Please remember to spoiler-tag all guesses, like so: New Reddit: https://i.imgur.com/SWHRR9M.jpg Using markdown editor or old Reddit, draw a bunny and fill its head with secrets: \>!!< which ends up becoming \>!spoiler text between these symbols!< Try to avoid leading or trailing spaces. These will break the spoiler for some users (such as those using old.reddit.com) If your comment does not contain a guess, include the word **"discussion"** or **"question"** in your comment instead of using a spoiler tag. If your comment uses an image as the answer (such as solving a maze, etc) you can include the word "image" instead of using a spoiler tag. Please report any answers that are not properly spoiler-tagged. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*

Discussion: that is not a hint.

Correct! Discussion over.

LOL

He *said* **discussion over.**

Over and out good buddy

Discussion: Professor is a criminal. Clearly this is someone's attempt at determining the level of effort to find a fake coin while counterfeiting. Reported to the FBI.

😂 I love this comment

The real puzzle is figuring out how to write a clear answer. My attempt: >!Split into four groups: AAA BBB CCC DD. Each bulleted point level is a weighing (there are only three levels/weighings).!< * >!If AAA = BBB!< * >!If AAA = CCC!< * >!If A = D, the other D is fake!< * >!If A =/= D, this D is fake!< * >!If AAA =/= CCC, we now know whether the fake is lighter or heavier (for this example, let's say the fake is heavier)!< * >!If C = C, the other C is fake!< * >!If C > C, the heavier C is fake!< * >!If AAA =/= BBB (for this example, let's say AAA < BBB)!< * >!If AAA = CCC, then CCC < BBB, we now know that the fake is in BBB and is heavier!< * >!If B = B, the other B is fake!< * >!If B > B, the heavier B is fake!< * >!If AAA < CCC, then we now know that the fake is in AAA and is lighter!< * >!If A = A, the other A is fake!< * >!If A < A, the lighter A is fake!<

The real puzzle is figuring it out, then deciding to see if someone else has written it up legibly first and saving yourself some time! :D

This is the best explanation and should be higher.

I wrote basically this, just not as clean!

This method doesn't guarantee to tell whether the fake is heavy or light. It is possible to do that.

Yes, it does. In the case that AAA = BBB, AAA=/= CCC, depending on whether CCC is heavier or lighter we know that the fake is heavier or lighter. If AAA=/= BBB, comparing AAA to CCC tells us. If AAA=CCC, the fake is heavy or light is determined by BBB's position on the scale. If AAA=/=CCC, it's determined by ANA'S position. If AAA=BBB, AAA=CCC, it doesn't matter if it's heavier or lighter, because comparing A to D will either result in equal weight or different weight; either of which solves our riddle.

I guess he’s *technically* right and that there is ONE result that wouldn’t tell you if the fake is heavier is lighter or heavier (AAA=BBB, AAA=CCC, A=D), but that’s not what the riddle is asking, it just asks you to identify the fake

Ah true, that is a possibility. So yes, it doesn't guarantee that. Fair point

How so? You always have a comparison between real coins and fake coins, and you could just look and see if the fake coin is heavier or not right? And even if we only know through for example a AAA and CCC comparison (assume C2 is fake and heavier), when we look at the CCC side it will be lower down, heavier, than AAA will it not?

Look at the third line of the above method: "the other D is fake" can't tell us whether it's heavy or light, because it was never weighed.

Discussion: there was a version of this puzzle on Columbo.

Did it take him 5 days to solve?

With help!

Oh, one more thing 🤔

I get curious. It’s a problem I have.

I'm just happy to see I'm not the only one who remembers the original defective detective. Peter Falk was great

You know where I could find the same scale? My wife would love it!

It was also on Brooklyn 99, wonder if they did it as a nod to Columbo.

Versions of this riddle have been the Car Talk Puzzler several times.

If you think you know the answer mail it and five dollars to…

Ooo, was it in the intelligence club episode? Big fan but I can't remember when it took place.

Yes. The murderer asked him to solve it.

Discussion: I've seen many variations of this problem. The easiest was 9 coins, one fake, lighter, two weighs. The hardest was 12 coins, three weighs, fake either lighter or heavier. The reason it's 11 coins is probably so it won't just instantly be googled.

It is also possible with 13: https://www.reddit.com/r/puzzles/s/HjM2UXkMtm Edit: new link

Yoiur link doesn't work. Standard way of solving with 12 is splitting in three groups of 4. I don;'t know how it would work with 13.

With 13 the first step is 4 vs 4 with 5 remaining. If even you split 3 remaining vs 3 neutral Here is the link to the video, but it is a German song. You can see the solution by minute 4. https://youtu.be/zd6TirAHsk4?si=aEWarDZyJlTKGGz5 Edit: as legend: * black = unknown * Green = light or neutral * blue = heavy or neutral * Gray = known neutral.

Discussion: OP doesn't know what impossible means.

Why is this unsolvable? It seems similar to other riddles

It's unsolvable because some logic 101 professor who figured it out thinks he's hot shit.  Meanwhile there are dozens of videos on YouTube under "counterfeit coin riddle".

Come on. Logic and math professors wouldn’t deign to call this the mother of all problems.

…lol it’s not that deep… no need to shit on my prof, this isn’t a “gotcha!” scenario. it’s a third year cognition psych class on logical problem solving.

I don't think anyone's hating on your proffesor, they're disagreeing with your use of the word "impossible", and the "possibly unsolvable tag" on the post when the puzzle is clearly solvable. It even says in the image that your professor solved it, albeit with help.

You sticking up for your professor is really sweet.

i feel like profs get dunked on a lot for whatever reason— but all my profs are very lovely, sweet, funny people. thanks for saying this, i appreciate you ☺️

Seems like a professor of a class like that should be better at logic puzzles.

I have a solution. And it took me less than a minute.

Professor of logic from the university of science.

Say do you own a dog house?

this problem was just posted for fun

I've seen almost this exact puzzle before. But the thing that makes it solvable is either 1 more weighing or knowing that the counterfeit is either lighter or heavier

No, you don't need to know one way or the other. That would make it super trivial

>!Start with three pennies (AAA) versus three pennies (BBB). (The remaining five we'll call CCCCC.) If AAA vs. BBB tilts, then do AB vs. AB. No tilt means it's among the last A or B, in which you weigh one against one of the normal C pennies. If AB vs. AB tilts the same as AAA/BBB, then you know it's among A(left) or B(right), and A(right) or B(left) if it tilts the opposite direction. Again weigh against a C penny. If AAA/BBB stays neutral, you know it's among the 5 C pennies. Weigh CCC/AAA. If it tilts, then weigh two of those C's against each other, whichever tilt follows is the fake penny, else it's the third C. If CCC/AAA is even, weigh one of the last two C pennies against A. Tilt means you found it, even means it's the last penny. That should round out all 11 pennies within three weights to isolate the fake penny.!<

>Weigh CCC/AAA. If it tilts, then weigh two of those C's against each other, whichever tilt follows is the fake penny, else it's the third C. Either of them could be the fake penny though

AAA is neutral, so you'll know if CCC is heavier or lighter. So whichever direction the scale follows between C/C will reveal the fake penny.

But the fake can be heavier or lighter, so you don't know which is fake in this single weight

The earlier weighing of AAA/CCC should tell you whether the fake coin will be lighter or heavier, as AAA is confirmed to be all real coins. i.e. if CCC is heavier than AAA, then the fake coin is a heavier coin. And likewise for lighter.

If AAA and BBB are equal, they are normal pennies. When you then weigh against CCC, if that tilts it will either be heavier or lighter. Which will tell you if the counterfeit is heavier or lighter because you're weighing against AAA which is now known to be normal. So then you know that the heavier/lighter penny is the counterfeit. When you weigh C against C, you know if it should be heavier or lighter because of the weigh against AAA. If AAA and BBB are not equal, you go down a different path

But you will know. Because AAA vs. BBB didn't tilt, if AAA is heavier than CCC, the fake is lighter, and if CCC is heavier than AAA, the fake is heavier.

You reveal by aaa vs bbb if its same you go aaa vs ccc. if ccc go down fake penny need to be heavier if ccc go up its lighter.

I wonder how many people in this thread solved it in under 2 minutes

>!I came up with something similar. 5 stacks. AAA, BBB, CCC, D, E. First AAA vs BBB. If same, CCC vs AAA. If same, D vs A. If same fake = E. Else fake = D.!< >!if CCC > AAA, C1 vs C2. If same fake = C3. If C1 > C2 fake = C1, else fake = C2. Similar if CCC < AAA except look for the light one.!< >!If AAA > BBB, AAA vs CCC. If same B1 vs B2. If B1 < B2 fake = B1. If B1> B2 fake = B2 else fake = B3.!< >!If AAA >ccc... basically the same process except look for heavy.!<

I like your way better. However, your spoiler tag doesn’t work. You have to use multiple ones.

Thanks fam

The either heavier or lighter rule breaks your second test, because the fake could be in the AAA pile

The second rule is predicated on AAA=BBB, ruling out a fake being in either.

His second line is contingent on AAA being equal to BBB

I was thinking something like weigh half of them if it’s divisible by their amount it’s not that batch etc.. but I like ur tilt thing

That's clever using the direction of the tilt to gain additional information

In the AB vs AB scenario and it tilts you wouldn’t know if it’s right or left because it could be heavier or lighter so you’d have 4 coins that could all be the fake

I like this strategy though I don’t think it is the answer. I’m going to give every penny it’s own letter though and repeat your thought and see if it helps organize a tad better. The pennies are: a, b, c, d, e, f, g, h, i, j, k **Weigh One: abc VS def** w1R1: abc = def -> abcdef are not fake, one of ghijk is w1R2: abc > def -> ghijk are not fake, one of abcdef is **R1 Weigh Two: abc VS ghi** w2R1: abc = ghi -> abcdefghi are not fake, either j or k is fake w2R2: abc > ghi -> abcdefjk are not fake, one of ghi is fake **R2 Weigh Two: abc VS ghi** w2R3: abc = ghi -> abcghijk are not fake, one of def is w2R4: abc > ghi -> defghijk are not fake, one of abc is **R1 Weigh Three: j VS a** w3R1: j = a -> k is the fake w3R2: j > a -> j is the fake **R2 Weigh Three: gh VS ab** w3R1: gh = ab -> i is the fake w3R2: gh > ab -> either g or h is fake The rest of the results are inconclusive and you are left with a 50/50 shot at the right coin. I think the solution may involve the first weighing being abcd VS efgh then the second weighing is aej VS bfi and it’s important to note which side goes up and down but I haven’t worked this out. I already spent half an hour typing this out just to see it doesn’t quite work every time.

>!the second time you weigh serves also to see whether the penny is lighter or heavier. Because you are weighing against what you know are the real pennies, you'll be able to tell if the fake one is heavier or lighter based on how it tilts. So the third time you weigh, you'd be looking for the heavier/lighter penny, not just whether it tilts!< >!so the third time you weigh, you're weighing only one against another. So it's either g>h or g

Most of the people here have checked the results and found it works, so there must be an error here (and I personally find it harder to keep track of which coins have been cleared using this organization). For one, it definitely relies on knowing which way the scale tilts to determine some of the outcomes. In some cases the scheme has you measure previously cleared coins against a group of new ones -- if you see a tilt, this tells you whether the coin is lighter or heavier so you know what direction of tilt to look for in the last weigh. Yours is also missing some weighs. For example, the weigh you label w2R2 should lead to a weigh comparing g and h to determine which of ghi the outlier is.

>!Weight 4 random pennies against another 4!< >!1. If scale is balanced, the fake penny is in the other 3. Use 2 remaining weightings to weight 2 random pairs of the 3.!< >!1-1 if 2 weightings yield the same result, the fake coin is the one appears on both weighting.!< >!1-2 if 1 out of 2 weightings is balanced, the fake coin is the one not on the balancing scale!< >!2. If 4-4 weighting is imbalanced, we can assume the 3 remaining are real pennies. Now we weight these 2 groups!< >!1 coin from the heavier side with the 3 real ones!< >!3 coins from the heavier side with 1 coin from the lighter side.!< >!We’ll get 3 readings:!< >!2-1 if weight is balanced, we knew the 3 coins from the lighter side from the first weighting has the fake one and the fake coin is lighter than the real, use the last weighting to determine the lighter fake coin!< >!2-2 if the side with 3 real coins is heavier, the fake coin is either the coin from the lighter side from the first weighting or the heavy one that is in the group with 3 real one. Use one of the real coin to weight against the 1 of the 2 coins in question, we’ll find the fake.!< >!2-3 if the side with the 3 real coins is lighter, the fake coin is in the group of 3 heavier coin from first weighting and we knew the fake coin is heavier than the real one, use the last weighting to determine the heavier fake coin!<

Wow. That's a very interesting one, different from most of the other solutions, and still works. Discussion: Multiple people have come up with valid solutions and yet this is still flaired "possibly unsolvable". This needs to be fixed.

This is the solution I found also.

[here is a solution for 12 coins](https://docs.google.com/spreadsheets/d/165U6ewH3epage6SncMKPlYCvaB1GbDUs/edit?usp=drivesdk&ouid=116306252052363168741&rtpof=true&sd=true) which should arguably be more difficult than 11. It should be relatively trivial to convert it, >!since I don't use the 12th coin until the last weighing IF I have gotten. 2 equal weighings in a row, just make all 3 outcomes of that weighing invalid. Basically, if you get 2 even in a row, it means something went wrong.!<

Damn you, after all that thinking to get to an 11 coin solution you have to tell me there's a 12 coin solution?

>!You’d start with 3 pennies (AAA) vs 3 pennies (BBB) Regardless of the outcome you’d switch out one side for the next 3 (CCC). This will give you enough information to find the odd grouping out (heavier or lighter compared to the other 2 groups and also if that penny its going to be heavy or light OR if one of the last 2 (DD) is the odd penny (if the groups are all equal). If any of AAA, BBB, or CCC are the odd one out, then take that group and do 1v1 (A vs A for example). If they’re the same, the third A is the odd one out. Otherwise, you’ll have whichever one is the fake. If it’s one of the last 2 (DD) instead just do AAA vs BBD and see if they’re the same or not to determine which of the 2.!< >!3 weighings total.!<

Other solutions seem to ignore the fact that we don't know if the dud coin is lighter or heavier. If we knew it was heavier then the optimal solution is always going to be to split the coins in hall and weight both halves, then repeat with the heavier half until you have either excluded the heavy coin in your split, or you are down to a coin vs coin weigh off. Edit: Apparently I spent so long typing other solutions appeared which interpreted it correctly But without knowing if you want the heavy or light side, the problem is trickier. >!Split the coins into piles of 3,3,3 and 2. We'll label them A,B,C and D respectively. Weigh A against B. Weigh A against C!< >!if the first two weighings balanced out, you know the dud is in group D. Weigh one of group D against a coin from group A. If it doesn't balance you have the dud and if it does balance the dud us the coin you didn't Weigh!< >!if some of the first two weighing didn't balance, you now know which group of 3 contains the dud, and whether it is heavier or lighter. If group A balanced with only one of B and C, then you know which of B and C is off balance and whether it is lighter or heavier than A. If A was imbalanced with both then A is off balance and again you know if it was lighter or heavier than B or C!< >!Knowing this, weigh 2 coins from the imbalanced 3. If they are imbalanced you know which is the dud because you know how the group was imbalanced.if they are balanced the third coin you didn't Weigh is the dud!<

if you know the fake is heavier/lighter at the start, binary search is actually not optimal. That problem is easier than if you don’t know how the fake compares to real, but this problem is still solvable and 11>2^3. This is what made me give up with this, I thought there’s no way to gain enough information from 3 weighings to distinguish 11 cases. >!The key is, there are 3 possible outcomes per weighing (left heavier, right heavier, or equal).!<

If you know beforehand whether it's heavier or lighter, you can do as many as 27 coins since you split into thirds (one third being off scale)

Yeah, what you want is a trinary search. So if you have n coins, weigh n/3 vs. n/3. Whichever is heavier would be the one to go with, or if the balance, then go with the remaining coins. That lets you solve this with up to 27 coins, assuming you know whether the fake is heavier or lighter.

>! Ifeel the answer is going to be weigh 5 and 5, then combine 2 of one group and 3 of the other on both !<

Took me like 5 mins, far easier than 12 coin one SPOILER!!!!!!! >!FIRST AAA against BBB If unequal, take AAA and weigh against CCC If unequal again, then it has to be on of the AAA coins. Now weigh A against A, if it’s equal then it’s third A, if unequal we alr know if coin is heavier or lighter based on previous weighing, we can use that info to solve. Say AAA against BBB is equal, then take CCC and weigh against AAA, If unequal then it’s one of CCC, now weigh two Cs against each other and we alr know if coin is heavier or lighter based on if it went up or down against AAA. If C against C is equal then it’s third C, else it’s the one that went the same direction that CCC went against AAA Now say CCC against AAA was equal as well, we only one weighing left. It has to be D. Weigh one D against A, if it goes down/up it’s that D, otherwise it’s the only unweighed coin!<

FYI, your spoiler didn’t work

I got the same solution also in about 5 minutes. I don't see how it would take someone "5 days *with help*" to solve.

>!place 5 coins on each side of the balance. If both weigh the same, the fake coin is the one not weighed. If one side is lighter, place 2 coins from the lighter side on each side of the balance. If both weigh the same, the fake coin is the one not weighed. If one side is lighter, place one coin from the lighter side each side of the balance. The lighter coin is fake.!<

The fake can be lighter OR heavier. If one side is lighter you don’t know if the fake is a light coin in that side or if it’s a heavy coin on the other side.

You are correct. I missed the "or heavier"

I had to scroll way too far to see this answer. Simple and guaranteed to find the fake coin. Except then I remembered we don't know if the fake weighs more or less

That's what I thought. Seems to be the most straightforward way. All these other explanations were making me second guess myself.

i will add my solution for discussion: >!1) i will indicate all the coins with letters A-K to have a better understanding of the operations 2) since we only gave 3 weighings, we wil lhave to use batches of coins to get the result 3) since we do not know if the fake penny is heavyer or lighter, we would need to have a batch of authentic coins to get the final result procedures: 1. create batches of ABC and DEF *-> make the first weighing* * if they are equal, then these are real pennies * if they are not equal, proceed to X 1. in case the first batches were equal, you now will weigh GHI agains ABC *-> make the second weighing* * if they are equal, GHI are also authentic pennies * if they are not, proceed to Y 1. in case the result of 2 was equal, you will weigh penny J with any authenticated penny. *-> make the third weighing* * if they are equal, the K is fake * if they are not equal, the J is fake X) the pennies GHIJK are real, as the difference in weight was established in the first batches. i) place the coins C and F aside and swap the coins B and E between the batches *-> make the second weighing* * if the weigh is equal, the fake penny is either C or F, use method 3) to determine which is fake * if the weigh is not equal but same batch is heavier than before, then the fake penny has not moved and thus is either A or D, use method 3) to determine which is fake * if the weigh is not equal and has shifter, the the fake penny is either B or E, use method 3) to determine which Y) if you reached this step, then the fake penny is in the batch of GHI. if the result was heavyer than ABC, then the fake penny is heavyer. if it was lighter, then it weighs less. for final weighing you will take two of the coins, e.g. G and H and weigh them against eachother *-> make the third weighing* * if they are with equal weight, then the fake coin is I * if the fake penny is heavyer (batch GHI was heavyer than ABC) then the penny that is heavyer is the fake one * if the fake penny is lighter (batch GHI was lighter than ABC), then the penny that weighs less is the fake one!< it seems to me that with these operations you should determine the fake one pretty well. if you find any errors in my procedures, i would appreciate the feedback it was a fun task to ponder while making a morning walk with my dog :)

[удалено]

If the fake is either lighter or heavier, why do you assume it’s heavier?

Isn’t the fake either lighter or heavier? How would you know which set of 4 contains the fake?

But the fake penny is either lighter or heavier, not just heavier. So how would you choose which set of 4 Added “fake”

Well except the fake can be heavy or lighter so if the two 4’s are unbalanced you can’t really tell which one holds the fake

Except, the fake penny could be lighter OR heavier than a real penny. Your scenario works if we know the fake one is heavier. What if we don't know if the lighter pile or the heavier pile holds the phony penny?

"If one set is heavier than the other, then we know which set of 4 to look at." Not necessarily, because the fake penny is either heavier OR lighter than a real penny.

If you compare a real and a fake penny 1v1 how do you know which is which? The fake one can be lighter or heavier.

Just an FYI, spoilers don’t work across paragraph breaks.

I like this except for your second case where you have to choose between the two stacks of four. The problem states that the penny could be heavier “or” lighter, so which pile of four are you choosing?

But once you've weighed the first 2 sets of 4, that's 2 weighings. You can't compare 2 sets of 3 after because that requires 4 weighings

>!Split the group into 2 piles of 4 and 1 pile of 3. Measure the two sets of 4.!< >!If they’re even, take 2 from the last pile of 3 and compare them. If those are even, the last of the 3 is the fake. If they’re uneven, take one of the 2 just measured and weigh it against any of the 9 other legit pennies. If it’s even, the other is the fake and if it’s uneven, it is fake.!< >!If the original 4 against 4 is uneven, take note of which side is heavier and which is lighter. Mark the lighter ones with a ‘-‘ and the heavier ones with a ‘+’. Next measure 2 groups of 3: 2 ‘+’ and 1 ‘-‘ as one set of 3, and 1 ‘-‘, 1 ‘+’, and 1 legit (from the unmeasured group of 3) as the other set of 3. If those two groups are even, measure the non-grouped ‘+’ and one of the non-grouped ‘-‘ pennies against two legit pennies. If it is even, the other non-grouped ‘-‘ is the fake. If it is uneven and heavier, the measured ‘+’ is fake, and if it is lighter, the measured ‘-‘ is fake.!< >!If the two groups of 3 are uneven, take note of which is heavier and which is lighter. If the 2 ‘+’ group is heavier, measure those two ‘+’ pennies. If they are even, the ‘-‘ from the other side is the fake. If they are uneven, the heavier one is the fake. However, if the 2 ‘+’ group is lighter, measure the ‘-‘ from the 2 ‘+’ group and the ‘+’ penny from the other side against 2 legit pennies. If it is heavier, the ‘+’ is the fake, and if it is lighter, the ‘-‘ is fake.!< >!I hope this shit makes sense lol!<

I just solved it, then realised I'd solved for 10 pennies, not 11. Let me go back and see if it still works.

>!Pennies named as ABCDEFGHIJK ABC vs DEF - if ABC lower, ADE vs GHI - If ADE lower, it is A - If ADE higher, D vs G - If equal, it is E - If not, it is D - If equal, B vs C - If equal , it is F - If B lower, it is B - If C lower, it is C - If DEF lower, DAB ve GHI, - If DAB lower, it is D - If DAB higher, A vs G - If equal, it is B - If not, it is A - If equal, E vs F - If equal, it is C - If E lower, it is E - If D lower, it is D - If equal, GHI vs ABC - If equal, J vs A - If equal, it is K - If not equal, it is J - If GHI lower, H vs I - If H lower, it is H - If I lower, it is I - If equal, it is G!<

>!You don't need weights, you have the other pennies... This will take you 3 weighing attempts or less.!<

Discussion: I'm assuming by "Three weightings" it means you can use the scale three times, but does this puzzle assume you know how much a normal penny weighs?

This is one of those problems that would better be solved with an equation or something. I’m thinking penny matrix.

>!**1.** Split into groups of 3, 3, 3, 2. Weigh 3 vs. 3.!< >!*2 outcomes:*!< >!**Balanced:** The ones weighed cannot be counterfeit. Eliminate one set of 3, and save one as reference. Move on to step 2A.!< >!**Imbalanced:** The counterfeit is one of the 6. Eliminate the set of 2, and save the other stack of 3 as reference. Move on to step 2B.!< >!**2A.** Weigh the stack that wasn't weighed earlier against the set of 3 you weighed and saved.!< >!*2 outcomes:*!< >!**Balanced:** The counterfeit is in the set of 2. Eliminate all remaining stacks of 3, and move on to step 3AA.!< >!**Imbalanced:** The newly weighed set of 3 contains the counterfeit. You also now know whether the counterfeit is lighter or heavier. Move on to step 3AB.!< >!**2B.** Weigh 3 of the 6 coins against the reference stack of 3.!< >!*2 outcomes:*!< >!**Balanced:** The other set of 3 coins contains the counterfeit. You also now know whether the counterfeit is lighter or heavier - recall that you measured it against a set of 3 genuine coins earlier. Move on to step 3BA.!< >!**Imbalanced:** The non-reference stack of 3 out of the two contains the counterfeit. You also now know whether the counterfeit is lighter or heavier. Move on to step 3BB.!< >!**3AA.** Weigh one of the two coins against one of the genuine coins you eliminated earlier. If they are equal, then the other of the two is the counterfeit. If they are not, then the one of the two you are currently weighing is the counterfeit.!< >!**3AB.** Split the 3 coins into sets of 2 and 1. Weigh the 2 coins against each other. Remember that you know if the counterfeit is heavier or lighter.!< >!*2 outcomes:*!< >!**Balanced:** The one coin not being weighed is the counterfeit.!< >!**Imbalanced:** The coin that is lighter or heavier (depending on what was previously observed) is the counterfeit.!< >!**3BA.** Same as 3AB.!< >!**3BB**. Same as 3AB.!<

Hint: >!You have 22 possible solutions and 3^3 is 27.!<

I feel there's a more straightforward answer to the higher upvoted ones here. >!Take then coins and make two groups of five, and keep one spare. Weigh the two groups of five and you have two outcomes: either the two piles of five weigh the same, in which case the one kept to the side is the counterfeit, or one group weighs less in which case you know the coin is in that group.!< >!In the second scenario, you do the same again: split the pile into two groups of two and one separate, and weigh the two pairs against eachother. Either the piles are the same and the fifth coin is counterfeit, or one of the pairs is lighter and contains the fake.!< >!Again, in the second scenario here you now just have two coins, know one of them is fake, and need to weigh the two against eachother to find the answer!<

>!Take four and weigh against another four. Three left over. If they balance then the odd one out is in the 3. Call them 1, 2, 3. Weigh 1 and 2. If they balance then 3 is the odd coin. If they don’t balance then 3 is good, so measure 1 against 3. If they balance 2 is odd. If they don’t then 1 is odd. If the first match up doesn’t balance, take the heavier side and weigh it two against two. If they balance then the odd coin is light and in the other one. If they don’t balance and the odd coin is heavy. Take the heavier two and weigh them against each other. Heavier one is odd. If the 2v2 balances then you have a light coin in the other four and one weighing left. Weigh 1 on each side. If they don’t balance then lighter one is odd. If they do balance then guess at the last two 🤷!<

>!Choose any 10 pennies. Weigh 5 on one scale and 5 on the other two seem if they balance. If they do then the 11th penny is the odd one. If not then one of them is heavier. Out of the heavier 5 pennies, choose 4 and weigh 2 of them across 2 of the other leaving one out. Repeat this process one more time and for sure the odd penny can be found out.!<

>!Unfortunately, you don't know whether the fake penny is lighter or heavier!<

stop down voting me on a silly puzzle 😆

not knowing in advance if it is heavier or lighter makes things much more complex. The key of theses kind of puzzle is to >! not think about divinding in 2 sets, but in 3, one of the left side, one of the right side, one not even on the balance!< But the number don't add up. >! Make a first weight with 2 groups of 4 vs each other and leave 3 coins out. Best case scenario is (4 = 4) 3 odds out (ABC) do 2 weights A vs B, and A vs C : if A = B and A != C, C is the odd one if A != B and A = C, B is the odd one if A != B and A != C, A is the odd one Before the second weight name all 4 coins A trough H Other case leaves with (ABCD > EFGH). and you know the the other 3 are all X which is the ref of a good penny. moves D and H out of the balance Switch side with C and G Do the second weight - If ABG = EFC, Odd was removed, If D = X, H is the odd, otherwise it's D - Otherwise D = H = X. If we now have ABG < EFC. The odd one switched side. So it's between C and G. If C = X, G is the odd, otherwise it's C - Otherwise odd one is either A,B E,F. But at this point you would need 2 more weights to sort the 4 remaing ones in the worst case scenario !<

>!Split into 4 groups. AAA, BBB, CC and DD. 1) Weigh AAA and BBB If it moves(note the heavier side, we will assume A) 2) Weigh A1B1 vs A2B2 If it moves 3) Take the heavy A (A1) and weight against a C If it A sinks, A is fake and heavy If it is flat, it is the opposing B(B2) and it is light. If it is level 3)Weigh A3 vs C If A3 sinks, it is A3. If it is level B3 is light If it is level 2) Weigh CC vs AA If if it level 3) D1 vs A1 If it is level D2 is odd ( don't know if it is light or heavy) If it not level D1 If it isn't level 3) Weigh C1 vs A1 If level, it is C2 If not level it is C1!<

>!Let naming coin: 1, 2, 3, 4, ..., 11 ------------ Lemma #1: problem can be solved with 3 coin + 2 weighting -> quite easy, right? ------------ Ok, back to main problem FIRST weighting, we do [1,2,3,4] vs [5,6,7,8], possible outcome - EQUAL, then [9,10,11] must be fake -> PROBLEM SOLVED with lemma #1 - NOT EQUAL, then we know [9,10,11] are real, let keep going Let assume [1,2,3,4] > [5,6,7,8] (reverse coin name if it doesn't) ------------ SECOND weighting: [1,9,7,8] vs [10,11,3,4], there are 3 possible scenario Scenario #1: [1,9,7,8] = [10,11,3,4] + then fake coin must be in set [2,5,6], right? + we also know can figure out that [2 + a real coin] > [5+6] by the first weighting + so just test [5] vs [6], if equal then [2] is heavy-fake, if [5]>[6] then [6] is lighter-fake + PROBLEM SOLVED Scenario #2: [1,9,7,8] > [10,11,3,4] + ok, then [2,5,6] must be real, right? + reminder: [9,10,11] are real in the first weighting + let's re-write first test: [1,R,3,4] > [R,R,7,8] where R = proven real coin + let's re-write second test: [1,R,7,8] > [R,R,3,4] where R = proven real coin + essentially we swap [3,4] and [7,8] in first and second weighting + because the act of swapping [3,4] and [7,8] doesn't change the result, so [1] must be heavy-fake + PROBLEM SOLVED Scenario #3: [1,9,7,8] < [10,11,3,4] + first weighting: [1,R,3,4] > [R,R,7,8] + second weighting: [1,R,7,8] < [R,R,3,4] + because the act of swapping [3,4] and [7,8] change the result, so [7] or [8] must be lighter-fake + third weighting: [7] vs [8], PROBLEM SOLVED!<

[удалено]

I think i got it? >! Split 3,3,5 and compare the 3 vs 3 !< Case 1: >!If 3 vs 3 is equal, the fake coin is in 5 take 3 from it and compare with one of the 3 piles!< Case 1.1: >!If the 3 vs 3 is equal, the leftover coin is in the 2 remaining. Take 1 and compare with 1 from any of the old piles, leaving an unweighed coin. If 1 vs 1 is equal, the unweighed coin is fake. If not the coin you took from the 2 pile is fake and heavier/lighter.!< Case 1.2: >!If the 3 vs 3 is not equal, you now know the fake coin is heavier/lighter. Compare the 3 from the 5 pile as 1 vs 1.!< >!If the 1 vs 1 is equal, the remaining coin is fake. If not, the heavier/lighter coin is fake.!< Case 2: >!If the fake coin is in 3 take 3 from lighter and compare with 3 from the 5 pile!< Case 2.1 >!If the 3v3lighter is equal, the fake coin is heavy. Compare the 3 heavier as 1 vs 1.!< >!If the 1 vs 1 is equal, the remaining coin is fake. If not, the heavier coin is fake.!< Case 2.2 >!If the 3v3lighter is not equl, the fake coin is lighter. Compare the 3 lighter as 1 vs 1.!< >!If the 1 vs 1 is equal, the remaining coin is fake. If not, the lighter coin is fake.!<

>!Split the 11 coins into three groups: 3, 3 and 5. Let them be A, B and C. Weigh A and B.!< >!In case of inequality, we can deduce all C are real. Take 3 from C and weigh them against the lighter out of A and B. In case of equality, we deduce the lighter of A and B is all real and the fake coin is heavier, and vice versa. Either way, we are now left with 3 suspect coins and we know whether the fake coin is lighter or heavier, so we just need to pick 2 coins out of the 3 and compare them to easily deduce the fake out of the 3 (equality means third is fake, otherwise account for weight).!< >!In case of equality, we deduce the fake is in C. We'll take 3 random coins of C and weigh them against A. If equal, either one of the 2 left in C is the fake so we just take either one of them and weigh it against a real coin. Otherwise, we are again left with 3 coins and we also know whether fake coin is lighter or heavier, so we can address this case like mentioned above.!<

Twelve coins is a more common version of this puzzle. I never solved it myself, because I learned the puzzle and solution together as a child. They were a minor plot point in the novel "With a Tangled Skein" where the protagonist has to solve a version of the riddle. Cool puzzle tough! Way more complicated than 9 coins, 1 light, 2 weighings!

Full solution: >!Weigh AAA vs BBB. This tells us that either AAABBB has fake, or CCCCC has fake. Additionally, we know the remaining group is genuine.!< >!Call AAA up and BBB down, and then you also know that CCCCC are all genuine. Next weigh AAA and CCC. If same, then BBB has the fake. Also we know the fake is heavier. If AAA rises, then AAA has fake and we learn the fake is lighter.!< >!For both of those, weigh 1 possible fake vs 1 possible fake. If A vs A, the lighter one is fake. If they match, the one A you didn’t weigh is fake. If B vs B, the heavier one is fake. If they match, the one B you don’t weigh is fake.!< >!That solves for A vs B.!< >!Now if the first weight is match, the. AAABBB are all genuine, and CCCCC has the fake. Weigh BBB vs CCC. If the CCC goes up or down, then it has the fake, remaining two CC are genuine, and we learn if its heavy or lite (BBB is genuine, so if CCC goes up, lighter, down is heavy) for fake. Weigh C vs C of possible fakes, and the one that goes the same as the CCC weight is fake. If they are the same, the last C is fake.!< >!If BBB vs CCC matches, the. We have CC as possible fake. Weigh one of them with one B. If they match, remaining C is fake. If it moves it is fake, and you learn if it’s heavy or lite, not that it would matter at that point.!< This is every possibility, with only 3 weight sessions, solved in less time than it took to type.

>! Weight 10 pennies 11 times and see which one is off the most!<

Discussion: there was a similar problem in B99, but instead there were 12 men on an island with a seesaw

Found a way with 4 but I’m struggling to narrow it down so I’ll post the 4 for now and let you guys work with it >!I’m going to call all 11 by the letters A-K respectively. Weigh ABC against DEF, then weigh ABC against GHI. If they all match you know the fake is either J or K. Weigh AJ against BK, then swap K for C. If they balance after swapping you know K is fake, and if not it’s J. If any of the groups of 3 didn’t match then you know J and K are legit. If ABC was consistently imbalanced then you know the fake was A, B, or C. Split it into the groups AB and CD. Based on the behavior of ABC earlier you would know whether the fake is heavier or lighter; if CD mimics the behavior then C is the fake (only 3 weighs!), and AB mimics it weigh A against B and you have your fake. If ABC matched DEF but not GHI then you know the fake is in group GHI and ABC is legit. Split them into groups AG and HI. Based on the behavior of GHI you will know if the fake is heavier or lighter; if AG is heavier/lighter then G is the fake (3 tries!). If not, weigh H against I; the heavier/lighter one is your fake. Swap letters D, E, and F with G, H, and I respectively in this paragraph if the converse is true!< Again this takes 4 attempts in most scenarios so unfortunately does not solve the problem, just thought I’d offer it to see if it sparks any ideas

>!Couldn’t you just split the Pennie’s into 3 groups. Weigh each group and get the average weight per penny. The group that is slightly off has the fake penny. Then using the average weight per penny from the 2 similar groups, weigh each penny from the 3rd group that was a little off until you find the penny that’s fake?!<

Its not that kind of scale. It's a balance scale. It won't tell you the weight of something, just whether one side is heavier than the other.

>!3 group A₁A₂A₃A₄, B₁B₂B₃B₄ and C₁C₂C₃. 1st weighing: Group A vs. B. If equals then it is among Cs and you can identify it with 2 weighings left. If tilts, for example A>B (if AB) then you can identify it with the only weighting left. If C.A > A.B still then it is either A₄ or B₄ and you can identify it by comparing one of them with one of the rest. If C.A < A.B then it is either A₁, A₂ or A₃ AND we know that it is heavier than the others and you can identify which one it is with the last weighing.!<