Counterfeit Coins: Newsletter Challenge (01/23/07)

Split the 12 coins into 3 groups of 4. Weigh any 2 groups against each other, test 1.

If they balance, the counterfeit is in the remaining group and all coins already weighed are genuine, case a. Weigh any 2 from the suspect group against 2 from the genuine group, test 2a. If the suspect group goes up you know the counterfeit is lighter than the rest. Take either of the 2 remaining suspect coins and weigh it against a known genuine coin, test 3a. The fake is the lighter of the 2.

If test 1 results in an imbalance you can now label each coin in the 3 groups either (G)enuine, (L)ight, or (H)eavy. Take 1 light and 1 heavy coin out of their groups and set them aside. Now put 1 of the light coins with the heavy and 1 heavy with the light and weigh them. Like this LLH -- HHL, test 2b.

If they balance, you know the fake was either the light or heavy that you removed from the group. Weigh the light coin against one of the known genuine coins, test 3b.

If they balance, you know the fake is the heavy coin you set aside. If the light coin is still light, it is the fake.

If test 2b was an imbalance; the 2 coins you set aside are genuine.

case 2bb LLH / HHL the H coin on the left is heavy or L coin on the right is light. Goto test 3ba

case 2ba LLH \ HHL one of the LL coins on the left is light or one of the HH coins on the right is heavy. Goto test 3bb

Test 3ba Weigh the suspect H coin against a G coin. If they balance, the L coin is a light fake, Otherwise the H coin is a heavy fake.

Test 3bb, we are left with L1, L2, H1, and H2

Weigh L1 and H1 against L2 and G

If L1,H1 = L2,G H2 is the heavy fake

If L1,H1 \ L2,G L1 is the light fake

If L1,H1 / L2,G L2 is the light fake. OR H1 is the heavy fake. This is the only possibility that requires another weighing. Damn, thought I had it.

I suspect I need to remove 1 coin from test 2b and substitute it with a known genuine coin. To tired to think it out right now.

slo

Sloco -

I think you missed one case here.

If your first test balances, AND the second test balances, then what? You have two untested coins, and you don't know whether they're light or heavy...

Now that you've had a rest, I bet you can fix this. :)

Actually I didn't mention the process if the 2nd test revealed a heavy suspect either. But it would be the same as for a light suspect as shown. You are right, I missed that case. I seem to remember I had it worked out in my head last night and thought it was a trivial matter so I wanted to get the rest written down before it left my head. It seems that while I was typing the idea got tired of waiting and left my head. I guess the crux of my problem (assuming I am somewhere near the right track) is finding the fake in 1 weighing from 4 unknowns. Well I'm off to a dinner party now but rest assured it will bug me all night. I apologize for the terrible notation.

slo

For clarification what counts as a "weighing" on the balance. As yet I cannot solve for all cases with less than 4 weighings.

This assumes I tell someone what coins to put in which pans. They do so all at once out of my sight. I only look at the balance after it has come to rest.

However, if I am allowed to load the scale myself or watch someone load or unload it one coin at a time until reaching the specified configuration, then I can do it. So is that allowed? I had assumed that each time a new coin was added or removed from a pan that constituted a new weighing.

slo

You said: "I tell someone what coins to put in which pans. They do so all at once out of my sight. I only look at the balance after it has come to rest."

That's right! Tricky, isn't it? But there IS a solution - more than one, actually, but they're all pretty similar.

Couldn't you simplify it by balancing 4 each side. Which will tell you if it is among the ones on the scale or the ones left over.

Then if the unbalance is on the scale you would swap any 2 which will determine where the odd one out is and from that you should be able to tell your final 4 and possibly heavy or light.

Then split into 2 of 2 this will tell which side it is on and finally by removing one from each side you will know where the odd coin is.

If the scale balances initially you split the remaining pile then do a single coin swap to make the final determination.

Seems to follow logic either way but just thought I would check with you to see what you thought.

You did a good job condensing the process but you may have oversimplified it to the point where it is unclear exactly what you mean.

"Then if the unbalance is on the scale you would swap any 2 which will determine where the odd one out is and from that you should be able to tell your final 4 and possibly heavy or light."

Swap any 2 with what? Each other, 2 coins from the other pan, or 2 coins from the remaining group? "and possibly heavy or light." How do you know which?

"Then split into 2 of 2 this will tell which side it is on and finally by removing one from each side you will know where the odd coin is."

Huh? If you don't already know for sure if it is light or heavy you wont find out here either. Each time you remove a coin from the pan, if you use the balance result it counts as another weighing.

"If the scale balances initially you split the remaining pile then do a single coin swap to make the final determination."

Again, swap the coin with what? After doing the coin swap you may have only narrowed it down to 2.

Split the 12 coins into 3 groups of 4

Weigh any 2 groups call them group 1 & 2

OC-1) If they balance then all the coins in group 1 & 2 are genuine; the suspect coin is in group three.

OC-2) Take any 3 coins from the third group and weigh them against 3 coins from any of the genuine coins in group 1 and or 2. If they balance then all the 3 coins are genuine the counterfeit is the remaining coin.

OC-3) Weigh the counterfeit against any known genuine coin you will know if it is heavy or light.

OC4) If the weighing from OC2 did not balance (assume the left side was lighter) you will know you are dealing with a lighter coin at this point. Weigh any two coins from this group against each other if they balance the counterfeit is the remaining coin and you already know it is lighter.

OC5) If the weighing from OC4 did not balance the lighter coin is the counterfeit.

OC6) If the weighing from OC1 did not balance you know the third group of coins is genuine. You can now mark each coin in the 3 groups either G genuine L for light or H for heavy. Remove one light and one heavy coin from the scale and place one heavy coin on the light side and one light coin on the heavy side and weigh the results (your groups should be LLH & HHL). If the scale balances then one of the coins you removed is the counterfeit.

OC7&8) Weigh either of the coins from the two coins you removed against a known genuine coin, if it balances then the un-weighed coin is the counterfeit and you know if it is heavy or light because it is marked. If it does not balance then you know it is counterfeit and it will be marked heavy or light.

OC9&10) If the weighing from OC6 did not balance then from the (LLH) group of coins weigh the LL coins against each other. If they balance then the H coin is counterfeit and heavy. If they do not balance then the lighter coin is counterfeit.

OC11&12) If the weighing from OC6 did not balance then from the (HHL) group of coins weigh the HH coins against each other. If they balance then the L coin is counterfeit and light. If they do not balance then the heavy coin is counterfeit.

use a balance beam scale

put six coins on each side of the balance

split the heavy side into two stacks of three

take any two coins of the heavy stack of three

it they balance the third coin will be the one

Imart23, what happens if the forgery is lighter?

Remember you not only need to locate the forgery within three measures you need to ascertain if it is lighter or heavier.

Ah, but what if the bad coin is actually lighter? Then you have wasted a trial weighing the 6 good heavy coins, since they will balance evenly against each other in two stacks of three and you will have to look for the lighter coin now.

In your method, it is then pointless to continue with the heavier 6 coins, since both stacks of three are the same. If you do choose one of the stacks, and pick two coins which will of course balance, the third coin would be a good one also because you made a wrong assumption that the "bad" coin was heavier. Only if you got lucky and the bad coin was actually heavier (50/50 chance) would your method work in three balancings. But that would not account for all conditions as stated in the problem. In fact you only tested for half the possibliities, where the bad coin was heavier.

I am with SLOCO. It takes 4 balancings (I won't call them weighings) to isolate the bad coin and to know whether it is heavier or lighter:

Take 12 coins divided into two stacks of 6 and balance them (first balancing)

Take the heavier stack and split it into two stacks of 3 and balance them (second balancing). IF they balance evenly they are all good coins and the bad coin is lighter and in the other six. Split the other six into two stacks of 3 coins and balance them (third balancing). The lighter stack will have the bad coin. Balance any two coins from the lighter stack (fourth balancing). If they balance evenly, the third coin is the bad, light one. If they do not, the lighter coin is the bad one.

Going back to after the second balancing, IF they do not balance evenly, THEN you will have proved the bad coin is heavier and you can proceed to balance two coins of the heavier stack of 3 and isolate the heavy bad coin in the third balancing.

Of course, you could have chosen the other weigh, or way, and guessed that the bad coin was lighter and choose to balance the 6 lighter coins in two stacks of three (second balancing), then choose the lighter stack of three to isolate the bad lighter coin in the same way by choosing any two coins, the third being the bad one if they balance evenly (third balancing), or the lighter of the two will be bad.

So what's it gonna be punk? Do you feel LUCKY?

If there is a way to do it in 3 that does not violate the terms of the question or how we understand it, I will be surprised!

I am with you on this. You will know if the coin in question is lighter or heavier on the first weighing of the 2 stacks of six. On your second weighing you will use the stack that is heavier or lighter and split them up into 2 stacks of three. After this chosing the stack that is heavier or lighter you will proceed as stated and you will then know which one is left whether it is lighter or heavier it wouldn't matter. Your method still works as far as I can tell.

NO, this is not the answer. See Guest #20 for the correct answer.

You cannot tell if the bad coin is lighter or heavier on the first weighng simply by knowing which group is heavier or lighter at this point. You may guess correctly, and then this will lead you to the right answer, but if you guess incorrectly, you will not solve it in three weighings. A fourth will be necessary.

The key is to mix up the groups, keeping track of the individual coins and knowing which groups they were in and doing a comparison as shown by the answer in #20.

This was a tough one. Did anyone get it without "cheating", Astronut?

Yeah - a few folks did! Let's give it another day, then I'd like to comment on some of the solutions.

yes, that is the correct and simplw answer!

Nope. It doesn't work if the bad coin is light.

MechTech -

Your OC9&10 starts with the same premise as OC11&12. Please give us an explanation of when you would use one rather than the other.

Thanks!

I see in my haste I skipped a step.

You must weigh either group LLH or HHL against 3 genuine coins to determine whether the counterfeit is light or heavy assume you chose LLH if it was lighter than the three genuine coins then weigh the 2 LL against each other to find the lighter counterfeit. If LLH balanced then weigh the 2 HH against each other to find the heavy counterfeit.

You may be closer than you think. There is no need to weigh them against genuine coins. The reason they are labeled to begin with is because you know from the first weighing that if the counterfeit is among them, its weight will be as labeled.

slo

I see my skipped step puts me up to 4 turns at the scale, back to the drawing board.

"OC6).... If the scale balances then one of the coins you removed is the counterfeit."

This does not make sense. If you switched coins, one from each group, and there was a difference to begin with (H or L), then the possibilities are there will be no change (genuine coins were exchanged) or a genuine and counterfeit exchange places, then the scale should tip exactly opposite since you are only shifting the weight difference (+ or -) from one side to the other.

I think you used 4 weighings in at least one of your other sequences as well.

The reason for the exchange is to eliminate the greatest number of possibilities with a single test. Whether a coin is labeled H or L does not make it light, heavy, or genuine. It just precludes the possibility of being the opposite. The H coins would more accurately be labeled "not light", and the L coins labeled "not heavy". We know there is 1 and ONLY 1 counterfeit coin. As you said "a genuine and counterfeit exchange places, then the scale should tip exactly opposite". We are counting on that. If both coins were genuine, no real change, thus no imbalance would occur.

slo

Remember that there were 4 coins in each group of L and H coins one each was removed and set aside, one H placed with 2 L and one L placed with 2 H.

I my original post I don't believe I have more than 3 weighing (OC1-OC2-OC3, OC1-OC2-OC4, OC1-OC2-OC5, OC1-OC6-OC7&8, OC1-OC6-OC9&10, OC1-OC6-OC11&12)

I see in my haste I skipped a step.

You must weigh either group LLH or HHL against 3 genuine coins to determine whether the counterfeit is light or heavy assume you chose LLH if it was lighter than the three genuine coins then weigh the 2 LL against each other to find the lighter counterfeit. If LLH balanced then weigh the 2 HH against each other to find the heavy counterfeit.

Unfortunately my skipped step puts me up to 4 turns at the scale, back to the drawing board.

Don

You do not need a separate weighing against 3 genuine coins because if the fake is in 1 group, the other pan ALREADY contains 3 genuine coins.

slo

Ok divide 12 coins into 4 coin groups.1st weighting

Best result 8 good coins 4 unknown (easy)

minumin result 4 good, 4 light, 4 heavy coins

2nd weighting 2 heavy and 1 light vs 2 heavy and 1 light

result 1:if equal one of the light coins not weighted is counterfeit (easy)

result 2: one of the2 heavy coin from the heavy side is bad or the light coin from the light side is bad.

3rd weighting the 2 heavy coins.

result 1 if equal light coin not weighted is bad

result 2 if unequal heavier of the two coins is bad

Could you give us the details on your first "easy" case - the one where the initial weighing comes out balanced?

The first balance you have 8 good coins 4 unknown. you have 2 weighting left so multiple answers. One weight 2 unknowns against 2 good. If equal weight one of the last unknowns against a good coin. If unequal weight one of the 2 possible bad coins against a good coin in both case if equal the unweighted coin is the bad one or if unbalance the last coin being weighted is bad.

Ah, yes - but if the last try is balanced, and the unweighed coin is bad, you don't know yet whether it is light or heavy! Not quite solved yet...

All the elements of a correct answer have now been given, just not all in 1 post. A.E Newman solved the part I missed when he said "2nd weighting 2 heavy and 1 light vs 2 heavy and 1 light" The important element that differed from mine being to have different total number of light and heavy coins on the scale at 1 time. The easy case that he missed was the same one I missed. But mechtech got that part right in OC-1, OC-2, OC-3, and OC-4. It is possible to get it done in 3 'balances'. That was a good challenge.

slo

Sorry miss the heavy or light portion of the Question.

weigh 3 of the 4 unknowns against 3 good if balance weigh bad coin against good coin to find if heavy or light.

if unbalance weigh 2 of the 3 bad coins against each other if unbalance the coin that matches the 2nd weighing results is the bad coin. if balance the 3 coin is the bad one see result of 2nd weighing for heavy or lightr.

Astronut!,

Please explain if by "balance scale" you mean a pre-weighted scale that allows you to accurately measure weight/mass, example "gram-scale", or do you mean a balance that only allows relative measures, say between groups or between a group and known free weights?

I think there is some confusion on this point as some are thinking you can find the exact weight of a group and others are thinking you can only find if a group or single coin is greater than, lesser than, or equal to (balanced evenly) the other group or coin. (My thought is toward the relative balancing, but please correct me if I am wrong).

Also, it seems that some are saying that one weighing is the same as weighing several different groups each one time. To me that would be three weighings.

Please clarify.

Thanks!

STL:

The puzzle uses a simple balance scale showing only relative measures. With coins on each side, there are three possible outcomes: left side goes down, right side goes down, or they balance. That's all the information you get.

It's a tricky puzzle!

(Does STL stand for St. Louis, MO?)

Yes.

Well, duh - it says "St. Louis, MO" right there under your avatar, doesn't it...

Guess my brain was on break.

Fun city, St. Louis! The Science Center is a blast.

Where in Illinois, AstroNut?

Warrenville, due west of Chicago, just a coin-toss from Fermilab.

See http://home.att.net/~numericana/answer/weighing.htm

Expanding on the question a little, we'll show that 3 weighings are enough to...

• Find an odd marble among 12 and tell if it's heavy or light.
• Find an odd marble among 13.
  If you are given an extra regular marble, you may also...
• Find an odd marble among 13 and tell if it's heavy or light.
• Find an odd marble among 14.

(Please, understand that the extra "reference" marble is in addition to the 13 or 14 "unknown" ones, but its presence makes the problem simpler to solve.) We'll also prove, in each case, that the above is the best that can be achieved.

First, let's deal with 12 marbles, call them ABCDEFGHIJKL:

First weighing: Compare ABCD and EFGH.

• If ABCD=EFGH, you know the special marble is among IJKL. Use your 2nd weighing to compare AI and JK (you know A is ordinary):
• If AI=JK, you know L is the odd marble. You may compare L and A in the 3rd weighing to determine if L is heavy or light.
• If AI and JK are not equal, you know L is ordinary. Use your 3rd weighing to compare J and K.
• If J=K, you know I is the special marble (the result of the second weighing tells you if it's heavy or light).
• Otherwise, the special one is either J or K and you can tell which: It's the heavier one if we had AI<JK on the second weighing, it's the lighter one if we had AI>JK.
• If ABCD is not equal to EFGH, you know IJKL are ordinary (we may use L as a known ordinary marble in the rest of the procedure). Also, you will always know from this first weighing whether the special marble is heavy or light, once you've determined which it is.
  Use the 2nd weighing to compare ABE and CFL :
• If ABE=CFL, you know the special marble is among DGH. You may use the 3rd weighing to compare G and H :
• If G=H, then D is the special marble.
• If G and H are not equal, the special marble is the heavier one if we had ABCD<EFGH, and the lighter one otherwise.
• If ABE and CFL are not equal, the special marble is among ABCEF and we may distinguish four cases :
• ABCD>EFGH and ABE>CFL :
  Either F is light or one of AB is heavy.
  Compare A and B in the 3rd weighing to find out.
• ABCD>EFGH and ABE<CFL :
  Either E is light or C is heavy.
  Compare C and L in the 3rd weighing to find out.
• ABCD<EFGH and ABE<CFL :
  Either F is heavy or one of AB is light.
  Compare A and B in the 3rd weighing to find out.
• ABCD<EFGH and ABE>CFL :
  Either E is heavy or C is light.
  Compare C and L in the 3rd weighing to find out.

I guess the keys to solving this, used in the explanation above, is:

1, always allow for balance left, balance equal, and balance right for each comparison unless comparing two.

2, use at least one known weight in the second balance; a known weight being one defined as genuine from the outcome of the first balance.

By George I think he's got it! (Guest #20 that is).

The next question is did he solve it himself or find the answer on the Internet (that's cheating of course!).

Obviously, giving a URL as a reference, he did not solve it himself.

Where is the fun in that?

I agree STL_Engineer. I has figured that the trick was to take the three possible outcomes of each weighing into account and that you need to start with three groups of 4 coins. I was half say through the second weighing when the Guest in post #20 posted the cheat sheet.

You spoil sport Guest!

For the Trekkers among us, this would be the "Kobayashi Maru" solution...

Thank you Guest for #20 answerer. Engineering is more than just knowing the answerer, it's knowing where to find the answerer that leads to* discovery. *That's when what you are looking for isn't there.

Easy one.

Step 1. Move the scale out of the way (and sell 2 of your "limited 3 Balances" to someone who needs more!!) and take out your pocket knife.

Assuming your are tone deaf:.

Step 2. Balance each coin on the tip of your finger and tap it on the edge with a pocket knife in front of your microphone attached to your digital storage oscilloscope. Record each tone from each coin as you tap it.
Step 3. Take the coin with the only unmatching tone and throw it away as a counterfeit.

If not tone deaf:

Step 2: "Ring" each coin as described above. Toss the bad one out when you hear it.

All 24 outcomes are covered this way. All struck coins "ring". The tone with which they ring is determined by the metal they are made of. The identical coin, being of the same diameter, thickness, picture, color etc. and with only the weight being different MUST be made out of a different alloy. This material will ring at a different pitch as it will have a different harmonic frequency. The counterfeit will stand out immediately when struck.

Old shop keepers could drop a 1964 silver Washington quarter on their counter or a 1965 copper nickel clad quarter on their counter and with their eyes closed and tell you which one was Silver and which was clad. Same picture, same size, different weight and different ring... try it sometime!

Oh.... and if I am naked in a locked room with only a scale, 12 coins, and a bear that comes out to eat me on the fourth weighing... I can use one coin to strike the others. So a scope, mic, AC outlet, or even a pocket knife is not required really.

Step 3. I weigh the counterfeit with one balance test against a known good one and tell you if it is heavier or lighter.

Why waste time balancing 3 times when 1 will do?

David - currently unemployed in the snowy north of michigan

Now that's thinking outside the box - very cool!

(The "bare with bear" scenario is slightly disturbing, but it works.)

Glad you like it. I did take advantage of the fact that an "only" was missing in the problem description.
If it had said "using ONLY a balance scale" it would limit my choices some. I would have had to strike the coin with the scale.... grin! So even if you are deaf as suggested by Sloco, no where is it specified that I can't use a scope and mic to record the tones and compare them by sight. And I must have sight, else how would I read the balance scale, or know the coins were identical? It is hard to write a problem like this and limit it to only one answer....

I did ask sloco how he would produce this counterfeit that differed in weight only and had NO other chemical or physical differences... I thought about it for a while.... and this poor electrical engineer could not think of a way or see how it could be done. But maybe some mechanical or chemical engineer out there can describe a way?

David (up with the bears in the snowy north of Michigan)

Regardless which answer is deemed correct, this is definitely the best. Bravo! It uses the balance as required by the challenge, answers all the questions. The only reason I cannot accept it as a fully complete and accurate answer is because of 1 unfortunate sentence in the challenge. "The only measurable difference is that the bad coin weighs either more or less than the others." It is probably not good for a coin's self esteem to be called 'bad' because he is a little too heavy or too light. But for our purposes the important part is "only measurable difference". As David correctly noted "The identical coin, being of the same diameter, thickness, picture, color etc. and with only the weight being different MUST be made out of a different alloy. This material will ring at a different pitch" however determining the pitch is a measurable difference, thus explicitly denied as a means of determining the fake. So presumably you are "naked in a locked room with only a scale, 12 coins, and a bear that comes out to eat me on the fourth weighing" AND DEAF.

slo

Some days you get the bear, some days the bear gets you!

While I accept that the challenge states "the only measurable difference" is the weight, in my mind, that statement cannot be true. A statement like that is often in the eye of the beholder. How many times has someone said "that cannot be done" only to find out that they were only correct when looking at the problem from one side? Some other engineer solves it by coming at the problem from a different perspective. The statement "the only measurable difference" comes from a person/group/engineering team/manager/etc. and their personal (or group) view of the problem. It is up to us to determine if each statement is correct, and what tools we might bring to "bear" on the problem. (Sorry- just couldn't resist it!!!)

However, if I am wrong... and once it a while it happens... I am will to take suggestions on how one would manufacture a coin where the ONLY difference it has is the weight and ALL other chemical and physical properties were identical. This would include the ring of the coin of course. Note the ring test would cover using ANY different materials and even cover a metal casting of the same material as the genuine coins. We have already stated the size and appearance is the same. Now that seems to me after some thought to be a REAL head scratcher... But I am open to being taught ; )


David (up in the snowy north of michigan)

No fault exists in your logic or conclusions.

So I will manufacture some for the sake of argument. "the only measurable difference" need not refer solely to the physical properties of the coins. It could mean simply that the ability to measure has been withheld from you. This is reasonable given the purpose of this challenge. Precedent has already been set by limiting the tools available to you and the ways you may use them.

On to the science of creating a counterfeit coin with the weight being the ONLY difference. I do not know how unique ring signatures are. Maybe they are like fingerprints or snowflakes. They say no 2 are alike but unless you've seen every example how do you prove it? 11 different examples of genuine coins may generate slightly different rings. Assume the coins in question are ancient. Roman coins maybe. They can vary in press angle, depth and offset resulting in non identical ring signatures while still looking like any other coin from the same mintage, for practical purposes identical. The counterfeit may fall within the range of the genuine coins thus making it indistinguishable. Traditionally the most important quality of precious coins has been their weight and composition. An ounce of pure silver is an ounce of pure silver even if it is struck a bit off center, a bit too deep or shallow. The weight and composition was being certified by putting Ceaser's face on it. A counterfeit may be made from the same metal, only slightly less of it than a genuine coin. A void like an air bubble in a cast coin that happens to be located at the center node might not affect the ring much but still affects the weight. It may also be possible that a counterfeit coin could be made from an alloy that is so close in density and physical properties to the original that it falls within the margin for error of the ring test. 1 counterexample does not make a proof.

That said, you still have my vote.

slo

Points well taken... that is one possibility that I hadn't considered. I had figured out the three sets of four thing...but seemed like if I thought harder, I could do quicker and do it with less ; )

You bring out an idea I hadn't thought of. Suppose the coin is a legitimate coin made in an identical way to the regular coin at the mint, but made out of an extremely precious metal and the only value of the coin was in the metal itself. Therefore the coin that was too heavy or light was "out of tolerance" rather than truly counterfeit.. In this particular case, my test would fail because 10 thousandth of a gram either way on the weight of the coin may be the goal, this coin is 11 thousandths off, and the tone would be indistinguishable. So certainly the balance scale might need to be used on well made, modern replica. Sigh... so there is a case were I might need to use all three weighings. Good thinking!

We can't assume they are ancient though. Ancient coins were nuggets of metal that had ceasars face pounded into them. These can vary widely in weight. It would be darn near impossible to find 12 the same weight. They were often cut in half with an axe or some such similar precision tool that a local peon had to make change. A piece of 8 was popular because it was easy to cut up. They were even trimmed on the edges and passed on. Etc. Etc. Etc. Even early machine age British coins were often underweight. The British finally did something about it around 1760 or so by passing laws to mandate weight fall in a certain range at least and the result was that most underweight coins were shipped to the colonies. Lucky us! Mostly because it then became a problem here, and that resulted in a US mint.

In order for weight to be a valid way to check the coin, we have to be in the late 1700's with machined coins. Even then, they varied somewhat as the early presses did not make exceptionally even sheets of metal. This was big complaint of the early congress who thought that the mint should be abolished for a while since the coins were too crude and expensive to make. By the time we get to the about 1800 the machinery is much better, and the weight of the coins starts to get quite consistent.. But probably still not good enough for a balance scale. A few more years had to pass.... In US coins, the death of the large cent causes the first wide-scale coin collecting by Americans. So around 1857, there is a lot of money to be made by making copies. Cast coins, electrotypes, mint employess stamping out rare dates after hours, the mint itself making coins that were intentionally sold to collectors as "rare" patterns that were never patterns, etc... all sorts of hijinks by mint directors and employees! Cast coins of this era have a seriously different ring...has to do with it being cast vs stamped...... but a properly machined modern replica would be perfectly indistinguishable... therefore the US law requiring they be labeled "copy"

When I used to collect coins, I did enjoy the history.....

David (up in the upper part of the UP of Michigan)

david

how about a coin with an air pocket unseen because it is internal?

1.divide coins into 3 piles of 4 each

2. weigh each pile against each other

one weighing will be balanced, the other two unbalanced.
the bad coin is in the pile common to both unbalanced weighings; it will then be obvious whether it is lighter or heavier than the other coins

Remember that a balance is just that.... it can be ratiometric ;-)

Your method narrows it down to a pile of four coins. You do know whether it is light or heavy - but how do you know which of those four coins is the bad one?

A classic that I did years ago only it was ping-pong balls. And I hope I have descrbed the procedure clearly enough. I had kept the solution in a word document so I have pasted copied the answer so I will not bother changing it to coins.

First step was to number the balls one through twelve.

The first weighing is four against four.

If the suspect ball is in this group, the scale will tilt. Every ball that is on the side that goes up is marked with a minus sign and the ones on the side that goes down you mark with a plus sign.

On the next weighing (#2) you remove two minuses and a plus. Switch two of the balls, a minus with a plus, and put one of the remaining balls that you did not weigh from the first weighing on the minus side. You are now weighing three against three, a neutral, a minus and a plus against a minus and two plusses. If the scale does the same thing, you can eliminate the two balls you switched which will leave you with one minus and two pluses. Weigh (#3) a minus with a plus against two other normal balls. If it goes up, it's the minus; if it goes down, it's the plus. If it balances, it's the plus that you had put aside.

If on the second weighing the balance tilts the other way, then you know its one of the two balls you switched from the first weighing, a minus and plus, that are the suspects. Use your third weighing of these two balls against any two other balls.

If the scale balances on the second round, then you weigh (#3) one of the two minus balls and a plus ball that you had removed after the first weighing against two normal balls. If it goes up, it's the minus ball. If it goes down, it's the plus ball. If it balances, it's the minus that you did not weigh.

If on the first weighing of four against four, the scale balances, your second weighing will be three of the four remaining balls and a normal ball, two against two. Mark the balls accordingly and weigh (#3) a plus and minus against two normal balls. Same as the previous step.

Of course if by the second weighing the scale did not tilt at all, then you know that the remaining ball (the first weighing used eight and the second three) is the one off-weight. Just weigh it,the third and last, against any other ball.

Basically the trick to this problem is in the second weighing where you remove three balls (it doesn't matter whether you remove two minus and a plus or vice versa), switch two and add one of the remaining four that weren't in the first weighing.

This is my all time favorite puzzle that maximizes the most information from the least number of actions. Glad it made to this forum.

A good challenge AstroNut, well done.

There are tow important things that this demonstrates. Firstly it is important to carry all the information that is ascertained be each weighing through to the end. Secondly there are three possible outcomes at each stage and that all three outcomes must be used to locate and determine the relative weight of the counterfeit coin..

It's not unlike the IF statement in Fortran as similar languages that looks like this

IF A-B 100,200,300

Where you have a statement than needs to be calculated and three pointers that govern the next line to execute, dependant on whether the result is negative, zero or positive. This can be a very powerful tool and if use correctly can be much more efficient than the more common If THEN GOTO type statement.

Sloco came pretty close in post #1 but to quote the famous Maxwell Smart, "He missed it by that much".

case 1. divide the coins in 3 groups of 4. Measure any 2 groups against each other.

If the scale balances the counterfeit is in the group not on the scale.

Now measure this group against one of the "genuine" groups.

If the "genuine" group goes up the counterfeit is heavier. If the "genuine" goes down the counterfeit is lighter.

Measure any 2 coins in the "counterfeit" group against each other. If the scale balances the counterfeit is the one not on scale. If the scale does not balance the counterfeit is either the heavier or the lighter depending on the result from the second weighing.

case 2. divide the coins in 3 groups of 4. Measure any 2 groups against each other.

If the scale does not balance measure the group not on the scale in the 1. weighing against either group on the scale in the 1. weighing.

If the scale balances the counterfeit is in the group not on the scale and the counterfeit is heavier or lighter depending on the result from the 1. weighing.

If the scale does not balance the counterfeit is in the group that was on the scale in both measurements and the counterfeit is either heavier or lighter depending on the result from the 1. and 2. weighing (The "counterfeit" group is in the same position on both measurements)

Measure any 2 coins in the "counterfeit" group against each other. If the scale balances the counterfeit is the one not on scale. If the scale does not balance the counterfeit is either the heavier or the lighter depending on the results from the second weighing.

I think this should cover all the possible cases. Any comments?

Forget about my previous post!!

I seem to have forgotten 1 coin in every group. I had 3 groups of 3 coins (not 3 groups of 4 as I said before). Sorry about that. This is tricky!

You weigh 6 coins on each side of the balance scale, this should reduce your choices by half, do the same again with 3 on each side, again reducing your choices to 3. now you are down to 3 coins, place two on the scale, if one is heavier you have found the fraudulent coin, if they are equal you know the bad coin is the one left out.

Don Kerfoot

Eng Tech IV

Manitoba Hydro

So what happens in the 12 out of 24 possible cases where the fraudulent coin is lighter than the others?

I have the answer, but the question is stolen from the this puzzle: ten balls equal diameter etc, but one is heavier or lighter than the others. A pair of balances and three weighings, how do you determine which is lighter or heavier than the other nine?

William.

Split the coins into groups of 6+6 and the lighter or heavier coin is in that respective group. Then separate the group of 6 (heavier or lighter coins) into 3+3 and repeat, then finally 1+1. The heavier or lighter coin will be one of the 1+1 weighings or the one you didn't put on the balance.

Same wrong answer as #6 and #19. See #20 for correct answer.

Interesting question. Key is use the information from the weightings each time.

Start with 4 left and 4 coins right. If equal wt, then has to be one of remaining coins. Weight one of remaining 4 against the other. If balanced, then weight one of the last two not weighted against one of the coins already checked. If balanced, then the last coin not touched is the counterfeit. If unbalanced, then the new coin weighted is the counterfeit.

Harder case is that the 4 coins against 4 coins does not balance. Note which side is heavy and which is light. Assume ABCD as coins on heavy side and DEFG are coins from light side. NOte that we still do not know if coin is light or heavy, but it makes no difference. Take two from the heavy side, say AB, and place on the light side. Leave one coin from the light side on the light side, say E. On the heavy side, leave one coin from the heavy weighting, say C, and add one coin from the light side from the previous weighting, say F. Finally add one good coin from previous weightings (any of the previous 4, call them O, not weighted but must be ok since we only have 1 counterfeit and we know it is in the 8 under study) to the heavy side. Note the balance.

We have AB E balanced against CF O. If the coins balance, then counterfiet is one of the coins not weighted from the first weighting, D G or H. WE also know that the counterfiet if D will be heavy, and if G or H it is light as determined from the first weighting. Therefore in the third weighting, weight G against H. If balanced, D is counterfeit and heavy. If unbalanced, then the light side between G and H is the counterfeit since we know from previous weightings that counterfiet would now have to be light.

If in the second weighting, ABE does not balance against CFO. Then the counterfeit has to be ABEC or F. We once again note the balancing from this weighting. Assume ABE is heavy side and CFO is light. We now know that E and C are ok (changed sides with no change in balance therefore are ok), and that AB or F is counterfeit. We also know that if A or be is counterfiet, the counterfeit needs to be heavy, and if it is F it needs to be light (info from previous weightings). We thus balance A against B. If the same, then F is the counterfiet and light. If unbalanced, then the heavy side of A against B is counterfeit and it is heavy.

One the other hand, if the second weighting of ABE against CFO switches with ABE being light side and CFO becoming the heavy side, then now know that the counterfiet is either E and is light, or C and it is heavy since these are the only two changed to cause the balance to flip sides. We thus weight either E or C against any of the other coins, O (since they are now known to be good). thus weight E against O and balances, then C is counterfeit. If unbalanced, E is the counterfeit.

3 weightings and presto, one counterfeit uncovered, light or heavy. Key is to carry the information from the previous weight as either light or heavy.

Your first case has a problem. If all three weighings are equal, then you know that the one remaining coin is bad - but is it light or heavy? Look at that case again...

You are correct. I was more worried about solving the challenge with 8 coins in two weightings as opposed to 4. solution is similar and here goes.

First weighting has 8c coins balanced, therefore one of four remaining. call them ABCD.

Weight AB against C. Place good coin from previous weighting with C. If balances, the D is culprit. Weight it against a good coin to determine light or heavy. If AB does not balance against CO, note scale. If AB heavy side, and CO light, then A or B is heavy, or C is light. Weight A against B, is balanced, then C is counterfiet and light. If A and B unbalances, then heavy side is counterfeit. Reverse for AB being light side and CO being heavy side. Weight A against B and if balanced, C is counterfiet and heavy. If unbalanced, light side is counterfeit.

Test 1 Split into 2 groups of 6 and weigh each group to determine which group is lighter (contains the fake coin)

Test 2 Take lighter group and split into 2 groups of 3 to determine which group of three contains the fake coin.

Test 3 take any 2 of the 3 coins from the lighter group in test 2 and weigh them, the lighter coin will be indicated by the out of balance scale, if the scale is balanced then the third coin from the group must be the lighter coin thus the fake.

This solves only half of the puzzle. Remember, the fake coin could be heavier!

a). Number the coins 1 through 12.
b). Weigh coins 1,2,3,4 against coins 5,6,7,8. If they balance, weigh coins 9 and 10 against coins 11 and 8 (we know from the first weighing that 8 is a good coin). If they balance, we know coin 12, the only unweighed one is the counterfeit.
c). The third weighing indicates whether it is heavy or light. d). If, however, at the second weighing, coins 11 and 8 are heavier than coins 9 and 10, either 11 is heavy or 9 is light or 10 is light.
e). Weight 9 with 10. If they balance, 11 is heavy. If they don't balance, either 9 or 10 is light. f). Now assume that at first weighing the side with coins 5,6,7,8 is heavier than the side with coins 1,2,3,4. This means that either 1,2,3,4 is light or 5,6,7,8 is heavy.
g). Weigh 1,2, and 5 against 3,6, and 9. If they balance, it means that either 7 or 8 is heavy or 4 is light.
h). By weighing 7 and 8 we obtain the answer, because if they balance, then 4 has to be light. If 7 and 8 do not balance, then the heavier coin is the counterfeit. i). If when we weigh 1,2, and 5 against 3,6 and 9, the right side is heavier, then either 6 is heavy or 1 is light or 2 is light.
j). By weighing 1 against 2 the solution is obtained.k). If however, when we weigh 1,2, and 5 against 3, 6 and 9, the right side is lighter, then either 3 is light or 5 is heavy.
l). By weighing 3 against a good coin the solution is determined.

1. place 6 coins in one dish and 6 in the other.

(the dud is somewhere there, so the scale will be out of balance. Say the left pan is heavier.)

2. Using left & right hands, remove the coins two at a time ,one from the left pan & one from the right, checking to see if the scale balances each time. while it is out of balance you know that all the coins removed are genuine. place them in a pile in fromt of the scales. As soon as you remove two coins ( one from the left & one from the right) and the scales go into balance you know you are holding one genuine coin & the dud but you don't know which is which.

(I'M TREATING THE ABOVE PROCESS AS ONE WEIGHING)

3. clear any remaining coins from the scales and place them in front of the scales with the rest of the genuine coins. Place the coin in your left hand to the left of the scales & coin in your right hand to the right just so as not muddle them up.

4. From the genuine pile take a coin and place it on the right pan & place the left-hand coin on the left pan.

a) If the scales balance they are both genuine and the dud is the coin you place to the right of the scales, furthermore you know that it is lighter than the genuine coins from observing what happened in process 1. above.

b) if it out of balance, the left-hand coin is the dud and it is heavier than the
genuine coins.

"(I'M TREATING THE ABOVE PROCESS AS ONE WEIGHING)"

Incorrect. Every time you change what is being weighed/balanced by removing two coins, that is a separate weighing. Please see how this CAN be done in only 3 weighings (only 3 different pairs of groupings) by the method in #20.

Split the stacks into 4 groups of 3

take group 1 and 2 and weigh them

if one is heavier than the other take group 3 and compare the two.

if they balance you are looking for a light coin in the group you removed.

if they don't balance then you are looking for a heavy coin in the group you didn't remove.

split the known bad group into individual coins

take coins 1 and 2 and weigh them,

if they balance coin 3 is bad if they don't use the results from above to weed out the too heavy or too light coin.

If the first test balances take off group 2 and put on group 1

if group 3 is too heavy or too light you will know what to look for.

split group 3 into individual coins

take coins 1 and 2 and weigh them

if they balance coin 3 is bad if they don't use the results from above to weed out the too heavy or too light coin.

if groups 1 and 3 balance then group 4 is bad, split them up into individual coins

weigh coin 1 and 2, if they balance then coin 3 is bad, if they don't then you will need a 4th measurement. Take those two coins and stick them in a vending machine that you wont have to deal with later and buy a bag of chips. If it refuses to take one of the coins throw it away and hit the coin return button.

HA!! You had me going, there, until you threw in the vending machine...

If you discover how to handle that last case, be sure to "weigh in" again!

No matter how I divide them up or re-divide them up I come up with 4 measurements or 2 coins on at least one case

Make four groups of 3 coins

1. Check 1&2

if bal=true then groups 1&2 ok

if bal=untrue then groups 3&4 ok

2. Check ok group&unresolved group

if bal=true then 3 groups ok

if bal=untrue then other unresolved = ok - check light or heavy

3. Check 2 of 3 unresolved coins

if bal=true then remaining coin is fake

if bal=untrue then offweight coin on scale is fake

You have the same unresolved issue as Batsu: If the first two weighings are equal, you can't discover whether the bad coin is light or heavy.

Reply to AstroNut Re; Re; #63

If check 2 bal=untrue then heavy or light coin must be 1 of 3 coins in that group.

Goto 3.

Don

Yes, if 2 bal=untrue you're OK.

But if 2 bal=true, then you have a problem. You have nine good coins and three coins that have never been weighed. With only one weighing left, how can you determine which coin and whether it's light or heavy?

There is only 1 bad coin

To isolate it

#1 Balance - 3 coins with 3 other coins, if they balance the bad one is not there and they are all good.

If they don't balance the bad one is on the scale and the other 6 which are not yet weighed are good.

#2 Balance 3 known good coins with 3 of the remaining 6 unbalanced coins. if they balance the 3 are also good. If they do not balance note position of offbalance of the 3 coins containing the bad one (over or under weight). also assume that the remaining 3 unbalanced coins are good as the bad one is on the scale.

#3 Balance number 1 and 2 of the 3 coins containing the bad one. If they balance then the 3rd coin is bad. If they don't balance then the bad coin is on the scale in the position of over or underweight indicated in balance #2

Regards

Don

Don -

If the bad coin is in the last set of three, it has never been on the scale, so looking at the results of "2" won't help. Here's the problem:

1) Weigh coins 1,2,3 against 4,5,6 - assume they balance.
The bad coin is among 7-12, no clue yet whether light or heavy.

2) Weigh coins 1,2,3 against 7,8,9 - assume they balance.
The bad coin is in 9-12, no clue yet whether light or heavy.

3) Now weigh 10 against 11:

• If 10 is heavier than 11, is 10 heavy, or is 11 light?
• If 10 is lighter than 11, is 10 light, or is 11 heavy?
• If they balance, then it's coin 12 - but is it heavy or light?
  

See my point?

Keep at it - you'll get it yet!

This works. I checked and double checked.

1) Number the coins 1-12

2) Note:

a. If any set of coins are imbalanced, it means all other coins that were not weighed must be genuine for the fake coin must be within the set of imbalanced coins.

b. If a coin previously suspected to be in a group heavier coins is then weighed and found within a group of lighter coins then it must be genuine for no coin can be heavy and light.

3) Weigh 1,2,3,4 vs. 5,6,7,8

a. If equal, weigh 8,9 vs. 10, 11

i. If equal, 12 is the fake. Weigh 12 vs. any other coin. It will be either heavier or lighter

ii. If 8,9 are heavier than 10,11 then weigh 10 vs. 11

1. if equal then 9 is the a heavy fake

2. If one is lighter, it is the light fake

iii. If 8,9 are lighter than 10,11 then weigh 10 vs. 11

1. if one is heavier it is the light fake

2. if equal then 9 is the light fake

b. If 1-4 are heavier, weigh 3,7,8 vs. 4,5,6

i. If 3,7,8 are heavier, weigh 5 vs. 6

1. if imbalanced, the light one is the fake

2. if equal then 3 is the heavy fake

ii. if 4,5,6 are heavier, weigh 7 vs. 8

1. if imbalanced, the light one is the fake

2. if equal then 4 is the heavy fake

iii. if equal weigh 1 vs. 2

1. the heavier of the two is the fake

c. If 5-8 are heavier follow the same procedure as if 1-4 were heavier but the results will be opposite

4) This does account for all possibilities. The numbers could be shuffled around with different combination but the technique is the same and works for all possible outcomes.

What if number 1 is the heavy one???

If coin #1 is heavy, the steps in this method would be:

3) b. iii. 1.

Ah, yes, the numbering was confusing me.

I was thinking you grab your Ohmeter and test the resistance between the opposing faces of each coin until you find the one that is "not like the others."

Then place this California Coin on the left side of the scale, any other "genuine" coin on the right side of the scale, and immediately know in one weighing whether it's a lightweight or just too dense.

Then put the other two admissions to the scale on eBay and call it a day.

HAH! You & "David in Michigan" are a good team. Reminds me of a story:

Physics teacher asks students to explain how to use a barometer to find the height of a skyscraper. Smart kid comes up with three methods:

1. Lower barometer over edge of roof with a rope; measure length of rope.
   
2. Drop barometer over edge of roof; time its fall; calculate height.
3. Sell barometer, bribe building superintendent to tell you the height!

Wonder how many more "outside the box" solutions we'll get before Tuesday...

This is an "oldie but goodie" and deserves its own thread!

See "Kobayshi Maru" entry on Wikipedia for more "outside of the box" solutions!

Captain Kirk was (will be?) no dummy. But then neither was Alexander the Great, according to legend. The Gordian Knot typifies "outside of the box" thinking, even in 333 B.C.! In this legend, Alexander the Great (a real person), "untied" a huge puzzle knot that had no seeable beginning nor end by cleaving it with his sword after hours of frustrating attempts. An oracle had prophesied that he who untied the knot would rule over Asia (current Asia Minor), and Alexander went on to conquer most of it.

I'm not positive this one would work. There is a distinct possibility that the coin is cast out of an identical metal, is lighter because of a casting bubble or two, and not an alloy. Silver or gold comes to mind. In my limited coin collection knowledge, one of the things I learned is that a popular way to counterfeit coins in the 1800's was to take a legitimate rare coin, and make multiple castings of it. The coin then has the identifiable marks that you would expect to find on an original coin. The coin can even be cast out of the correct metal. But it is counterfeit because it was not stamped at the mint anywhere near the date on the coin. There used to be books out there -probably still is- that show pictures of known cast coins so that buyers can be aware of them. The value of these coins is in the rarity and not the metal anymore. The 1933 $20 gold piece that recently sold for a couple million comes to mind. I believe cast metal is likely to give an identical resistance value assuming it is of the same purity as the originals. It has the conductance of gold, or silver, etc. - same as the coin it duplicates. However, the metal molecules themselves are not lined up the same and therefore cast coins do not "ring true" as compared to rolled metal, punched, stamped, mint made coins. The processing of the metal that makes the final coin affects its resonant frequency.

David (up in the snowy upper part of the Upper Part of Michigan)

You really got me interested in the whole idea of identification by acoustic resonance. I imagine it has applications in verifying welds, structural integrity, homogeneous compositions, cheap non destructive verification of alloys. I would like to do some experiments but I do not have a scope. Can you point me to any experimental data or information about the process?

Specifically I would like to know what to look for in the waveform. What aspects of the waveform are unique only to the particular example under test. What aspects are unique to all genuine copies. What aspects might vary in subsequent tests on the same sample due to variations in test set up. Such as relative significance of ;where the item under test is held or struck; holding fixture design, material, grip strength; striking instrument material, design(rigidity), mass, velocity, incident angle, damping, ability to get out of the way of subject vibrations; mic placement and properties; room acoustic affects, air temp, humidity; sympathetic vibrations.

In the example of dropping the coins on the bar and knowing the difference by sound I would bet that alot of the things that make it possible are not present when striking with a knife. The knife strike is mostly a singular rapid event. More accurate data is available because you can record (mostly)the free space reverberation, natural resonance of the struck coin. But when dropping on the bar you get multiple impacts, some bouncing, sliding, rolling. The length of the ringing is affected very much by the damping of the bar, no longer an exponential decay as in the free space example. You hear more from the bar than the coin. You might think of these 'noises' as interference with the coin resonance that you are trying to hear. I believe it is primarily these affects that make the sound identifiable by ear. Dropping change on a table is a familiar sound. I believe this familiarity atunes us to qualities that differ even very slightly from what we expect. The ear is a remarkable instrument not yet fully understood. Off topic, I would like to know why a simple 30 year old tube amplifier with 10%THD sounds remarkably better than some of the most sophisticated modern transistor amps with better than 0.001%THD. Hint: it is related to odd vs. even harmonics. I think there may be some breakthroughs there for someone willing to invest the effort.

slo

I think what you mention could be done in some instances. It became an interest of my in my coin collecting days when an engineer suggested it. You can catch me at dafinzel@up.net if you want to discuss further. I found it a real help when it came to coins... Never had a counterfeit pass this test that I owned or had access too.

David (Up in the snowy upper part of upper michigan)

Based on SLOCO's post #24 the following is one of the correct answers. Note that OC-1 to OC-5 are the same as the original post to save you some reading.

Excellent challenge AstroNut, it sounded simple but was tricky indeed.

Split the 12 coins into 3 groups of 4

Weigh any 2 groups call them group 1 & 2

OC-1) If they balance then all the coins in group 1 & 2 are genuine; the suspect coin is in group three.

OC-2) Take any 3 coins from the third group and weigh them against 3 coins from any of the genuine coins in group 1 and or 2. If they balance then all the 3 coins are genuine the counterfeit is the remaining coin.

OC-3) Weigh the counterfeit against any known genuine coin you will know if it is heavy or light.

OC-4) If the weighing from OC-2 did not balance (assume the left side was lighter) you will know you are dealing with a lighter coin at this point. Weigh any two coins from this group against each other if they balance the counterfeit is the remaining coin and you already know it is lighter.

OC-5) If the weighing from OC-4 did not balance the lighter coin is the counterfeit.

OC-6) If the weighing from OC-1 did not balance you know the third group of coins is genuine. You can now mark each coin in the 3 groups either G genuine L for light or H for heavy. Remove two light coins and place two heavy coins and one light coin on each side of the scale and weigh the results (your groups should be LHH & HHL). If the scale balances then one of the two light coins you removed is the counterfeit.

OC-7&8) Weigh either of the light coins from the two coins you removed against a known genuine coin, if it balances then the un-weighed coin is the counterfeit and you know if it is light because it is marked. If it does not balance then you know it is counterfeit and again it will be marked light.

OC-9&10) If the weighing from OC-6 did not balance (assume the LHH side went down) then from the (LHH) group of coins weigh the HH coins against each other. If they balance then the L coin from the HHL group is counterfeit and light as this side went up. If they do not balance then the heavy coin is counterfeit.

OC-11&12) If the weighing from OC-6 did not balance (assume the HHL side went down) then from the HHL group of coins weigh the HH coins against each other. If they balance then the L coin from the (LHH) group is counterfeit and light as this side went up. If they do not balance then the heavy coin is counterfeit.

Let a,b, c and d be 4 heaps of 3 coins each.
b1,2 would then mean coin 1 and 2 from heap b
a1b1 would mean coin 1 from heap a and coin 1 from heap b
^ means the scale
= > < means equal heavier lighter heaps

1. a1,2,3 c1^ b1,2,3 d1
2 If a1,2,3 c1 < b1,2, 3 d1, then c2,3 and d2,3 are fine - goto 2.1
3. If a1,2,3 c1 > b1,2,3 d1 , then c2,3 and d2,3 are fine - follow the same arguments as for 2. to get counterfeit
4. If a1,2,3 c1 = b1,2,3 d1 , then problem is with c2,3 or d2,3 - goto 4.1

2.1 a1,2 ,3 ^ c2,3 d2
2.1.1 If a1,2,3 < c2,3 d2 , so counterfeit is lighter and is in a1,2,3 and we a1 ^ a2 to get it. - end
2.1.2 If a1,2,3 = c2,3, d2 , so c1 or d1 must have been counterfeit
2.1.3 c1 ^ c2 , if c1 < c2 then c1 is counterfeit and lighter , if c1 = c2 then d1 is counterfeit and heavier - end

4.1 c2,3 d2 ^ a1,2,3
4.1.1 If c2,3, d2 < a1,2,3 , then counterfeit is in c2,3 d2 and is lighter and we c2 ^ c3 to get it. -end
4.1.2 If c2,3,d2 > a1,2,3 , then counterfeit is in c2,3 d2 and is heavier and we c2 ^ c3 to get it - end
4.1.3 If c2,3 d2 = a1,2,3 then counterfeit is d3
4.1.4 d3 ^ a1 , if d3 will either be heavier or lighter -end

Drop each onto a smooth concrete floor from the same height every time, one at a time and the culprits will sound different, than weigh them, once. Done like dinner!

CR4 members: This has been a blast! Best regards to all who posted here; lots of good minds doing innovative thinking.

Guest 22 (David in Michigan) went "outside the box," as did Sandman.

Guest 20 solved it by doing internet research.

510 inspector solved it from memory.

Sloco solved it teamwork style, by picking up tips from A.E.Newman's post and giving due credit.

6275, mechtech, Guest 47, Guest 66, and Guest 84 all found working methods.
(Guest 84 invented a cool notation system, too.)

NOTES: I first encountered this puzzle in the mid-sixties. The fastest solution I've ever seen was done in about an hour, by my current boss, WHILE DRIVING A CAR. Scary. His breakthrough was to recognize the setup as a trinary system:

• Each coin has three possible identities: light, heavy, or good.
• Each weighing has three possible outcomes: right heavy, left heavy, or balanced
• Each weighing uses three sets of coins: left pan, right pan, not on scale

From there he developed a working method in no time. Brilliant! I guess that's why he makes the big bucks. :)

Like I said, this week has been enormous good fun. Thanks for playing!

(GO BEARS)

I solved it from memory because I remembered that I had solved it....about fifteen years ago.

Trust me to be away from the office when the first decent challenge question for ages turned up. However I think I can still offer a slightly simpler solution which has not been covered in the rest of the posts.

First notice that if you know that the coin is one of three, and, that it is heavier of lighter, you only need one weighing to identify the culprit.

W1.) ABCD against EFGH
W2.) irrespective of result 1: IJKD against ABCH

Result 1.) Both balance: L is duff and one more weighing reveals heavier or lighter.
Result 2.) Balance stays the same: D or H is duff; you know which one may be light and which may be heavy; one more weighing of one against a good one reveals the dud.
Result 3.) Balance swaps: A, B, or C is duff and you know if it's lighter or heavier.
Result 4.) W1 unbalanced W2 balanced: E, F, or G is duff and you know if it's lighter or heavier.
Result 5.) W1 balanced W2 unbalanced: I, J, or K is duff and you know if it's lighter or heavier.

Oh, my! I like that one. Very elegant. Thanks, Randall!

Line 9 of this solution says "IF 3) DOESN'T BALANCE: The heavier coin - 9 or 10 - is bad." That isn't a solution is it? Which is it, 9 or 10?

pikeslayer

In getting to line 9, step 3) was weighing 9 against 10. If it doesn't balance, then either coin 9 went down (it is heavier) or coin 10 went down (it is heavier). Whichever coin went down, that's the bad one.

I have nothing to say really. I just thought it might be fun to have post #1 and 100 in the same thread.

slo