't use the milkman as my movements are not predictable enough (that's a bad choice of phrase if ever I heard one ). It sounds like it would be much 'greener' to use glass containers delivered to the doorstep , but I wonder.So many factors in that equation. I'm not even sure that they perform a social function these days ,and Postman Pat is more like the version in Viz now. Gawd , nostalgia and rose-tinted spec are depressing. No matter though , Spring Heeled Jack is equipped to escape. He is 'Penny-Dreadfully' still on topic though (even if he may be un-balanced).

Hi Kris and Fyz,

Thanks for your responses (361, 362) and encouragement. As you can see I have followed your advice and registered with CR4. I fear though that it may prove to be an addictive activity, so I should rely on my slow/unreliable dial-up internet to clamp a brake on any undue exuberance.

Fyz's remarks regarding the "improved" arrangement apparently pertain to post 20, rather than my proposed solution in post 360, and there was already a suggestion in post 100 for using 20/40/60/80/100 coins on each side so that the 'balancing' weight required is 1 to 5 grams. No doubt my 'guest' post has remained essentially unread so let me repeat the details of my proposal under the 'browser' status, on the off chance that somebody sees it.

I had suggested using a straightedge as a scale. Suspend it from a point just above its centre so that it hangs 'level'. We need ten scale markings SYMMETRICALLY arranged on either side of the centre (not necessarily at uniform intervals). Suspend the ten boxes at these markings using strings of equal weight. Then find the position on the light side of the scale where a 5 gram weight makes the beam level again. It will be over the light box.

I trust that would meet the requirement of 'one weighing', and most of the semantics of the original problem. This really has nothing to do with counterfeit coins or boxes, just ten objects of which just ONE whose weight is known to be LESS by a KNOWN amount is to be detected by a single weighing (placement) using any suitable scale or balance. There is a better known problem of finding the defective coin in a set of twelve in three weighings on a balance, given that it is either lighter OR heavier than the others. In our case we can't deal with the "lighter OR heavier" option without another weighing.

A word about the practicality of my proposal may be in order. Since a typical coin weight would be in the range of 2 to 5 grams, the total weight of the boxes or objects would be say 2 to 5 kg in the original problem. We are trying to detect a difference of 0.1%, so the set-up should be sensitive and error-free to level of about 0.01% of 5kg if we take the unfavourable case. A somewhat high precision requirement!

We may just manage to do with using an ordinary wooden metre scale as the balance. It can be suspended using a small hook exactly above the 50 cm mark, and levelled by adding a pin or two at the light end. The weights can then be hung at multiples of 10 cm on either side since the graduations (hopefully accurate) are already provided. If the scale bends too much the spacing can be reduced. A little numerical work will show that a 10 mm height of pivot above the line of suspension points (assuming no bending) will cause the scale to tilt about half a degree for a 5 gram unbalance at 10 cm. Bending will make things worse. So we should lower the pivot height to just above the scale top, for the tilt to be noticeable. Maybe just loop the suspension string under the scale, and depend on bending deflection to provide 'stability'. It can also be shown that the 500 gram weights at the extreme ends should be equidistant from the pivot within +/- 0.25 mm or so for the tilt caused by a 5 gram unbalance to be distinguishable from inherent errors in the system. It's not very difficult to make appropriate assumptions and do the calculations, but it will take time, and the above values are rough estimates only (arithmetic not checked). It would be great if someone who can try it out in practice will provide some feedback.

The merit of KennyT's solution in post 58 is that it will detect any 'noticeable' weight deficit, not just 5 grams, but I think the load positions with respect to the pivot will need to be VERY accurate. However the spirit of the original problem seems to suggest that Hendrik's solution in post 20 is the most suitable one, despite all the semantic confusions.

I'm totally new to this kind of discussion, but it has been an interesting one.

Regards, =TeeSquare=

Welcome back TeeSquare ! I have to confess I was only peeking back here for a second , the 'answer' having been given. I decided quite sometime back that I liked KennyT's solution best. Like a lot of questions , this has got people going largely because of how the question is phrased. Personally I think that is deliberate , in order to give room for varied angles of discussion. I don't know if you've seen , but last weeks question on 'Who wants Chocolate' was very good. It had loads of room for interesting stuff to be thrown into the pool (both technical and cultural).This weeks question on ice seems to have a very apparent answer , but again leaves room for lots of interesting input.

I am probably the worst example you will find for going off-topic , but many people (such as Fyz) give great analysis of questions.You're right about CR4 being addictive - a lot of people must have partners who are tearing their hair out !

Welcome

I missed 20 40 60... because I was looking for something specific. 20 40... has a minimum 'gap' of 20 and a total number of 300 coins per side. You can find a ratio that is about 18% better than that.

Fyz

P.S. Might I recommend the thread "Acclaimed hardest logic puzzle". Although, as posed, isn't the most difficult I've seen; I've suggested a constraint that makes it more interesting, but it still only addresses a limited area of logic issues.
I'm considering whether it is worth further adjustment it to make it a more complete puzzle...

Your solution seems to be based on each coin being 5 grams. It was stated there are 100 coins in each box and there is an entire box of bad coins and the weight difference is a total of 5 grams. It seems your answer is off by a factor of 100.

Must admit this Coins and Poxes business is dead -- expecting anyone to actually read through the 360th post (let alone the 386th or this one) is really asking too much, even if it has new content, considering that the official answer has long since been unanimously discredited, and at least two of the members' solutions have received widespread acclaim. Obviously I got into this game far too late in the day!

By way of flogging this dead horse a bit, since Fyz continues (even in post 389!) to promote the idea that the suggested 20/40/... coin arrangement (as modification to solution No. 20) can be improved upon by his 11/33/... suggestion to get larger intervals, may I point out that if there are 11 defective coins on one side, the weight 'interval' is only 0.55 grams (from zero), though the subsequent ones will be 1.1 grams, whereas the 20/40/... has all 1 gram steps. But it's all irrelevant now.

Following the suggestion by Kris guru, I managed to briefly look into a few of the previous CR4 challenge problems. One which interested me particularly was on bouncing balls -- and it does not appear to have been resolved convincingly at all. No doubt fatigue sets in by 200 posts or so and the posters lose their concentration and interest, but I want to work that one out for my own satisfaction in due course, possibly using plain Newtonian mechanics without invoking sound waves etc. However I'm not likely to post frequently, as I have a general phobia about computers/internet and go 'on-line' only briefly, then read some 'saved' pages later on.

The discussions are admittedly interesting and enlightening, despite the load of repetition and irrelevant verbiage. It's nice to see that are still people willing to think about some of the fundamentals of science and engineering even if some blunders crop up, as I find the modern breed of 'engineers' increasingly reluctant to tackle a problem involving some technical challenge for which a 'book' solution or computer programme is not available. We should perhaps be cautious about making harsh criticisms of 'different' ideas as I think we tend to learn more from our mistakes or at least the lessons sink in better. And it's also great to learn from others' mistakes.

=TeeSquare=

One of the things I like about the Challenge Questions is all the incidental stuff. Sometimes the 'answer' is given by somebody very quickly , but lot's of interesting stuff gets posted (usually related , but often completely unrelated. All interesting). There are a few die-hards who carry on for weeks , long after the majority have left . Lots of the questions have no definite answer. It's almost impossible to make sense of it all unless you're in on the act from the start (probably one reason why many people post stuff that has already been posted). There was a very good continued discussion (I think 'Blowing in the wind' or something like that , about a boat ) recently with some excellent analysis after most people had lost interest. To this day , I still hold that the answer to the bouncing balls question is , well , balls. (speaking of which , my 'Guru' part is balls as well - simply denotes number of posts , which is almost embarrassing in itself).

I'd be surprised if somebody didn't return to discuss further. ( I've been seduced elsewhere , and had simply left my topic subscription alert on)

As long as your computers security software is reasonable , there's no need for too much concern about being on-line at CR4. Assuming you know the fundamentals of protecting personal information etc you should be OK. CR4 has lots of pages devoted to the topic of security software. Putting a few keywords into the 'Search..General' box on the right of screen may deliver useful threads to read. Browsing the section on 'Software and Programming' may also be of interest (it's not all highly technical , and most people will happily explain stuff)

A bit more irrelevant verbiage and this thread is going to crack the 400 hundred mark. Does anyone know what the record is. This one must surely be well up there.

The only irrelevant thing I can see is your post. I can't be &*&** to answer your question.

You're totally right Kris; completely irrelevant, but look back through the preceeding 400+ posts and tell me with a straight face that they are all more relevant. In your &*&* way, you too have contributed to the irrelevance!

Hi TeeSquare

Clearly, even though the question was poorly presented and the official answer rotten, there is still some nutrition in the horse-meat.

The point is that, as we know there is definitely one box of fakes, there can't be a nominally zero weighing. The undecidable point is always half-way between measurements - in the case that worries you, that is exact balance or 0-gms, which is still 0.55-gm away from either "expected" measurement. Another way of looking at this is that the 1.1 gm difference between +0.55 and -0.55 grammes is just as usable as the 1.1gm difference between (say) 0.55 gm and 1.65 gm.

If we don't already know for certain that there is a box of fakes, then the most robust solution using standard equipment would be +/- 20, 40, 60, 80 and 100 coins as you suggest. (Of course, that situation might also defeat KennyT's method if there was variation in the weights of the boxes of good coins.)

My proposed method and differentials are certainly correct for the problem as posed - I hope they are now clear as well.

Regards

Fyz

Hi Fyz -- Some more quibbling (and inching towards the tercentenary)

Your statement "... My proposed method and differentials are certainly correct for the problem as posed - I hope they are now clear as well." still doesn't clinch the issue for me.

I won't be able to find the difference between the +0.55 and the -0.55 condition unless I actually use a 0.55 gm weight to achieve 'balance'. Suppose I have only five 1.1 gram weights and place one on the light side. If it becomes heavier, I must move it to the other side and then if that becomes heavier, I still don't know whether it was box1 or box6 (11 coins from each) which was the culprit.

With this method (post 20), unless 'balance' is achieved we don't get the answer (unlike KennyT's solution where the magnitude of underweight is immaterial). The 1.1 gram difference between the two unbalanced conditions has more mathematical than practical significance. It's not 'usable' as you stated.

If the coin distribution is 20/40/... we need only five 1 gram weights to find the fakes (with all the proper assumptions of course). I like to believe that a 1 gram weight is more 'robust' (and easier to come by) than a 0.55 gram weight.

I suspect the last word on this has not yet been uttered.

Regards =TeeSquare=