You’re the realm’s greatest mathematician, but ever since you criticized

the Emperor’s tax laws, you’ve been locked in the dungeon with only a marker to count the days. But one day, you’re suddenly brought

before the Emperor who looks even angrier than usual. One of his twelve governors has been

convicted of paying his taxes with a counterfeit coin which has already made its way

into the treasury. As the kingdom’s greatest mathematician, you’ve been granted a chance to earn

your freedom by identifying the fake. Before you are the twelve identical

looking coins and a balance scale. You know that the false coin

will be very slightly lighter or heavier than the rest. But the Emperor’s not a patient man. You may only use the scale three times before you’ll be thrown back

into the dungeon. You look around for anything else

you can use, but there’s nothing in the room – just the coins, the scale, and your trusty marker. How do you identify the counterfeit? Pause here if you want

to figure it out for yourself! Answer in: 3 Answer in: 2 Answer in: 1 Obviously you can’t weigh each coin

against all of the others, so you’ll have to weigh several coins

at the same time by splitting the stack

into multiple piles then narrowing down

where the false coin is. Start by dividing the twelve coins

into three equal piles of four. Placing two of these on the scale

gives us two possible outcomes. If the two sides balance,

all eight coins on the scale are real, and the fake must be among

the remaining four. So how do you keep track of these results? That’s where the marker comes in. Mark the eight authentic coins

with a zero. Now, take three of them and weigh them

against three unmarked coins. If they balance, the remaining

unmarked coin must be the fake. If they don’t, draw a plus on the three

unmarked coins if they’re heavier or a minus if they’re lighter. Now, take two of the newly marked coins

and weigh them against each other. If they balance, the third coin is fake. Otherwise, look at their marks. If they are plus coins,

the heavier one is the imposter. If they are marked with minus,

it’s the lighter one. But what if the first two piles you weigh

don’t balance? Mark the coins on the heavier side

with a plus and those on the lighter side

with a minus. You can also mark the remaining four coins

with zeros since you know the fake one

is already somewhere on the scale. Now, you’ll need to think strategically so you can remove all remaining ambiguity

in just two more weighings. To do this, you’ll need

to reassemble the piles. One method is to replace

three of the plus coins with three of the minus coins, and replace those

with three of the zero coins.>From here, you have three possibilities. If the previously heavier side of

the scale is still heavier, that means either the remaining

plus coin on that side is actually the heavier one, or the remaining

minus coin on the lighter side is actually the lighter one. Choose either one of them, and weigh

it against one of the regular coins to see which is true. If the previously heavier side

became lighter, that means one of the three minus

coins you moved is actually the lighter one. Weigh two of them against each other. If they balance, the third is counterfeit. If not, the lighter one is. Similarly, if the two sides balanced

after your substitution, then one of the three plus coins

you removed must be the heavier one. Weigh two of them against each other. If they balance, the third one is fake. If not, then it’s the heavier one. The Emperor nods approvingly

at your finding, and the counterfeiting Lord

takes your place in the dungeon.

Lmao I dunno why but this was easy

I thought I had got it when I divided the 12 coins into halves… until I realized the counterfeit could be heavier OR lighter. UGGGGGGGHHHHHHHHH

Brooklyn 99 has this riddle, but it's people on a island instead of coins. This one makes much more sense and the series doesn't have the solution, so this video was very useful.

Well easy for checking with the scale you put one coin on one side and if its not equal once then you know one of the two is your counterfeit but you will need another chance to use it to write on those two then measure each one once taking up your remaining two chances if the first one is balanced then you must use your last chance to identify the last coin the second one is the counterfeit just to be sure but how to take it down to only two is the hard part

I’m the realm smartest mathematician?

illusion: 100I hope that’s washable marker. If it was permanent you’d probably go back in the dungeon

Plot twist:the mathematician was put to jail again for vandalising the coins

Hardest so far

Me:(watches solution thinking) won't you be able to tell the fake coin by holding it since lighter or heaver than the normal coin?

My solution:

Weigh all coins in your hands, if one feels heavier or lighter than the others, weigh one normal feeling one to the heavier/lighter one, if they balance then keep weighing them in your hands until you find another different feeling one then repeat the process, if they imbalance, keep the weird one on the scale and mark the normal one with a dot or something so you can keep track, then get an unmarked coin and weigh it against the odd one, if they imbalance again then the odd one is the counterfeit, if they balance then the one with the dot is the counterfeit.

And before everyone says "BUT YOU CAN ONLY WEIGH IT THREE TIMES!!" He said that you can weigh with the scale 3 times, he says nothing about weighing in your hands.

What if I weigh 6 vs 6. Then take the pile which is heavier and weigh 3 vs 3.

Now from the heavier pile, I weigh 1 vs 1 if both are equal then the remaining coin is the counterfeit. If not, and one is heavier then that is the counterfeit one.

See keep 6,6 on both sides,one side will be heavier n then put 3,3 one side will be heavier. Then put 1,1 n one will be heavier or else the one that u have out of the balance is the fake

Captain Holt needs to see this

#NOINE NOINE

Very nice logic puzzle

Unfortunately I couldn't solve because I'm too lazy

6-6 2-2 1-1

There is a easy way to do this.

3 stacks of 4

Put 2 stacks of 4 on the scale

If one is heavier than the other, take the lighter stack

If they are the same weight, take the extra stack.

Take the stack you took and weight 2 and 2. Take the lighter stack

Take the 2 coins and put one on each side. Take the lighter coin.

the real mystery is how on earth that marker hasn’t died yet

Divide 12 to 6 and 6, then use the scale to see which stack of 6 is lighter. Then divide the lighter stack of 6 to 3 and 3 and use the scale for the second time. So now you will know in which stack of 3 is the false coin. So from thoese 3 coins take 2 at random and use your last scale. If the coins are the same weight, then the one you didnt pick from the stack of 3 is the false coin. If you picked the false coin at random from thoese 3 then you will just see which is lighter from the 2. So yeah i didnt rly need the marker.

Easier way would be:

1. Make three groups of four coins and then weight two of them – if they are of equal weight, then the third group contains a false coin;

2. Divide that into groups of two and weight them, one must be lighter;

3. Divide lighter group and weight those two coins – and boom, you get the counterfeit coin… without a single marker-drawn line

King's logic: You can take however long you want to think about how to do this, BUT I'LL ONLY ALLOW YOU TO USE THE BALANCE THREE TIMES WHICH WILL LITERALLY TAKE SECONDS AS A TIME.

So wait… why can’t you just take the 4 that hold the counterfeit, split them in half instead giving you 2 on each scale, then split the 2 holding the counterfeit as a 1 v 1 and find it that way? Then you don’t even NEED the marker or all that extra stuff with the pluses or minuses…. Seems easier than what this video explained 😂

Well if no one knows which one is the fake and know that it just exists, just pick up a random one and say that’s the fake one.

Easiest way (uses scale just 3 time) put 6 coins on both sides. Now remove coins one by one. Just as the scale balances, mark your last removed coins. Scale is balanced, it means rest of the coins are real. Now weigh the marked coins each with the real coin. If balanced then real coin if not, you got the fake coin.

you just have to divide the coins in two piles of 6 coins each. see which one is slightest. and saparete this one in two piles of 3 conis each, use the scale. the slightest one you use the scale again comparing two coins, the slightest one is the false, if the balence the one out of the scale is the false. so no need of the marker.

Before listening to the answer:

Divide the coins by two and weigh them, 6 on each side. Since the fake is either lighter or heavier you’ll see either one side is slightly higher/lower. Mark the coins on the side that is higher/lower and put the normal ones on the other side on the table so as to not get confused. Weigh the six marked coins 3 on each side, same thing one side will be higher the other, lower. Then you weigh two of the three coins from the heavier/lighter side why? Because if you are weighing two genuine coins that means their weight would be the same and the last one is the fake.

After:

OH! So different!

SPOILER ALERTOurs was slightly different: •Make four piles of three

•Check two against each other

•If they're

a) equal leave any one on and put a third pile on.

If that's equal the fourth is false

If it isn't the new one is false

b) not equal put any other one on. If they're still not equal then the one you left there before is fake. Else it's good.

• use the same method of observation. When you saw the unequal one you knew whether it was lighter or heavier when you changed it out. Use this to differentiate between the three coins like the video

Idk why but this felt like the easier option

I did it the wrong way and somehow found the right one how well take 12 coins divide it by 6 = 6 so make them into groups now there are 2 groups now take both groups put 1 pile on each scale now lets say the left is the most lightest so the right is put to the side now take the left pile and divide them 6 coins divided by 3 = 3 so now 2 groups and 3 in each put both piles on the scale if one group lets say the right this time the left group goes with the other 6 now 3 coins left and one more try with the scale all you have to do is take 2 coins put both on the scale of they balance the on not on the scale is the fake coin if the left or right is light then it is the fake

Wow got the wrong way to do it and still got it right

Actually, it isn't the only way. you can weigh 6 and 6 coins in the first round, and where you can locate the fake coin in one of the two piles. then you weigh it with 3 and 3, at last round you can just weigh any 2 of the coins left.

Can't you just divide the 12 coins into two piles of 6, one pile will be heavier, so you divide the lighter pile into two piles of 3, again one will be lighter. You have three coins left, you weight two of them, if one is lighter you know that's the one, if they're exactly the same, it's the one you didn't weigh. BOOM, didn't even need the markr

I feel bad for the lord

Quase consegui. Mas havia a pequena probabilidade de eu errar em um caso específico

Por que ele não inventa um plano pra depor o rei?

Wouldn’t a simpler answer without the marker be to split them in half, weigh, split the lighter six in half, weigh, then do the last step with three coins

YOU OWE ME WHY WOULD YOU THROW ME BACK INTO JAIL

I know how to solve this problem!

Just don't anger the king.

The real problem is that I’m the realms greatest mathematician

this is the easiest question seen in this series

But who says that he is guilty. What if all coins weigh the same

Awesome Riddle

Easier way to do it actually split it six and six usually counterfeit coins are lighter so I’ll go with that. See which is lighter take the 6 on the heavier side mark so you know there real. Put three and three mark the heavier ones so you know they’re real the put 1 on each from the remaining 3 equal? The one out is the counterfeit ones lighter? That’s the counterfeit

Alternate solution:

Step 1: Weigh four against four. If it balance follow the video in dealing with the 4 remaining coins. If the 8 coins don't balance, label them H for the heavier end and L for the lighter end and unweighed coins are real

Step 2: Weigh HHL against HHL. If balance, step 3 will be weighing one of the remaining 2 Ls against the coins you already know is real from step 1. If unbalanced, the two H on the heavier side and the one L on the lighter side is suspicious.

Step 3 for consecutive unbalanced weighing:

The HHL combination outlined in step 2 contains one fake coin. Weigh the two H against each other. The heavier side is fake. If neither is heavier, the L coin is fake

But don’t you use this more than 3 times

I found a easy solution.

Put the half in balance, one of them wilI be with the fake

Put the half of the half with fake, get the 3 with the fake.

Now you have 3 coins, put 2 of them.

The fake can be in the balance, so you will know that is this one due the weight difference.

If the fake isn't in ballance, it is the one that is out of the ballance.

Solution without the marker:

First step: weight 3 piles of 4 coins to determine the fake pile

Second step: weight 2 coins of the fake pile: if they are balanced, the fake coin is among the 2 last

If not, the fake one is among the 2 on the balance

Third step: take the two considered coin and change one with a real coin, then weigth them: if it’s balanced the fake coin is the replaced one, if not, it’s the not replaced one

You are welcome my lord

Yo this is the riddle that was on that one episode of Brooklyn 99, just with a different premise.

My solution is slightly different but pretty much the same principle, 1st weigh, divide the coins equally into 6 and narrow down the lighter one, then proceed to do it another time with 3 coins on each side, narrow down again, and for the final one put one on each scale and if its balanced, the one you didnt weigh is the counterfeit, if not, the higher scale is the counterfeit.

I have solved this one in an interview

wouldn't the king be mad you ruined the coins?

“Did you just deface my currency? No prison for you you’re going straight to the electric chair”

Easier solution : divide in pile of 3 coins. You get 4 stacks, say A,B,C,D. now weigh A Vs B. Case 1: both balance equal. Then all 6 coins in A and B are good coins. Weigh either of A and B with C, you will find that counterfeit coin is in C or D and whether the fake coin is lighter or heavier In the third attempt, place one coin of fake stack in each balance and identify the fake coin

Case 2: A and B are not equal. Check either of them against C. The fake stack will be lighter or heavier both the times so you identify the fake crack and repeat last step of case 1.

Bot what happens if they are all counterfeits but one because they are all remainers in the pay of the EU.

There’s another way. On the first scale use weigh two groups of 4. Which ever one is heavier or if it’s equal take the lighter pile and then weigh 2 and 2 coins from the pile that contains the counterfeit coin. On the third use, you weigh the final two coins against each other and you will find the counterfeit coin

So far this is the only solution of your videos I’ve figured out by myself lol

They made it way harder than it needed to be

You could also weigh two piles of four. If they balance equally, give those eight coins to the king. Then, get the other pile of coins which would have to contain the fake and put two coins on each side. Whichever side is heaviest holds the fake so you give the lighter ones to the king and then put the remaining two on the scale. Whichever coin is the heaviest is the counterfeit coin. You can then mark it with a marker so it will be clear that its fake. If they don't balance equally on the first go, you could then give the lighter coins and the other group of coins you didn't weigh to the king. You could then, as done in the first scenario, put two coins from that pile on each side and then give the two lighter coins to the king. Then, you could weight the two remaining coins and mark the fake one!!!

JUST HOPE THAT NONE OF THEM FALL ON THE FLOOR💀💀💀

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

"At least one of you are real…" (wait…

Well, I’ve figured it out without any marker. Is it bad?

Use your hands.

If you weigh 6 against 6 on the first weigh, one will be lighter and you can mark all 6 of the heavier one as real

then weigh the 6 as 3 against 3 and again mark the heavier three as real

then take two of the remaining three and weigh them against each other- if they balance, the third one you didnt weigh is the counterfeit and if not, the lighter one is the counterfeit.

I feel this is a faster method and would work just as well

There is a much simpler solution. By dividing into 6 then 3 and the 3

This was pretty easy compared to the rest.

Well this problem has many solution, first scale 6 by 6 and take the 6 which is lighter, then split that 6 into 3 and 3 and take the three which is lighter. Then take 2 out of the 3 and weigh, if they balance the third is the counterfeit but if one is lighter that is the counterfeit. Ur welcome

I was gonna say: split the stack in half, whichever side is lighter, split that, then split the 6 into two, then take the 3 on the lighter side, then if they match, take the counterfeit, if one side is lighter, its the fake.

Shortcut -take 2 coins on one side and other 2 coins on other side and see which set of 4 coins weigh unequal

who would win?

A sock

ORA emperor with some coins

But wait… If you pick 8 random coins and put half in each side of the scale, there'll be just 2 possible outcomes, even or uneven. If it's even, the counterfeit coin is among the 4 coins you didn't weight, and if it's not even, it's obviously among the lighter 4. In both cases, you'll know amongst which 4 coins the counterfeit one is after a single measurement. Then you just repeat it, put two of those in each side, and one side will definitely be lighter. Then you take those two lighter side coins, put one in each side and the lighter one will be counterfeit.

The actual riddle is how nobody realized this and used a ridiculously more complex with a marker instead.

Is this the b99 riddle?

I figured out an easier solution without using the marker:

1. Divide the coins into three stacks of four coins

2. Weigh one stack of four coins against another one (then the lighter stack will be known either on the scale or the one out)

3. Divide the lighter stack of four coins into two stacks of two coins and weigh them against each other (the lighter stack of two coins will be known)

4. Weigh the left two coins against each other and then you will have your fake coin!

Ok, Or weigh 6-6, liter pile then 3-3, and finaly just 1-1, if they are equal then 3rd coin is fake, or liter is on scale! Use marker to draw dickbutt for emperor!

Isn’t it easier to divide it into 4 piles?

Or another way is(without the marker)

1)split into 2 sets of 6 and weigh it…eliminate the heavier one

2)split the light set into 2 sets of 3 and weigh…..eliminate the heavier on

3)now that the light 3 set and choose any two at random

-if it balances then other one is the counterfeit

-if not then the lighter one is

U have just saved your ink …your welcom

I thought you knew whether the counterfeit was heavier or lighter and I was like…this is too easy XD

This reminds me of Archimedes's eureka event. I mean the problems to solve aren't the same but the situations are pretty much alike.

Divide all 12 coins into 2 groups of 6 each. Weigh the groups, the lighter group has the fake coin. Take your lighter pile of 6 coins and divide it into 2 groups of 3 in each pile. Take the 2 piles and weigh them. The lighter pile has the fake. Now take the 3 coins and put 2 on the scale. If the scale is the same, the remaining coin is fake. If one side of the scale is lower, the other side is the fake. Like if you understand

Bruh i think you used the scale too much

I think you made this way more complicated than it needed to be. Doesn't this work just as well:

1) Weigh 6 v 6, whichever is lighter has the counterfeit coin.

2) Take the 6 from the lighter side and split them into four groups: 1, 1, 2 and 2. Weigh the two stacks of 2 coins.

3a) If they last step is balanced, the counterfeit is in the two stacks of 1. Weigh them to see which is lighter.

3b) If the last step is not balanced, the counterfeit is in the lighter side you and you can split and weigh it.

But if you divide the 12 coins in to two groups 6 and 6 then weight them you will have 2 tries left and then you can divide the heavier group into 3 and 3 then weight you will have 1 try left THEN only waight 1 and 1 if the are equal means the other one is the bad one

I think this is the ultimate hard answer

There is a very easier answer for this

Just devide them to two portions of 6, then take the lighter one devide it to two portions of 3.

So now you have a portion of three coins lighter than the other one. That means the light coin is one of those three. Just take any two of these three and weigh them. If they balance then the third one is the light coin. If one is lighter then its the light coin.

And then you get put back into the dungeon for drawing on the coins

The Korean subtitles are wrong. I tried to solve it by using the scale twice, but I couldn't solve it even after using a few days.

I actually solved it lol

But there’s a simpler way…

First round: Split the 12 coins into 2 groups of 6 coins. Weigh both and keep the lighter group of 6.

Second round: Split the lighter group of 6 coins into 2 groups of 3 coins. Weigh both and keep the lighter group of 3.

Third round: Choose any 2 of the 3 remaining coins and compare them. If one is lighter than the other, you have the counterfeit. If they weigh the same, the counterfeit is the 3rd coin you didn’t weigh.

without marker:

12 coins in front of you

1. put three random coins on each side of the scale

2 A) either they are equal: put them aside on a stack and TAKE 2 COINS (OUT OF THE OTHER 6 in front of you) on each side of the scale. if both sides are equal, REPLACE ANY COIN WITH 1 (OF THE 2 COINS LEFT) and put the coin you replaced to the other six ones you know are equal.

2 Ax) one side is heavier: it is the coin you took in (think you can remember without marking wich one out of the three was the last one you put in).

2 Ay) equal: it is the one left in front of you.

2 B) or one side is heavier. then put the other six coins away and remove one coin from each side of the scale and put it in front of you. now it's the same like in 2 A)

P.S. hope it's not written too cumbersome. I'm not a native speaker as you might see…

Brain.exe has crashed

😪I almost got i right

The time you think about the plan takes longer than weighing them single by single

You could also take the 12 coins and divide them into two equal piles of 6 coins each and weigh them.

One of the two piles will be heavier.

Take that pile and take 4 coins from it, weighing 2 coins on each side. If one one of them is heavier, then take again divide it with 1 coin on each side and you have the fake coin.

If both sides of 2 coins each weigh the same, then take the third pile of 2 coins and weigh 1 on each side.

P. S. : I just realised that the question says that the fake coin could be heavier or lighter too, so this answer is only partial. But i already typed it so thanks for reading it. 😅

4:07 when the fudge did he get the time to draw that, nevertheless explain it? impatient king would've stopped him less than halfway and thrown him in jail…

You dont need the marker by the way.

I would divide them 6 and 6, then take the side with the fake coin and divide them 3 and 3. then take the side with the fake coin and divide them 1 and 1 and leave one out.

Looks like I'll be in that dungeon forever

Why doing that when you can divide in two packs of six see wich is weighting divide it in two of three see wich is weighting and choose two from those if it weights you found it if it doesnt its the one you did not use. No need for the marker

Recursion explained at its best

Me: hopes the king doesn’t care about all the marks

King:doesn’t care

Dont use the scale judge by your hands

I DID IT! I SOLVE IT! OMFG!!!!

Hmm how bout i do some random things with the scale and coins then and pick a random coin and pretend thats the counterfeit one amd give it to the king, he'll never know becayse he's too inpatient to try again and just rolls with it.

Why don’t you go 6 and 6 then take the heaviest one and do 3 and 3 then take the heaviest and randomly take two and put then on each side so if they balance it’s the last coin that you didn’t use and if they don’t balance then u have your answer

U could also do 6 and 6 than 3 and 3 and than 1 and 1