My Solution: 1) Think for 10 minutes and tell the King you're ready. 2) GO mental with the marker and mark all the coins after randomly weighing them. 3) Pick one coin and tell the King that's the counterfeit. King doesn't know either, so you have survived. Cheers.
but the counterfeiter would still counterfeit ,and the king would realize that you made a mistake then throw you in the dungeon (If you don't run away)
0:52 But the Emperor's not a patient man **Weigh pieces of money one by one in 2 minutes** Emperor : Too slow! **Think in 2 hours of a possibility to only use balance 3 times** Emperor : Fast enough
I just split the 13 in half twice taking the lighter halves. So I had 12 and turned it into 6, I had 6 and turned it into three. I weighed two and and if equal then the fake is the one that I didn’t weigh but if one was lighter than the lighter one was the fake. Took me like 1 minute to solve
The hardest aspect of this and what people are missing is that the counterfeit is either heavier OR lighter, so you need to figure out that detail. You can’t assume the counterfeit is lighter/heavier from the beginning.
What's fascinating is that by marking the coins with + or - marks, you effectively negate that issue. Having 9 coins, 4 marked with a - and 5 with a +, you can discover the fake one exactly like how you could do so if you knew the fake was heavier/lighter and you had 9 coins. This is why this riddle should be solvable with 14 coins instead of 12, IF you have a "safe" real coin from the beginning.
I have a solution. 1) weigh 6 against 6 2) Take the lighter stack and weigh 3 against 3 3) take the lighter stack and weigh 1 against 1. If they balance, the remaining coin is the fake. If they are unbalanced, take the lighter coin.
I refused to watch the solution when this video came out and I have been thinking about this riddle for years. Today I finally figured it out. I can't express how satisfied I feel.
There’s a dark twist in any of the solutions to this puzzle… There will be one case out of all possible weighings where the scales have always balanced, and you have decided the last coin is counterfeit by elimination…but you haven’t actually weighed that last coin. If you encounter this particular case, you are sending the treasurer to his death without really knowing if that last coin is fake. This is exactly the case you would reach *every time* if this whole scheme is actually a set-up by the king to get rid of the treasurer…
This is not true. All three tries will only balance if there is no fake coin. This is because the solution not only allows you to determine the fake coin, but also determine whether it's heavier or lighter. If the first two weightings balance, you would have a single coin left over. You will then use the third attempt to compare this coin with any other coin to determine whether it's heavier or lighter.
With balanced results it only takes two weighings to identify the fake. The first balanced 4v4 eliminates 8 coins. Then second weighing is 3 good coins vs 3 of the remaining unknown. Since they balance then those three are now good so then you have left the singular unweighed coin as the only possibility. Problem solved. But since you have and extra turn, you could even weigh the bad vs one of the good coins to see if its lighter or heavier.
Im a bit late but in ye olde times there were also like ranks of coins I mean by that copper silver and gold it’s just like the normal times Ofcourse but still 100 gold coins there could mean 1000 silver coins and 10000 copper coins
@@therealveridicalyt497 considering there’s atleast 15 people in the kingdom (12 treasures, 2 guards, you), and the treasury having 12 coins, either coins are very valuable or the income tax is just THAT high
Nah. I have the simpliest solution. All you have to do is mark the coins randomly, pick up literally any coin, and say it's the fake. If the king asked you to do this calculation, that means he DOESN'T KNOW. So simply say whichever one is the counterfeit and get your tush out of jail. He'd never know because he will cease to never know because YOU are some genius and HE is not.
Either the king or one of his servants knows which one is fake, if they didn't then they wouldn't have been able to take the false coin out of the stockpile and put it into the group of 12 coins in the first place.
Maths : the scale won’t carry 9 coins so the counterfeit is an odd 1 out but you don’t know so you takes just hands test them all and the counterfeit coin will weigh different so boom the answer is simple : just weigh the coins and find the counterfeit
I truly admire ppl whose brains can comprehend things like this. Even watching the explanation, though, I resign myself to the fact that I would take the sweet release of death rather than trying to figure it out.
Well, I will tell you. I solved this puzzle a long time ago and the solution still went over my head (even though it was exactly the same). Even if you know the solution, just looking at an explanation doesn't do you any good. In order to fully understand it, you need to pause the video, map out what the commentator is saying and understand it in depth. Else, its just a sleeping pill.
The emperor is insane. If he gave me an unlimited number of tries, then i would be able to solve this problem faster than i would if he were to only give me three tries. But that’s just my 2 cents.
Technically, if one weighing turn is defined as "as long as you haven't finished stacking and haven't finished removing the ongoing stack", you can just add coins one by one on each side until one pair makes an unbalance. Once it does, you weigh one of the previous, real pairs' coin against one of the pair which was unbalanced. If they're equal, then the remaining coin of the unbalanced pair is the fake. If not, then you picked the fake one to test.
Same here, I think of case 1= 4,4 coin, if all balance then fake in left 4 can be find out in 2 more ways; case 2= lets consider 4,4 is not balanced, then again we left with 4 coin so found out in 2 more ways.
The strategy actually works for 13 coins as well, with the only difference being at 2:03, where you have 2 unknown coins left instead of 1, but there’s still one more weighing left which can determine which of the two is fake, by weighing one against a real coin.
With 13 coins, there's a series of weighings that will tell you which coin weighs different but depending on which coin it is you may not be able to tell if it is lighter or heavier-just that it does not weigh the same as the others. The algorithm of the 3 weighings that i'm familiar with is slightly different than what this video shows.@@kunalkashelani585
@@kunalkashelani585 I didn't check what the other person proposed. However, i want to add that the problem, if i'm not wrong, should be theoretically solvable for 13 coins too, by adjusting the procedure, with just 3 weighings still. That is because even with 13 coins, there's a total of 27 possible answers (coin 1 heavier, coin 2 heavier...coin 13 heavier, coin 1 lighter, coin 2 lighter...coin 13 lighter, all coins equal. 13+13+1 = 27) . Now, every time we use the scale we are able to discern between 3 possible scenarios based on if the scale is balanced or unbalanced to the left or right. If we were able to exclude one third of the possible answers every time we used the scale, by equally distributing them among the scale outcomes, we would first go from 27 to 9, then 9 to 3, and then 3 to 1.
@@kunalkashelani585 Yeah, i was thinking about it in bed tonight, and I think one would need an extra reference non-counterfeit coin, in addition to the other 13 to check
all this is nice, but in the end the kingdom is still being ruled by an avaricious dictator who over-taxes his subjects and sends people to jail just for criticizing him. you may have solved this puzzle, but the true puzzle still awaits its solution.
Alternatively, if you know the counterfeit is lighter, you could split the pile in half, then weigh them. Take the lighter side and then split it in half again, take this batches lighter side and put two of the coins on the scale. If they weigh the same, the third is counterfeit, if one is lighter, there's your counterfeit. This is a lot easier for me to remember than their solution, but obviously it requires you to know about this kingdom's counterfeit coin production.
had the same solution, but I think it's because we don't know if the counterfeit coin is heavier or lighter than the real coins which is why it became this complicated.
@@Zarrocification also apparently it takes less time to wait for the mathematician to get out of the dungeon, be given orders with several parameters, give him time to think of this whole strategy and also to wait for him to start stacking and re stacking and marking every single coin. Yes that should take less time than just weighing them right there.
Another elegant solution is to label each of the 12 coins one of the 24 three letter word, each letter being L,R or O (except for the LLL,RRR and OOO) words. Call two words conjugate if one can be the other after swapping L for R and vice versa. Hence LLR and RRL are conjugates. Now label each coin so that no two of them have conjugate words (since each of the 24 words have exactly 1 conjugate, this means that you can use exactly 12 such words which happens to be the number of coins!). So if a coin is labelled LRO then it will be on the Right, Left then Off the scale in the first, second and third weightings respectively. It's easy to see that on each weighting, there are exactly 4 coins on the right, left and off scale. Record the leaning side of the Ballance on each of the 3 weightings record O if it ballances (notice the recordings can't be one of RRR,LLL or OOO as how we labeled the coins). Suppose the recording of the leaning sides are ROL then if there is a coin labeled ROL then it is the Counterfeit coin and it is heavier than the rest. Otherwise there is a coin labeled LOR (a conjugate) this is the counterfeit coin and it is lighter than the rest.
Before watching the solution I think this is how it goes: 1. Put 6 coins on the other side of the scale, 6 on the other. Other stack will be slightly heavier, other lighter. 2. Now put the lighter stack aside and divide the heavier stack in 2, so it's 3 on the other side and 3 on the other. 3. Oops, just realized the fake coin can be in the lighter stack too, so this doesn't help you at all.
Paul Denino I will just find the two coins that are balanced (which will help me identify the real coin) then measure one coin to the other real coin it will took time but it wi be faster than planning it thoroughly
No, that’s not where you’re at fault. The counterfeit will be in the lighter stack. If you’re going to make a coin heavier with the same material then that just means that the value of the coin is more than it’s face value. So you put the heavier stack aside and split the lighter stack in half. Now you take the lighter one and out of them you take any two and put them in the scale. Now you’ve figured it out!
A much more elegant (and somewhat easy solution): 1) Divide pile into four groups with 3 coins each 2) take any two piles and weigh them: a) If the two piles weigh equal, this means the fake coin is in the remaining two piles b) else the fake coin is in one of the two piles you just weighed For case b) # mark the lighter pile "-" and heavier pile coin with "+". # weigh the "+" pile against any unused pile of 3 coins. # If these two are equal, then the fake coin is in the "-" pile of three coins else it's in the "+" pile # since we're now only left with a faulty pile of 3 coins, take one coin each from faulty pile and weigh both against each other # if they're equal, the remaining coin from faulty pile is the faulty one # else the heavier one with "+" sign is the fake. The same can be checked for other cases as well
so y'all gonna lock me up for no good reason, then call me to do your homework, and y'all really have the nerve to tell me you're impatient. Please lock me up forever
Even if we assume the master mathematician comes up with the problem instantly, it would still take longer to actually fetch the mathematician instead of having a servant brute force search the false coin.
It was very neat to get this and its variations in my college's Data Structures and Algorithms course. We got this version, one where there were two counterfeits with the same heavier weight, one where there were two counterfeits with differing heavier weights, and one where there was one counterfeit lighter and one counterfeit equally heavier.
Liz smith I did by making them into 3 groups and randomly weighing 2 of them. That way you will always know which group has the fake coin. Then it's just process of elimination by splitting the fake group in half until you get the fake coin. And it works because I only use the scale 3 times
You would have to use the scales twice to work out which group it's in, and then twice to work out which of the four it is, which is why that doesn't work.
alienzen If you're talking to me then no I only need to use it 3 times. 1st time: Weigh 2 groups of four coins. Then I will know which group of four contains the fake coin. 2nd time: Split the fake coin group in half so I can weight two coins on each side. Then take the lighter group of two. 3rd time: Weigh the two coins and the lighter one is the fake.
goodrapmusic I just told you that you cannot know which group it is in by only weighing once. You need to pay more attention. You have a 1 in 3 chance of knowing which group it is in, ie if the two you are weighing balance. 2 out of 3 times you will need to weigh a second time. Since you do not know whether the fake coin is lighter or heavier. This was clearly stated in the video. *pay attention*
I came up with a little diffrent solution, pls correct me if it doesnt work that way. So lets say, that the first 4 on each side are balanced. Than you can just take 2 of the remaining 4 and compare with 2 of the real coins: If its still balanced you know its one of the other 2 and you can just take 1 and compare to a regular coin and decide which 1 is odd. If it is unbalanced you do the same procedure with this 2. This way you dont need to mark them. Now if the 4 one each side are not balanced, you should mark them with a + and a -, depending if they are on the heavier or lighter side. Now you take away 2 possibly lighter coins and make group the remaining coins into to groups of 2+ and one -. If it is balanced you know 1 of the minus coins not on the scale are odd and you can just compare 1 from them with a normal 1 and see if this is the odd 1 or the other must be. If it is unbalanced you know the odd one is either the minus coin on the lighter side or one of the 2 plus coins one the heavier side. Now you can put the minus coin with one of the plus coins together and compare to 2 regular coins. if its equel the remaining plus coin is the odd one. if it is lighter then the minus coin is the odd one. and if it is heavier than the plus coin on the scale is the odd one. My english is not the best, but I hope y‘all understand my thoughts.
"But the emperor it's an impatient man" *Also the emperor when he realizes that this procedure takes 10 times the amount of time you would have used if only you were allowed to use the scale however tf you want* "-_-"
@@dillon8124 They wouldn't need to draw out or explain it though, that's only for the viewer to understand how it works. ETA: All they need to actually do is draw the symbols and weigh the coins until they identify the fake. They don't need to explain their reasoning as they go. It's faster this way than using the scale however you want.
Simpler solution if 4-4 stacks are balanced 1. From 3rd stack take 2 coins, if they balance, fake one is in the rest of 2 coins 2. Take one of them, balance it with one of those two which are already balanced 3. If they balance, the last one is fake one, if don't, the one you took is fake And vice versa if they are unbalanced in 1st step
The first balnce solution is really easier. But the problem with the first imblanace is, that you dont know which of the 4 coins stack has the fake coin. There are 8 possible fake coins, if the first ist imbalance, while there are only 4 if it was balance. So the "And vice versa if they are unbalanced in 1st step" woulndt work.
My solution was acctualy a bit different I will try to explain: 1st weightning is the same and we use same logic scales are balanced If scale is not balanced we can use same notation as used in the video and mark our coins + + + + and ~ ~ ~ ~ 2nd weightning we put + + ~ on the one side and + + ~ on the other side If we get equal weights we just weight one of the ~ coins with 0 coin and we can easily deduce answer If one side is heavier we can now mark heavier coins ++ ++ ~+ and lighter coins +~ +~ ~~ coins +~ and ~+ are safe as counterfit cant be both heavier and lighter than the rest so we are left with coins ++, ++, ~~ now we can weight ++ ~~ with 0 0 If the scale balances the ++ coin we did not weight is counterfit If one side is heavier we end up with coins +++ and ~~+ or with coins ++~ and ~~~ either way the one with all signs the same must be the counterfit :) If anyone can find a misstake in my solution please tell me.
*I solved it without ever using the balance scale.* Since the Emperor has no idea which one the counterfeit coin is, you can just pick any coin and pretend it's counterfeit.
Kalyani Kataki plot twist you never slept you have insomnia, you thought your mom was the king and you stole her jewlery to do the math. You're now grounded for halucinating.
Usually this riddle doesn't just ask you to find the fake coin, but also to determine whether it's lighter or heavier. 2:03 You still have one more weighing left after this case, and you should use that weighing to compare it to any of the eleven genuine coins. In every other case you've already marked the coin with a - or a +, so you also know.
I actually found a different, functional solution to this riddle! This only works if you know whether the counterfeit is heavier or lighter so take this with a grain of salt. The example has the counterfeit coin being heavier: Make two piles of six and put them on the scale. Whichever side is heavier, take the coins from the other side and put them away, you don't need them after this. Then make the pile that is still on the scale and divide it into two, and put it on the scale. Whichever side is heavier, keep it. The others are cast aside. The side that is heavier has one taken from it and marked with a plus. The. The two left on the scale are weighed. If the scale is a balance, it's the coin marked with a plus. If the scale is imbalanced, the counterfeit is the heavier coin. May not work perfectly if you don't know whether the counterfeit is lighter or not but that's my take on it. Have a good day!
15% of the comments: Jokes 80% of the comments: People who think they got it but didnt think that they dont know if the fake coin is heavy or light 5% is the rest
One easier algorithm is also there, divide in 3 groups each of 4 coins. Case1 :- Then lets say weighing G1 with G2 balances, the we know all 8 coins are pure, remove all of them from balance now take G3, divide them in half, which side weighs heavier means they are pure remove them from balance and now you left with 2 coins and 1 trail left. Case2:- Now lets say the G1 and G2 don't balances then remove the heavier side coins and divide the remaining 4 coins then weight again and remove the heavier side then same left with 2 coins and 1 trail. Congratulations🎉 you solved the problem without marker this time ❤❤
Me in math class: Sir, uhm how come there is a marker in the story? clearly the setting is from the age wherein markers are not invented yet, or is it post apocalyptic or something?
@@healy_ly you can even do it in reverse order by simply placing one coin on each side at a time (you can say that you want to place them all but they are precious so you place them one by one) and once they become unbalanced stop and weigh the last pair with any other coin and...
Can't you weight 6 and 6, then divide the lighter 6 into 3 and 3, weight those, and then from the lighter 3 weight 1 and 1, leaving one out, so either you see one of the coins on the scale being fake, or the one you put aside
I have an easy method. Stack them into 3 stacks of 4. Weigh 2 of them. (Balance 1) If they're unbalanced, the lighter stack holds the coin. Divide the lighter stack into 2 stacks of two and weigh. (Balance 2) And repeat for the final balance. If the stacks from the first try balance, then the final stack holds the fake. Use the second and third balances the same.
Yes, but the reason for the three use limit is the fact that the king is impatient. If he's impatient then he should want it done in the least amount of time rather than the fewest scale uses. He doesn't even need the mathematician, since he could have anybody weigh each coin against the others in less time than it would take to summon the mathematician from the dungeon.
The point is, that if there were thousand coins (as in real treasury), weighting each individually would take much more time (over an hour) than doing prety much the same thing as here in which case you'd need to use the weights 7 time if you were very clever. So say, 15 times would suffice without doing it in such a complicated way, and figuring out such relaxed solution takes like 5 minutes. Also, if you were looking for the false coin several times, it would be profitable to optimize the process. Welcome to algoritmization and computer science, that's what it's all about.
AJ Tomecek but if there were 1000 coins, it would take too much to make piles, and probably no space, a the marker would run out and you would be just stuck
I thought a different thing but I don't know if this is cheating, gimme your opinions, but now, let me explain: There are 2 ways of thinking this trough so I'll divide it. Situation 1 Step 1) Put 2 coins on the scale, if they're even, mark both with a "0". Step 2) Complete both sides with the remaining coins (One coin with the 0 + 5 others vs the other coin with 0 + remaining 5). Step 3) One of the sides wil be distorted, so I'd take one coin from each side at the same time, and when the scale got even I'd know one of the two coins I took was the problematic. Step 4) Mark one of this coins with an X and the other with an Y. Step 5) Put one of the "0 coins" on the scale and compare with "X coin", if it's even, Y is the false, if it's not, then X is the counterfeit coin. Situation 2 Step 1) Put 2 coins on the scale, if they're not even, mark one with A and the other with B. Step 2) Put A on the scale with an unmarked coin, if it's even, B is the weird coin, if it's not, then A is the counterfeit coin. P.S.: I'm not sure if it's cheating because I don't think taking the coins out would be considered using the scale, but like I said I'm not sure :P
i first heard of this puzzle back in 2010 and thought it was impossible later during my UG me with my friends look up for the solution and were in awe...To this day it was the hardest puzzle i've ever encountered.
KEVIN cheng because the lord HAD BEEN paying his taxes with fake coins, meaning1/12 oh the kings revenue is phony and useless until the perpetrator is rooted out.
I think mine would've worked too. Weigh 2 equal amounts of coins. Mark the coins in the coin pile that weighs less or more. Now, weigh the marked ones in 2 equal groups. Mark the coins in the coin pile that weighs less or more (again). Now, weigh 1 marked twice coin and 1 other marked twice coin separately. If their weights are both equal, take the leftover coin not on the scale. Otherwise, take the coin that weighs less or more on the scale.
When you measure two piles of 6 , one will be heavier and other one will be lighter . it's not an either or situation . If one thing is heavier the other is lighter .
70% of the comments are geniuses with "aww this is easy, just divide the coins into 2 piles of 6s...." and another 20% even more geniuses that even after watching the soution they're like "there's even more simple way, just divide the coins into two piles...."
The funny thing is that it would take less time to weigh all of the coins in equal piles until you found the one than it would if you marked all of them as you went
The king offered a reward and The Swordsman Apprentice ratted him out, it was the sword maker all along and they found the little Tool & Die coin molds used for making them :-) so quit asking questions or you'll be next!
Madisøn the Tøp! Crybaby he showed us him showing the emporer his work so i guess you need to work it out and not all fat emperors are stupid you know :)
So the King is impatient when it comes to how many times I have to weigh the coins, but not the the million hours I'll be having to figure how to do it with only three trails!
the trick is, in this situation, it wouldn't take that long. since you are the best mathematician in all the land, it would be very quick. it's still kind of flawed but it's just for the sake of the math.
A much more elegant (and somewhat easy solution): 1) Divide pile into four groups with 3 coins each 2) take any two piles and weigh them: a) If the two piles weigh equal, this means the fake coin is in the remaining two piles b) else the fake coin is in one of the two piles you just weighed For case b) # mark the lighter pile "-" and heavier pile coin with "+". # weigh the "+" pile against any unused pile of 3 coins. # If these two are equal, then the fake coin is in the "-" pile of three coins else it's in the "+" pile # since we're now only left with a faulty pile of 3 coins, take one coin each from faulty pile and weigh both against each other # if they're equal, the remaining coin from faulty pile is the faulty one # else the heavier one with "+" sign is the fake.
I did it in a different way, but using some of the same underlying principles: I divided the coins in 3 sets of 4. I measure 2 of those 3 sets against each other. If they have the same weight, then I conclude that the false coin is among the last set that I didn't weigh. Then I take 2 of these 4 potential coins that I weigh against 2 verified coins. If there is a difference of weight, then it means that the counterfeit is among the 2 potential ones I just weighed. Then I take 1 of these 2 that I weigh against a verified one, and if there is still a difference, then it's that one potential coin. If there is not, it's the other potential one. If there is no difference of weight, then it means that the counterfeit is among the 2 potential ones I didn't weigh yet. I do the same process as above and find the counterfeit coin in 1 measure. Now let's imagine that the 2 sets of 4 coins that I originally weighed didn't have the same weight. Then, I write a "+" on all the coins that are on the heavier platter, and a "-" on all the coins of the lighter platter. Then, I take 2 "+" coins and 1 "-" coins that I put aside. On the platters are left, on one side, 3 "-" coins and on the other 2 "+" coins. Like this: - - - vs. + + I add 1 verified coin ("0") to the platter with the "+" coins. Like this: - - - vs. + + 0 Then I switch 1 "-" coin and 1 "+" coin between the platters. Like this: - + - vs. + - 0 And I proceed to measure. (This is my 2nd measure.) 3 things can happen from there: 1) The two platters weigh the same: this means that the counterfeit coin is among the 3 coins (2 "+" and 1 "-") that I put aside after the first measure. I just have to weigh the 2 "+" coins against each other. If they weigh equally, then it's the left "-" coin. If they don't, then it's the heavier coin. 2) The two platters don't weigh the same, and the balance changed (the heavier platter became the lighter one and vice versa): this necessarily implies that the counterfeit coin is still on the balance, but it changed side. Since I switched 2 coins, it necessarily means that the counterfeit coin is one of them. I just have to take 1 and measure it against a verified coin: if there is a difference in weigh, then it's the 1 switched coin I just measured. If there is no difference, then it"s the other one. 3) The two platters don't weigh the same, and the balance is unchanged: this necessarily implies that the counterfeit coin is still on the balance, among the coins that didn't change side and are not verified (which are 2 "-" and 1 "+"). I just have to take the 2 "-" coins and weigh them against each other. If there is no difference in weight, then it's the "+" coin. If there is a difference, then it's the lighter one. It might seem more complicated but I like this one (because I found it lol). It took me a couple hours though. Frustrating but I'm happy to have done it!
My solution is almost identical, except if there is an imbalance at the first weighing, I put aside + - and my second weighing is + + - - vs. + - 0 0. If left is heavier than right, I know it's one of + + -, if right is heavier than left I know it's one of - - +, and if it's balanced it's one of the + - I put aside. The third weighing then works exacly like you describe it. What I like in my solution is that it has some "symmetry" in the second weighing, by that I mean if you swap + and - it's still the exact same thing.
or you could split them into 2 stacks with 6 coins each. the heavier or lighter side contains the imposter coin. (assuming it shows up heavier, meaning the imposter coin is heavier -- this will work if it appears lighter as well) do this again with the side with the imposter coin weighing 3 coins against the other 3. again, the heavier side contains the imposter coin. now you are left with 3 coins and you can just weigh any random 2. the trend of the heavier side being the imposter coin should continue, so if the 2 coins you picked contains the imposter, it will show up on the heavier side of the scale. if both are balanced, the last unweighed coin is the imposter coin.
![[MV] A leap of faith...ความเชื่อใจครั้งสุดท้าย (From "Petrichor the series")](http://i.ytimg.com/vi/U1piZH2CNXA/mqdefault.jpg)
the emperor is an impatient man.
*literally lets you think for eternity*
@Julious Lucas *doubt*
If this whole scenario isn't a huge red flag to how the kingdom is being run, you probably need to stay in the dungeon.
You can just ZA WARUDO apparently but you don’t abuse it
*wowza*
0MemeMan0 r/unexpectedjojo
My Solution:
1) Think for 10 minutes and tell the King you're ready.
2) GO mental with the marker and mark all the coins after randomly weighing them.
3) Pick one coin and tell the King that's the counterfeit.
King doesn't know either, so you have survived.
Cheers.
look at rule 6 at 1:11
Idan Raviv ignoring rule 6 would work
Idan Raviv didn't say u had to guess right
XD Perfect
but the counterfeiter would still counterfeit ,and the king would realize that you made a mistake then throw you in the dungeon (If you don't run away)
0:52 But the Emperor's not a patient man
**Weigh pieces of money one by one in 2 minutes**
Emperor : Too slow!
**Think in 2 hours of a possibility to only use balance 3 times**
Emperor : Fast enough
If ask me I came up with an answer a lot faster and I am not even bright
So other prolly would come up with an answer faster
*Methmatican
I just split the 13 in half twice taking the lighter halves. So I had 12 and turned it into 6, I had 6 and turned it into three. I weighed two and and if equal then the fake is the one that I didn’t weigh but if one was lighter than the lighter one was the fake. Took me like 1 minute to solve
sock man is fast
@@Idiotc 12...? you mean 12 right not 13 because then you would have been splitting a coin in half
@@Idiotc yeap, the marker is useless :P
The hardest aspect of this and what people are missing is that the counterfeit is either heavier OR lighter, so you need to figure out that detail. You can’t assume the counterfeit is lighter/heavier from the beginning.
Of cause but that isn't the lesson of the video if you get what I mean.
Exactly
no that's the easiest part. The hard part is solving the problem knowing that there's an extra piece of info you need to find out.
What's fascinating is that by marking the coins with + or - marks, you effectively negate that issue. Having 9 coins, 4 marked with a - and 5 with a +, you can discover the fake one exactly like how you could do so if you knew the fake was heavier/lighter and you had 9 coins. This is why this riddle should be solvable with 14 coins instead of 12, IF you have a "safe" real coin from the beginning.
@@alonkob2127 But the 5 coins will always be heavier. Because there's more of them.
“Youre the realm’s greatest mathematician.”
Ok, youve already lost me.
Mohammad King you only have 3 shots.
@@Muhammed_English314 you only get 3 shots tho
I have a solution.
1) weigh 6 against 6
2) Take the lighter stack and weigh 3 against 3
3) take the lighter stack and weigh 1 against 1. If they balance, the remaining coin is the fake. If they are unbalanced, take the lighter coin.
@@themarshy83 it also can be heavier.. you don't know that
@@nadian848 lmaoo that’s what I was thinking too
I'M the realm's greatest mathmetician?
Well, that's the first problem.
edit: woah 8k ya'll? Thanks!
Get me four thrippets of wine!
Hahaha
LOL
Lol
Relatable
I refused to watch the solution when this video came out and I have been thinking about this riddle for years.
Today I finally figured it out. I can't express how satisfied I feel.
That’s insane!
For a moment there I thought u were lying until I saw the release date of the video and not your post
congrats!
Nice!
The king had to wait years lol
There’s a dark twist in any of the solutions to this puzzle…
There will be one case out of all possible weighings where the scales have always balanced, and you have decided the last coin is counterfeit by elimination…but you haven’t actually weighed that last coin.
If you encounter this particular case, you are sending the treasurer to his death without really knowing if that last coin is fake. This is exactly the case you would reach *every time* if this whole scheme is actually a set-up by the king to get rid of the treasurer…
Except, if there is no counterfeit, you would have singled out one coin in two steps of weighing and have one left to check it.
Third dark twist: your marker was permanent, you are 😵
Lol
This is not true. All three tries will only balance if there is no fake coin. This is because the solution not only allows you to determine the fake coin, but also determine whether it's heavier or lighter.
If the first two weightings balance, you would have a single coin left over. You will then use the third attempt to compare this coin with any other coin to determine whether it's heavier or lighter.
With balanced results it only takes two weighings to identify the fake. The first balanced 4v4 eliminates 8 coins. Then second weighing is 3 good coins vs 3 of the remaining unknown. Since they balance then those three are now good so then you have left the singular unweighed coin as the only possibility. Problem solved. But since you have and extra turn, you could even weigh the bad vs one of the good coins to see if its lighter or heavier.
“The coins were already in the treasury”
Yet, you are presented with only 12.
This king is sure rich
Maybe they saw about where the coin ended up but not exactly where it was?
*he is totally so rich*
@@beepbop6542 that is true, not to mention what rich meant in older times
Im a bit late but in ye olde times there were also like ranks of coins I mean by that copper silver and gold it’s just like the normal times Ofcourse but still 100 gold coins there could mean 1000 silver coins and 10000 copper coins
so many died to lower the number to 12, each one with only 3 tries to find it. i salute them.
And then the King realized that the marker was permanent and... threw you back to jail...
my thought exactly lol
ikr
LOL DED
Anonymous71475 Good point, he doesn't want ink on his coins, he would most likely force you back into the dungeon or have you decapitated.
lolololo
“You’re the realm’s greatest mathematician.”
Guys, I think we have a bigger problem here than a counterfeit coin.
😂
This might’ve been stolen
Very similar to another comment but I’ll give you the benefit of the doubt
@@ContentDeleted10332 it doesn't have to be stolen this comment is a classic for Ted ed videos.
Well clearly all the other mathematicians didn't know what's addition and subtraction
Also, this is not a mathematical problem. It's a logical conundrum.
Who else never gets the answer/doesn’t try, but loves the answer because it’s so fascinating?
ME-
ME2-
hai hai:3
Me
Meh
“6) There's no bribing the guards or any other trick.”
they know our ways.
Bribe the king
@@therealveridicalyt497 considering there’s atleast 15 people in the kingdom (12 treasures, 2 guards, you), and the treasury having 12 coins, either coins are very valuable or the income tax is just THAT high
Buy a helicopter and scape
The romanian teacher ways dont work here
Nah. I have the simpliest solution. All you have to do is mark the coins randomly, pick up literally any coin, and say it's the fake. If the king asked you to do this calculation, that means he DOESN'T KNOW. So simply say whichever one is the counterfeit and get your tush out of jail. He'd never know because he will cease to never know because YOU are some genius and HE is not.
Ravioli ravioli, give me the formuoli 😂😂😂😂😂
It's how scammer works lol
what if he ask for explaination ?
The Lord has been using fake coins for awhile so he likely wanted the solution to find all the counterfeit coins.
Either the king or one of his servants knows which one is fake, if they didn't then they wouldn't have been able to take the false coin out of the stockpile and put it into the group of 12 coins in the first place.
"What are you doing drawing on the coins with a marker?! I said I was an impatient man! GET BACK IN JAIL"
This is a logic puzzle as well. Really TED?
Why are you sitting there confused?!!!!! Back in jail!
Lol
Hi dude i watch your vids! They’re awesome
Maths : the scale won’t carry 9 coins so the counterfeit is an odd 1 out but you don’t know so you takes just hands test them all and the counterfeit coin will weigh different so boom the answer is simple : just weigh the coins and find the counterfeit
I truly admire ppl whose brains can comprehend things like this. Even watching the explanation, though, I resign myself to the fact that I would take the sweet release of death rather than trying to figure it out.
if we are being honest here, all of these solutions are non practical. they are all made out scenarios to fit the puzzle.
@@gameskyjumper1721 its more a game for us, and actually problem solving training which is fun.
May we share ?
Well, I will tell you. I solved this puzzle a long time ago and the solution still went over my head (even though it was exactly the same). Even if you know the solution, just looking at an explanation doesn't do you any good. In order to fully understand it, you need to pause the video, map out what the commentator is saying and understand it in depth. Else, its just a sleeping pill.
I solved it, but when i heard the answer, even if it's same, it sounded much more complex.
I'm pretty much only subscribed for these riddle videos.
KarlMalowned32 Like me you can just tap the bell on the playlist.
KarlMalowned32 I thought I was the only one. I feel bad though, I think I should watch all of their videos for added knowledge.
Same haha
KarlMalowned32 Exactly
KarlMalowned32 yes, we are all in the same boat here.
The emperor is insane. If he gave me an unlimited number of tries, then i would be able to solve this problem faster than i would if he were to only give me three tries.
But that’s just my 2 cents.
the point is that u must do it very quickly with only 3 tries, if u require more time than it takes to do 3 quick balance checks then off with ur head
Actually your 11 cents and a counterfeit coin.
oh yeah yeah
TWrecks 2 cents...ba dum tss
The scale breaks or becomes unreliable as use increases.
If no one can tell the fake...
Me: *takes random coin after three random weighs*
Also me: THIS IS THE FAKE!
lol
Hahshaaa
There's a clever answer.
That could work!
It could work
Technically, if one weighing turn is defined as "as long as you haven't finished stacking and haven't finished removing the ongoing stack", you can just add coins one by one on each side until one pair makes an unbalance.
Once it does, you weigh one of the previous, real pairs' coin against one of the pair which was unbalanced. If they're equal, then the remaining coin of the unbalanced pair is the fake. If not, then you picked the fake one to test.
Here's an easier solution. The King can't tell which one is fake, so, pick one at random and tell him that's the fake one. He can't prove you wrong!
Oh, and what country is this? I think I'm paying higher taxes.
A mind even greater than of a mathematician.
Hats off to you!
But when he has somebody else weigh all 12 coins individually...
Sir Francis if he were able to get someone to do that wouldn't he just do it in the first place?
Let's say it takes too long to weigh things on the scale and that the maths guy is thinking on the way there, so he could take less time.
The king is secretly also a genius. He gave him exactly the amount of measurements needed.
Nah, he's just a tyrant who wanted his marker back, and tried to set the mathematician up to fail.
"I've figured it out!"
...Watches solution...
"I didn't figure it out."
Me with all of these videos, pretty much.
Same here, I think of case 1= 4,4 coin, if all balance then fake in left 4 can be find out in 2 more ways; case 2= lets consider 4,4 is not balanced, then again we left with 4 coin so found out in 2 more ways.
@@SujeetKumar-cd4cc we arent left with 4 coins we are left with 8 since they can be heavier or lighter
@@shalevasor911 oh yeah, got it now. thanks
Me too
Me i just watch them to see how to figure them out i have never solved one the right way NOT ONCE
I'll always have a respect for this video since it was the very first TedEd video I ever found. Back in 2017.
the king then says "tis a witch!" and sends you to the gallows
Lmao :') brilliant
Haha, that has a certain probability as well. :D
He'll need larger scales to prove that, though.
Aidan Collins iiiii
laugh my fucking ass off
Answer:
Step 1) verify that you have green eyes
Step 2) The king will give you permission to leave
What
lil Chico the green eyes riddle
This joke never gets old
Aniket Kangutkar a meme
THIS JOKE IS SO OLD
Why does the character who is supposed to be you look like a sock?
You are a sock puppet
Bold of you to assume that I don't look like a sock.
He looks more like a muppet with the beard if he is drawn
Clearly the best reply.
lolololololol
The strategy actually works for 13 coins as well, with the only difference being at 2:03, where you have 2 unknown coins left instead of 1, but there’s still one more weighing left which can determine which of the two is fake, by weighing one against a real coin.
No you can't. If you have two unmarked coins and only one chance to go, there's no way you can identify the coin and mark it heavier or lighter.
With 13 coins, there's a series of weighings that will tell you which coin weighs different but depending on which coin it is you may not be able to tell if it is lighter or heavier-just that it does not weigh the same as the others. The algorithm of the 3 weighings that i'm familiar with is slightly different than what this video shows.@@kunalkashelani585
@@kunalkashelani585 I didn't check what the other person proposed. However, i want to add that the problem, if i'm not wrong, should be theoretically solvable for 13 coins too, by adjusting the procedure, with just 3 weighings still.
That is because even with 13 coins, there's a total of 27 possible answers (coin 1 heavier, coin 2 heavier...coin 13 heavier, coin 1 lighter, coin 2 lighter...coin 13 lighter, all coins equal. 13+13+1 = 27) .
Now, every time we use the scale we are able to discern between 3 possible scenarios based on if the scale is balanced or unbalanced to the left or right.
If we were able to exclude one third of the possible answers every time we used the scale, by equally distributing them among the scale outcomes, we would first go from 27 to 9, then 9 to 3, and then 3 to 1.
@@naiko1744 I get your point, but mathematically it is not possible to identify the faulty coin, if the number of coins are 13.
@@kunalkashelani585 Yeah, i was thinking about it in bed tonight, and I think one would need an extra reference non-counterfeit coin, in addition to the other 13 to check
Of course the emperor lets you draw all over the coins
Damn. He would let us draw on his damn golden coins. Wow he sure is an patient man. But he isnt when someone uses counterfit coins to pay his taxes
•Mɪʟᴏ ᴀɴᴅ Vɪxɪᴇ • Or when you criticize his tax laws.
and change the weight of the coin with extra ink on the coin which make it impossible to guess correctly
wizardystaff Ink isn’t very heavy.
And yet you can only use the scale three times :P
The answer: The king gets mad because you draw all over his coins so even if you're right he'll throw you in the dungeon anyway
VirtuallyGlace c;
*EASY
*A-ALI INTO PLAYS*
Holly does...
well if it’s not permanent you might be able to erase it
And as punishment the king order you to eat all of the coins
all this is nice, but in the end the kingdom is still being ruled by an avaricious dictator who over-taxes his subjects and sends people to jail just for criticizing him.
you may have solved this puzzle, but the true puzzle still awaits its solution.
Isaac Mendez thats just society
Isaac Mendez establish a communist society
Deep bro
Isaac Mendez establish a democratic captalist society based on personal freedoms.
The final solution?
Alternatively, if you know the counterfeit is lighter, you could split the pile in half, then weigh them. Take the lighter side and then split it in half again, take this batches lighter side and put two of the coins on the scale. If they weigh the same, the third is counterfeit, if one is lighter, there's your counterfeit. This is a lot easier for me to remember than their solution, but obviously it requires you to know about this kingdom's counterfeit coin production.
This is the solution I came up with, too. Weird that they went the complicated route.
had the same solution, but I think it's because we don't know if the counterfeit coin is heavier or lighter than the real coins which is why it became this complicated.
@@seanty1317 nono they stated its lighter
@@ZurilasZone 1:12 Wrong, it says the counterfeit coin "is very slightly heavier or lighter"
@@bucketofparts because you don't know if it's heavier or lighter, this solution doesn't work
When the emperor realizes that it took me longer to figure it out then to balance the scale 11 times
True lol
If they already have a scale and the 12 coins .. why did they need a mathematician?
because the king is not a patient man :D
and apparently bringing you up from the prison took less time when simply weighing all of the coins
The mathematician realised the scale could be used to detect a counterfeit
@@Zarrocification also apparently it takes less time to wait for the mathematician to get out of the dungeon, be given orders with several parameters, give him time to think of this whole strategy and also to wait for him to start stacking and re stacking and marking every single coin.
Yes that should take less time than just weighing them right there.
Why is internet full of people who can't get sarcasm 😂 this guy above me
The concept of scales wasn’t known to commoners or kings and knights.
Me, an intellectual: "So this is how you solve Captain Holt's island riddle..."
ai , a man of quality
He couldn’t figure out the how the Monty Hall problem worked, of course he couldn’t solve this one.
Monty Hall problem of Captain Holt solved
Ooh
wait...
Another elegant solution is to label each of the 12 coins one of the 24 three letter word, each letter being L,R or O (except for the LLL,RRR and OOO) words. Call two words conjugate if one can be the other after swapping L for R and vice versa. Hence LLR and RRL are conjugates. Now label each coin so that no two of them have conjugate words (since each of the 24 words have exactly 1 conjugate, this means that you can use exactly 12 such words which happens to be the number of coins!). So if a coin is labelled LRO then it will be on the Right, Left then Off the scale in the first, second and third weightings respectively. It's easy to see that on each weighting, there are exactly 4 coins on the right, left and off scale. Record the leaning side of the Ballance on each of the 3 weightings record O if it ballances (notice the recordings can't be one of RRR,LLL or OOO as how we labeled the coins). Suppose the recording of the leaning sides are ROL then if there is a coin labeled ROL then it is the Counterfeit coin and it is heavier than the rest. Otherwise there is a coin labeled LOR (a conjugate) this is the counterfeit coin and it is lighter than the rest.
Great explanation. Thank you
That’s actually a clean and slick solution.
This does not work for this riddle, because the counterfeit coin can be either lighter OR heavier
Before watching the solution I think this is how it goes:
1. Put 6 coins on the other side of the scale, 6 on the other. Other stack will be slightly heavier, other lighter.
2. Now put the lighter stack aside and divide the heavier stack in 2, so it's 3 on the other side and 3 on the other.
3. Oops, just realized the fake coin can be in the lighter stack too, so this doesn't help you at all.
Paul Denino I will just find the two coins that are balanced (which will help me identify the real coin) then measure one coin to the other real coin it will took time but it wi be faster than planning it thoroughly
Atrid Leiyell this method doesn’t work since we don’t know whether the counterfeit is heavier or lighter than the rest.
I had that solution but it needs an extra step in it resulting in four or five weighs...
First I thought that too, then I relised that it can be lighter or heavier, so you still don't know
No, that’s not where you’re at fault. The counterfeit will be in the lighter stack. If you’re going to make a coin heavier with the same material then that just means that the value of the coin is more than it’s face value. So you put the heavier stack aside and split the lighter stack in half. Now you take the lighter one and out of them you take any two and put them in the scale. Now you’ve figured it out!
So apparently the emperor doesn't have time for you to measure all the coins, but does have time for you to think about how to do it in 3 weighings
And explain it on a white board with illustrations
The real question is:
How did they know he was the one who used the counterfeit
Xavier Phillips exactly lmao
Well your in jail, not knowing what's going on outside.
@FIRETEAM 2121 I know what you did there.
The king check all the coins before hand but then the guy just grabbed it and threw it in with the other coins s9 he could possibly keep going
Whistleblowers.
A much more elegant (and somewhat easy solution):
1) Divide pile into four groups with 3 coins each
2) take any two piles and weigh them:
a) If the two piles weigh equal, this means the fake coin is in the remaining two piles
b) else the fake coin is in one of the two piles you just weighed
For case b)
# mark the lighter pile "-" and heavier pile coin with "+".
# weigh the "+" pile against any unused pile of 3 coins.
# If these two are equal, then the fake coin is in the "-" pile of three coins else it's in the "+" pile
# since we're now only left with a faulty pile of 3 coins, take one coin each from faulty pile and weigh both against each other
# if they're equal, the remaining coin from faulty pile is the faulty one
# else the heavier one with "+" sign is the fake.
The same can be checked for other cases as well
But what if the coin is heavier
@@HeckYouEpicallyI have checked and acc to me many cases in this solution take 4 attempts, not 3
@@ExplainDigital Ok
It will work if it's case b
This is the solution I came up with when I watched it too.
here i thought i was smart since i instantly got a solution. but i didnt take in consideration that the fake could be either heavier or ligther
DeathlyDiJ same here
DeathlyDiJ sameeee
DeathlyDiJ same
Same here..
Same :(
so y'all gonna lock me up for no good reason, then call me to do your homework, and y'all really have the nerve to tell me you're impatient. Please lock me up forever
😂😂😂
😂😂😂
✌️😌 So true
In front of him, nobody will say this 😂😂
@@CNV71 Tbh I would say everything they said in Bonor's comment like- Don't call me to do your darn HW 😤🔪
If he's so impatient wouldn't he know that just weighing each one separately would be faster than all this fancy math shit
Emperors are always portrayed to be little shits.
Airam Roque do you have any proof that says otherwise?
Well, you're the master mathematician. Doing this would take around... 5 seconds?
Yeah
Even if we assume the master mathematician comes up with the problem instantly, it would still take longer to actually fetch the mathematician instead of having a servant brute force search the false coin.
It was very neat to get this and its variations in my college's Data Structures and Algorithms course. We got this version, one where there were two counterfeits with the same heavier weight, one where there were two counterfeits with differing heavier weights, and one where there was one counterfeit lighter and one counterfeit equally heavier.
figuring all that out would take more time than just dingo all the weighting in rapid succession. Impatient my ass.
Lucho-Core tru dat 😂😂
Was just about to say that.
To be fair, most of this would be done in his head, in rapid succession.
IKR?
Lucho-Core true
Ami the only one who figured it out but in a slightly different way...
Liz smith I did by making them into 3 groups and randomly weighing 2 of them. That way you will always know which group has the fake coin. Then it's just process of elimination by splitting the fake group in half until you get the fake coin. And it works because I only use the scale 3 times
You would have to use the scales twice to work out which group it's in, and then twice to work out which of the four it is, which is why that doesn't work.
alienzen If you're talking to me then no I only need to use it 3 times.
1st time: Weigh 2 groups of four coins. Then I will know which group of four contains the fake coin.
2nd time: Split the fake coin group in half so I can weight two coins on each side. Then take the lighter group of two.
3rd time: Weigh the two coins and the lighter one is the fake.
goodrapmusic
I just told you that you cannot know which group it is in by only weighing once. You need to pay more attention. You have a 1 in 3 chance of knowing which group it is in, ie if the two you are weighing balance. 2 out of 3 times you will need to weigh a second time. Since you do not know whether the fake coin is lighter or heavier. This was clearly stated in the video. *pay attention*
how will you know which group has the fake coin? the fake coin can weigh heavier or lighter.
"there is no bribing the guards" with what? the possible counterfeit money in front of you? of course not!
Especially when the king is in front of you
"Hey John, you want this coin? All you have to do is let me go, it's only a 1 in 16 chance to be fake. The king? Nah he can't hear us we're invisible"
@@susamogus8331 lol
I came up with a little diffrent solution, pls correct me if it doesnt work that way.
So lets say, that the first 4 on each side are balanced. Than you can just take 2 of the remaining 4 and compare with 2 of the real coins:
If its still balanced you know its one of the other 2 and you can just take 1 and compare to a regular coin and decide which 1 is odd.
If it is unbalanced you do the same procedure with this 2. This way you dont need to mark them.
Now if the 4 one each side are not balanced, you should mark them with a + and a -, depending if they are on the heavier or lighter side.
Now you take away 2 possibly lighter coins and make group the remaining coins into to groups of 2+ and one -.
If it is balanced you know 1 of the minus coins not on the scale are odd and you can just compare 1 from them with a normal 1 and see if this is the odd 1 or the other must be.
If it is unbalanced you know the odd one is either the minus coin on the lighter side or one of the 2 plus coins one the heavier side.
Now you can put the minus coin with one of the plus coins together and compare to 2 regular coins.
if its equel the remaining plus coin is the odd one.
if it is lighter then the minus coin is the odd one.
and if it is heavier than the plus coin on the scale is the odd one.
My english is not the best, but I hope y‘all understand my thoughts.
OH MY GOD YOU‘RE A GENIUS 😱😱😱 can i have your number papi chulo 🥵
"But the emperor it's an impatient man"
*Also the emperor when he realizes that this procedure takes 10 times the amount of time you would have used if only you were allowed to use the scale however tf you want* "-_-"
It really wouldn't though
@@DespOIcito it would definitely take less time to balance each coin than draw out and explain 4:09
@@dillon8124 They wouldn't need to draw out or explain it though, that's only for the viewer to understand how it works.
ETA: All they need to actually do is draw the symbols and weigh the coins until they identify the fake. They don't need to explain their reasoning as they go. It's faster this way than using the scale however you want.
@@DespOIcito but they would have to think of the solution. And if my life was at risk, I'd be double and triple checking my solution before starting.
@@lisahenry20 which would barely take any time because you're an expert mathematician...
Me:*calls a random coin counterfeit since the king wouldn't question the best mathematician's choice*
I mean, you're not wrong...
Well ozo, but actually ulu
What if he asks to prove?
Say that 1+1=2
@@arthurdabest8569 1+1=11
I didn’t know Thanos was so good at math
ROFLMFAOOO
@@junodisarapong6635 what even is the full form of that
In the comics actually, Thanos was an exemplary student. He aced every single test and he had a thirst for knowledge
Mr. Anonymous huh interesting
qweryt or looked like that before the gauntlet.
I loved this one! My approach didn't really work, but the solution makes perfect sense! 😁👍🏻
Simpler solution if 4-4 stacks are balanced
1. From 3rd stack take 2 coins, if they balance, fake one is in the rest of 2 coins
2. Take one of them, balance it with one of those two which are already balanced
3. If they balance, the last one is fake one, if don't, the one you took is fake
And vice versa if they are unbalanced in 1st step
The first balnce solution is really easier.
But the problem with the first imblanace is, that you dont know which of the 4 coins stack has the fake coin. There are 8 possible fake coins, if the first ist imbalance, while there are only 4 if it was balance. So the "And vice versa if they are unbalanced in 1st step" woulndt work.
So the Lord aint got time to wait for twelve measures but he does to wait for the drawing of the probability chart in the board at the end...
Diego Rey yeah cuz he’s testing your math skills
I just gave up so quick. And I know I'm not the only one.
def not the only one
CHeeTah FalconZT yeah me too
CHeeTah FalconZT caugth me
me
Laaetri same lol
My solution was acctualy a bit different I will try to explain:
1st weightning is the same and we use same logic scales are balanced
If scale is not balanced we can use same notation as used in the video and mark our coins + + + + and ~ ~ ~ ~
2nd weightning we put + + ~ on the one side and + + ~ on the other side
If we get equal weights we just weight one of the ~ coins with 0 coin and we can easily deduce answer
If one side is heavier we can now mark heavier coins ++ ++ ~+ and lighter coins +~ +~ ~~
coins +~ and ~+ are safe as counterfit cant be both heavier and lighter than the rest so we are left with coins ++, ++, ~~
now we can weight ++ ~~ with 0 0
If the scale balances the ++ coin we did not weight is counterfit
If one side is heavier we end up with coins +++ and ~~+ or with coins ++~ and ~~~ either way the one with all signs the same must be the counterfit :)
If anyone can find a misstake in my solution please tell me.
The emperor was too impatient to let him weigh them each separately, yet patient enough to hear out the long explanation near the end. Wot???
i mean, the video is only 4 minutes sooo,
even a toddler can wait that long
*I solved it without ever using the balance scale.* Since the Emperor has no idea which one the counterfeit coin is, you can just pick any coin and pretend it's counterfeit.
Check Mate
GG you win.. Eatshit king.. haha
Check Mate lol
Just make up a really complicated explanation. The king won't understand it, he's not the smartest mathematician of the land.
Check Mate LOOL so true
plot twist, the king lied and you are back in jail
Now the next riddle is how to get through bars without being noticed.
Plot plot twist.... You are the true King and the other one is a counterfeit. Now all you have is a marker and ....
Plot Twist..your Mom wakes you up saying "Wake up get ready for school, you're not a king". All this was a dream
Kalyani Kataki plot twist you never slept you have insomnia, you thought your mom was the king and you stole her jewlery to do the math. You're now grounded for halucinating.
We should make a movie lol xD
Usually this riddle doesn't just ask you to find the fake coin, but also to determine whether it's lighter or heavier.
2:03 You still have one more weighing left after this case, and you should use that weighing to compare it to any of the eleven genuine coins. In every other case you've already marked the coin with a - or a +, so you also know.
This last case can even accommodate a case where _none_ of the coins are fake - if the last coin balances with another coin.
I actually found a different, functional solution to this riddle! This only works if you know whether the counterfeit is heavier or lighter so take this with a grain of salt. The example has the counterfeit coin being heavier:
Make two piles of six and put them on the scale. Whichever side is heavier, take the coins from the other side and put them away, you don't need them after this. Then make the pile that is still on the scale and divide it into two, and put it on the scale. Whichever side is heavier, keep it. The others are cast aside. The side that is heavier has one taken from it and marked with a plus. The. The two left on the scale are weighed. If the scale is a balance, it's the coin marked with a plus. If the scale is imbalanced, the counterfeit is the heavier coin.
May not work perfectly if you don't know whether the counterfeit is lighter or not but that's my take on it. Have a good day!
this is the solution I came up with as well, I think it works and is better
@@natedavis5545 I wouldn't call it better, it is a solution for a simpler version of the riddle. Which is nonetheless still pretty cool.
I thought that the counterfeit could be lighter too?
The video does not say the counterfeit coin is heavier... It showed all the possible scenarios.
Teegan Schuler come on they said take with a grain of salt
15% of the comments: Jokes
80% of the comments: People who think they got it but didnt think that they dont know if the fake coin is heavy or light
5% is the rest
TakingThisWay you forgot the 0.5% that's people like you 😅
I ended up on the 2nd group... My solution only works if the fake coin is KNOWN to be either lighter OR heavier, not unknown.
Barney Holt then you will have 100.5%
me too in the second group :)
TakingThisWay haha I'm in the majority, womp womp
Just hand him any of the coins
Dusty_Sniper wow
👏👏👏
tea
Rule 6
He had to prove it in the end
One easier algorithm is also there, divide in 3 groups each of 4 coins.
Case1 :-
Then lets say weighing G1 with G2 balances, the we know all 8 coins are pure, remove all of them from balance now take G3, divide them in half, which side weighs heavier means they are pure remove them from balance and now you left with 2 coins and 1 trail left.
Case2:-
Now lets say the G1 and G2 don't balances then remove the heavier side coins and divide the remaining 4 coins then weight again and remove the heavier side then same left with 2 coins and 1 trail.
Congratulations🎉 you solved the problem without marker this time ❤❤
Me in math class:
Sir, uhm how come there is a marker in the story? clearly the setting is from the age wherein markers are not invented yet, or is it post apocalyptic or something?
Teacher: that won't be on the test so you don't need to know
Okay Mr Smart Guy, Chalk Then.
100th like
Ariel Salaza it's modern monarchy
wow you're sooo cool and smart honestly I'm astonished by your intellect
Here’s another way to solve this:
1. Check if you have green eyes
2. Ask the counterfeit coin to leave
@@Muhammed_English314 This solution doesn't guarantee that you can do it in only 3 scales. Only if you're lucky
@@healy_ly removing coins does not count as weighing them
Mohammad King oh
@@healy_ly you can even do it in reverse order by simply placing one coin on each side at a time (you can say that you want to place them all but they are precious so you place them one by one) and once they become unbalanced stop and weigh the last pair with any other coin and...
Can't you weight 6 and 6, then divide the lighter 6 into 3 and 3, weight those, and then from the lighter 3 weight 1 and 1, leaving one out, so either you see one of the coins on the scale being fake, or the one you put aside
“Did you just deface my currency? No prison for you you’re going straight to the electric chair”
but its medieveal tomes
deluxe sure then he'll just drop you in the moat infested with crocodiles
What if... that marker has erasable ink Lol
@@mithmoonwalker If the mathematician has a marker, the electric chair has already been invented.
I have an easy method. Stack them into 3 stacks of 4. Weigh 2 of them. (Balance 1) If they're unbalanced, the lighter stack holds the coin. Divide the lighter stack into 2 stacks of two and weigh. (Balance 2) And repeat for the final balance.
If the stacks from the first try balance, then the final stack holds the fake. Use the second and third balances the same.
"Can you solve the counterfeit coin riddle?"
Me: "Well... Back to the dungeon..."
How does the marker still have its ink through that whole Time he was in jail?
mohammed sachit hello
something called Magic
It's a good marker :)
Just mark it and close the cap and yeah. Markers can last for at least th ree months if you put the cap on and use it every ten seconds
mohammed sachit he got more
i feel like it would be quicker for me to just weigh all the coins against each other rather than take the time to come up with this plan
AJ Tomecek it would be easier but impossible since you're only allowed to use the scale 3 times and you have 12 coins.
Yes, but the reason for the three use limit is the fact that the king is impatient. If he's impatient then he should want it done in the least amount of time rather than the fewest scale uses. He doesn't even need the mathematician, since he could have anybody weigh each coin against the others in less time than it would take to summon the mathematician from the dungeon.
Calliope Pony lol I didn't even think about that. Yeah, he doesn't even need the mathematician, he could just have had anybody to do it
The point is, that if there were thousand coins (as in real treasury), weighting each individually would take much more time (over an hour) than doing prety much the same thing as here in which case you'd need to use the weights 7 time if you were very clever. So say, 15 times would suffice without doing it in such a complicated way, and figuring out such relaxed solution takes like 5 minutes. Also, if you were looking for the false coin several times, it would be profitable to optimize the process. Welcome to algoritmization and computer science, that's what it's all about.
AJ Tomecek but if there were 1000 coins, it would take too much to make piles, and probably no space, a
the marker would run out and you would be just stuck
I thought a different thing but I don't know if this is cheating, gimme your opinions, but now, let me explain:
There are 2 ways of thinking this trough so I'll divide it.
Situation 1
Step 1) Put 2 coins on the scale, if they're even, mark both with a "0".
Step 2) Complete both sides with the remaining coins (One coin with the 0 + 5 others vs the other coin with 0 + remaining 5).
Step 3) One of the sides wil be distorted, so I'd take one coin from each side at the same time, and when the scale got even I'd know one of the two coins I took was the problematic.
Step 4) Mark one of this coins with an X and the other with an Y.
Step 5) Put one of the "0 coins" on the scale and compare with "X coin", if it's even, Y is the false, if it's not, then X is the counterfeit coin.
Situation 2
Step 1) Put 2 coins on the scale, if they're not even, mark one with A and the other with B.
Step 2) Put A on the scale with an unmarked coin, if it's even, B is the weird coin, if it's not, then A is the counterfeit coin.
P.S.: I'm not sure if it's cheating because I don't think taking the coins out would be considered using the scale, but like I said I'm not sure :P
I'm no mathemetician, but I'm pretty sure this method would take longer than weighing the coins individually
Stuffed Cinema He is a mathematician. He solves a problems in minutes or maybe seconds
Ritik Chawla even if you're a mathemetician, I doubt you'd be able to solve a problem like 4x(1/3)^3=5(1/2)^x
in a few seconds.
Stuffed Cinema well it is possible?
Greatest mathematician in the world
King Cadmos has not in the world, just the county
how did they know it was a false coin in the first place then!!!?
mr jeeseeks ygs
Lol..................true
mr jeeseeks the lord could have admitted to it after
why would he do that?
Rick
am i the only one like to hear the solutions than solving them
Me too
obviously not, you are one person of billions so most likely not and I just realized my mistake but I won't fix it.
No
You're not
jaya priya no
i first heard of this puzzle back in 2010 and thought it was impossible later during my UG me with my friends look up for the solution and were in awe...To this day it was the hardest puzzle i've ever encountered.
looks like I'm going back into the dungeon... #savesirkoala
Sir Koala ok
#savesirkoala
#savesirkoala
#FreeMyNiggaKoala
Sir Koala ok *sends a assassin to set sir koala free and say **#savesirkoala*
I'm sure the impatient king wouldn't mind waiting 20 minutes for you to think up a strategy
what you do is you tell the emperor "look so you want to find the counterfeit coin or not? right. i'll use the scales 12 times then.
Georgia Ewington
👏🏾👏🏾👏🏾👏🏾
Georgia Ewington ok then I'm The Emperor and say that scale is in bad condition and will break after 3rd use...
sapphiro buy new scales
Georgia Ewington Precision scale is difficult to make and it will take my finest artisan one month to built. I need result now.
give me minute to make a call. theres one in the next town. about 10 mins from here
Simpler solution:
Step 1: confirm you have green eyes
Step 2: ask the king to leave
If the king's so rich, why does he care about one coin
because if someone else finds out HE payed with a fake, HE would be called out and framed
KEVIN cheng Maybe the coins are worth £400 each
Caring about every single coin is what made him rich
It's not the amount, it's the offense
KEVIN cheng because the lord HAD BEEN paying his taxes with fake coins, meaning1/12 oh the kings revenue is phony and useless until the perpetrator is rooted out.
Me the greatest mathematician
Impossible
The horses name was friday
nakul sindhwani LOL
The house is on the North Pole so the bear is white
Omg
1234 subs with no videos challenge q😂😂😂
And then he died
I think mine would've worked too.
Weigh 2 equal amounts of coins.
Mark the coins in the coin pile that weighs less or more.
Now, weigh the marked ones in 2 equal groups.
Mark the coins in the coin pile that weighs less or more (again).
Now, weigh 1 marked twice coin and 1 other marked twice coin separately.
If their weights are both equal, take the leftover coin not on the scale.
Otherwise, take the coin that weighs less or more on the scale.
EXACTLY! That's the solution I came up with also. So much easier than this over complicated answer in the video.
When you measure two piles of 6 , one will be heavier and other one will be lighter . it's not an either or situation . If one thing is heavier the other is lighter .
I didn't understand anything but I watched to the end
Kathy the Bestest same
I feel like this wouldn't be the mathematician's job.
weezact7 YEAH THEY NEED A GOLDSMITH FOR THIS DUDE
70% of the comments are geniuses with "aww this is easy, just divide the coins into 2 piles of 6s...." and another 20% even more geniuses that even after watching the soution they're like "there's even more simple way, just divide the coins into two piles...."
Shady Maze ginuses?? what
u are either exceptionally dumb or blind... my guess is u're a mixture of the two
Shady Maze lol wrong coment I replied to so sorry haha
I understood the solution right after he explained.So I think I am genius 😂😂
“Youre the realm’s greatest mathematician.”
Or the emperor could get any random sod to weigh the coins. Or weigh it himself lol
what kind of king only has 12 coins??????
Karl Wilson it was twelve example coins. Not just 12 coins the governor gave that they know of.
Nope, only ones that make no sense.
John Hasben not the jokes fault you are unintelligent
Ok what's the punchline to this joke? You can't mask your stupidity as "it's just a joke" all the time.
not all jokes have a punchline. some are humorous witticisms. perhaps when youre older they wont all go over your head.
Me: ignoring the riddle
"How has my marker not run out?"
Maaaagiiiccc
« What are you doing, drawing on my tax money with a marker?! BACK IN JAIL »
The funny thing is that it would take less time to weigh all of the coins in equal piles until you found the one than it would if you marked all of them as you went
how the hell did the king linked the coin with the impostor?
Roberto De Gasperi Now you are asking the real question
The king offered a reward and The Swordsman Apprentice ratted him out, it was the sword maker all along and they found the little Tool & Die coin molds used for making them :-) so quit asking questions or you'll be next!
There were only 12 coins in the treasury, obviously ...
Quite a broke kingdom...
Moreover, he paid taxes with only ONE counterfeit coin. If you're gonna go counterfeit, go big or go home.
if the emperor doesn't know which coin it is either, couldn't you just give him any coin then write down anything that looks strategical.?
Madisøn the Tøp! Crybaby You should be the realm's greatest mathematician.
Madisøn the Tøp! Crybaby he showed us him showing the emporer his work so i guess you need to work it out and not all fat emperors are stupid you know :)
Madisøn the Tøp! Crybaby Maybe the emperor has got some other mathematicians to check if it would actually work.
give them some of the coins to keep their mouths shut xD
you will be sending an innocent man to prison then... and letting a guilty one walk!
So the King is impatient when it comes to how many times I have to weigh the coins, but not the the million hours I'll be having to figure how to do it with only three trails!
Superb comment Bro..
the trick is, in this situation, it wouldn't take that long. since you are the best mathematician in all the land, it would be very quick. it's still kind of flawed but it's just for the sake of the math.
but it will still take you longer because you have to write on every damn coin
Al Seraidy Mohammed Took me 2 minutes.
A much more elegant (and somewhat easy solution):
1) Divide pile into four groups with 3 coins each
2) take any two piles and weigh them:
a) If the two piles weigh equal, this means the fake coin is in the remaining two piles
b) else the fake coin is in one of the two piles you just weighed
For case b)
# mark the lighter pile "-" and heavier pile coin with "+".
# weigh the "+" pile against any unused pile of 3 coins.
# If these two are equal, then the fake coin is in the "-" pile of three coins else it's in the "+" pile
# since we're now only left with a faulty pile of 3 coins, take one coin each from faulty pile and weigh both against each other
# if they're equal, the remaining coin from faulty pile is the faulty one
# else the heavier one with "+" sign is the fake.
I did it in a different way, but using some of the same underlying principles:
I divided the coins in 3 sets of 4. I measure 2 of those 3 sets against each other.
If they have the same weight, then I conclude that the false coin is among the last set that I didn't weigh. Then I take 2 of these 4 potential coins that I weigh against 2 verified coins.
If there is a difference of weight, then it means that the counterfeit is among the 2 potential ones I just weighed. Then I take 1 of these 2 that I weigh against a verified one, and if there is still a difference, then it's that one potential coin. If there is not, it's the other potential one.
If there is no difference of weight, then it means that the counterfeit is among the 2 potential ones I didn't weigh yet. I do the same process as above and find the counterfeit coin in 1 measure.
Now let's imagine that the 2 sets of 4 coins that I originally weighed didn't have the same weight. Then, I write a "+" on all the coins that are on the heavier platter, and a "-" on all the coins of the lighter platter.
Then, I take 2 "+" coins and 1 "-" coins that I put aside.
On the platters are left, on one side, 3 "-" coins and on the other 2 "+" coins.
Like this: - - - vs. + +
I add 1 verified coin ("0") to the platter with the "+" coins.
Like this: - - - vs. + + 0
Then I switch 1 "-" coin and 1 "+" coin between the platters.
Like this: - + - vs. + - 0
And I proceed to measure. (This is my 2nd measure.) 3 things can happen from there:
1) The two platters weigh the same: this means that the counterfeit coin is among the 3 coins (2 "+" and 1 "-") that I put aside after the first measure. I just have to weigh the 2 "+" coins against each other. If they weigh equally, then it's the left "-" coin. If they don't, then it's the heavier coin.
2) The two platters don't weigh the same, and the balance changed (the heavier platter became the lighter one and vice versa): this necessarily implies that the counterfeit coin is still on the balance, but it changed side. Since I switched 2 coins, it necessarily means that the counterfeit coin is one of them. I just have to take 1 and measure it against a verified coin: if there is a difference in weigh, then it's the 1 switched coin I just measured. If there is no difference, then it"s the other one.
3) The two platters don't weigh the same, and the balance is unchanged: this necessarily implies that the counterfeit coin is still on the balance, among the coins that didn't change side and are not verified (which are 2 "-" and 1 "+"). I just have to take the 2 "-" coins and weigh them against each other. If there is no difference in weight, then it's the "+" coin. If there is a difference, then it's the lighter one.
It might seem more complicated but I like this one (because I found it lol). It took me a couple hours though. Frustrating but I'm happy to have done it!
My solution is almost identical, except if there is an imbalance at the first weighing, I put aside + - and my second weighing is + + - - vs. + - 0 0. If left is heavier than right, I know it's one of + + -, if right is heavier than left I know it's one of - - +, and if it's balanced it's one of the + - I put aside. The third weighing then works exacly like you describe it. What I like in my solution is that it has some "symmetry" in the second weighing, by that I mean if you swap + and - it's still the exact same thing.
The things you could do with a balance weigh and a dozens of coins
This is how I did it too! though i did ++- vs +-0 lol
cool to see :)
Wait wasn't the limit 3 weighs only
Excellent! Thanks for taking the time to write this up and validate my solution.
You know, I'm here just wondering how the marker didn't run out of ink..
yeah
ClaireBear It's a magic marker
ClaireBear you only mark 12 coins,,,, what sort of markers are you buying?
TwentyOneOstrich's plus marking all those days in prison for years and years
You: "but the king is not a patient man"
Me: *thinks for 5 minutes and is thrown on the jail without even trying*
or you could split them into 2 stacks with 6 coins each. the heavier or lighter side contains the imposter coin. (assuming it shows up heavier, meaning the imposter coin is heavier -- this will work if it appears lighter as well) do this again with the side with the imposter coin weighing 3 coins against the other 3. again, the heavier side contains the imposter coin. now you are left with 3 coins and you can just weigh any random 2. the trend of the heavier side being the imposter coin should continue, so if the 2 coins you picked contains the imposter, it will show up on the heavier side of the scale. if both are balanced, the last unweighed coin is the imposter coin.
this solution only works if you know it's lighter or heavier beforehand, which you don't