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.

Nothing like a MARKER in ancient times

cant you just do this

cant you just divide the coins into 2 piles of 6 coins each and then weigh both piles and whichever pile weighs more you take of the scale and divide into 2 piles with 3 coins each and you take the piles and weigh them on the scale then whichever pile is heavier you put them in 2 piles with 1 pile of 2 coins and 1 pile of 1 coin if the 2 coins balance with 1 coin the single coin is the imposter

> emperor is an impatient man

> lets you write out all of the possibilities on the board

Well another method -:

1. Divide 12 coins into 2 grps of 6

2. After balancing u've got the side having 6 coins with fake coin .. Divide that into two grps of 3 and weigh

3.now select any two coins and weigh

The riddle assumes there is only enough TIME to use the scale 3 times, due to impatience. The solution cannot be more time consuming than that.

Me watching the solution and still not getting a damn thing.

The way I'll think you solve it which is not what they said but is also a viable option if I haven't missed any thing is

First divide the coins into six coin put them on the scale whichever side is lighter you take that get rid of the rest

Second weigh the six coins that were lighter 3 on each side then take the side that's lighter

Third put 1 on each side of the scale if they weigh the same the coin you did not put on is the counterfeit if one of the coins on the scale is lighter than the other it is the counterfeit

“You are the realm’s greatest mathematician.”

bish i almost failed math when i was grade 7Instructions unclear

My cat is in my washing machineThis la the same riddle as the on from the islanders un Brooklyn 99?

Three words:

Sniff the marker

The real question is, why is my chin like thanos???

I did it different I slit the coins into 2 groups of 6 then split the lighter one into 2 groups of 3 then after I would take put 1 on both sides from the lighter one and the one that goes up is counterfeit but if it’s even the one from the lighter side u didn’t put up is the one

Not only does the solution allow you to identify the fake coin, you will also know if it is heavier or lighter than the rest. If you only needed to identify the fake coin, this could still be done with 13 coins.

Now do this exact same riddle with 39 coins with 4 tries.

This is over complicating it.. a better method is splitting the coins into 3 groups of 4 then weighing two of the groups and if they’re even then the other coin group that you didn’t weigh has the fake and you keep splitting till you find the last coin. It’ll take 3 try’s and you’ll find it no need to mark anything

Not sure why my first guess doesn't work… Divide equally and weigh 6 on each side. Then using the light stack (the fake coin is lighter the video says), divide again and weigh 3 on each side. The lighter side again contains the fake coin. Finally weigh any two of the remaining three coins on each side of the scales. If they balance, its the one you didn't weigh. If they don't, it's the lighter one… Not sure why this isn't a perfectly good answer…?

Use the scale as a weapon and smack the emperor over the head with it

He kinda looks like a beer mug. Like, the lines and being colored slighty yellow. “But he has a nose going past the line collum things” well thats beer foam shut up.

The king beheads you anyway, for he was impatient waiting for you to figure this all out before using the scale for hours instead of just weighing the 12 coins which would have taken a few minutes at most.

Emperor: I WANT YOU TO FIND THE COUNTERFEIT COIN CAUSE ITS BAD FOR ECONOMY…

Me: Oh okay let me weigh them th..

Empeoror: YOU CAN ONLY WEIGH THE 3 times

You didn't say the second method, so I don't know if mine is right, but here's what I did.

#1 Same as video

#2 Weigh (- + +) against (- 0 +) leaving (- – + 0 0 0) on the side.

#3 If scale is balanced weight the (- -) from the side against each other.

If scale doesn't change weigh the (-) on the left or the (+) the right with any (0).

If scale changes weigh the (++) from the left against each other.

You can weigh it by balancing the marker on your finger and then you can put the coins there.

What about weight four and four. If balance weight the four left out (2 and 2) and then weight the 2 lightest(1 an 1). You find the coin

If not balance then weight the four lightest(2 and 2) and weight again the lightest (1 and 1). You find the coin.

Isn't easier so?

I have a much similar option why not put one coin on each side slowly say your just adding them on then if theirs an imbalance do the plus and minus then weigh them against the proven real coins if the balance then the other one is the fake

Perhaps the real solution is to convince the emperor to let you have a fourth try. Then, you save the time you would’ve spent trying to devise a solution for only three tries.

you can also go 6 at a time, then see if the are different in weight, if the right side is heavier than the left side (or lighter) and vice versa then you know its among the six, while marking the other six then weigh 3 to 3, continue the process and then 2 to 2, and after, when you cannot split the three just use the one you know is real.

You can also half twelve 6 & 6 way both on the scale the low one has the coin take the liter one off and then half the 6 to get 3 & 3 way both the low one will have the coin but if it is balanced then it's the coin that is not on the scale

There's a better answer

This is a really confusing riddle wow! But there is also another way to solve it than that without the marker.

For me I kinda did it differently at first (but also I thought the fake was lighter than the rest) I said you have 12 coins split the coins in two piles which would mean 6 on each side. When there is a imbalance the lighter side has the fake. Then you take the ones that are lower and write a checkmark on them, because you know they are real. The ones that were lighter take 4 of the 6 and split it up so 2 is on each side, if they are balenced then you know they are real. If not then you know the 2 coins that are off of the scale are real and you can put a checkmark on those, the side that is lighter you weigh them against each other. It would work the I other way around to if you knew it was heavier than the rest.

Or just pick up the coins one by 1 since there’s obviously not that much and the narrator seems to know how many and the one that’s lightest just measure it on the scaled against another to be sure,or just give him a coin to make up for it

Cant you just halve every pile of coin each time?

Example:

1st: 6 – 6 (one of them has the fake coin)

2nd: 3 – 3 (same as the above)

3rd: 1 – 2 or 2 – 1 (since fake coin either heavier or lighter than the real one it should make a big difference in weight right?)

My idea was just weigh 6 against 6. Take the heavier stack, and weigh three of those against the other three. Take 2 from the heavier stack, and weigh them against each other. If they're equal, the reaming coin is the fake. If one is heavier, than that one is fake

what kind of emperor taxes the kingdom so heavily and yet only has twelve coins in his treasury?

Woah their buster, their is a very easy way.

Put 6×6 on the scale.

Put light side away

Have 2 balances left.

Weight 3×3

Put lighter side away

3 left

Put 1 on each side

If they weight the same then the last one is counterfeit

If it is imbalanced, then the heavier one is it

Btw this only works if the coin is heavier

Or lighter just reverse the scenario

4:03 but i thought that we dk if the fake is heavier or lighter??

Captain Holt needs to watch this video.

This took me five months to solve

Me:just lock me up at least free fooddddddd

Another math question, ain’t I right?

why don’t you just throw a magnet and if the coins that doesn’t stick, it is fake boom easy solveWe can separate 12coins to 2 parts,6 coins each . Put them onto the balance scale . Put the heavier part aside as the fake coin must be inside the lighter part. Then, separate the lighter part to 2 parts,3coins each. Do the same way and we can find out that the fake coin must be inside the lighter part. Last, randomly put 2 coin from the lighter part onto the balance scale. If they have the same weight, rest of the coin must be the fake coin. If they are not the same weight, then the lighter one must be the fake coin. And we don't need to use the marker!!😂😂

Or, establish a federal reserve and just print.

"Can you solve the counterfeit coin riddle?"

Me: "Well… Back to the dungeon…"

The king is never finding that counterfeit coin, because i’m horrible at math 🙂

Alternative method:

– split the coins into two groups of 6

– whichever is lighter you know has the counterfeit

– choose two from the lighter group

– if the two coins are equal, you know the remaining coin is the counterfeit

– if one is lighter, then that one is the counterfeit

Can’t u do 2 groups of 6 take odd one into 2 groups of 3 then if they match its the extra one and if they are different the different one?

You lost me at coin

"Just the coins, a balance scale, and a marker."

Table:

Am I a joke to you?You do know you can just halve it twice to get 3 coins and just weigh two of them and that will still work just pick the easier way

How about this? I learned this in physics class with a different example.

You weigh 6 and 6. Then 3 and 3 and then weigh 1 and 1 if they balance the last coin is the fake. If they dont the lighter or heavier one is a fake.

Group each coin into groups of 3 so there should be 4 groups

Possibility 1:

The first stack weighs the same so put that in one pile with each coin having the number one the second stack weighs has one that is lighter so that side is out but take the one that was heavier and write the number three on it so no you have three coins left so you take 2/3 of the coins and place one coin on each side of the balance if both coins weigh the same that means the only one that was not weighed is the fake

Possibility 2

The first stack has a side that is lighter so that is out that also means 9 coins are real so you write a zero on them so now you take 2/3 of the coins and put one coin on each side if they both weigh the same that means the last one is fake if not the one that is lighter is fake

Please like my thumbs really hurt

Won't the emperor be mad if you marked all the coins with pluses, minuses, and 0s?

But the Emperor is greedy, once he finds out that a coin is real he will take it immediately (even after one weigh). How do you solve the question now?

I feel like the king didn’t want to have the counterfeit coin out of his wealth. I mean, if he really wanted to, he wouldn’t limit your use.

“You better help me, but even though you have all the time in the world to do this, and since this’ll benefit me greatly, I’m going to give you only 3 chances to get it! If not I won’t let you help me anymore, and this coin will be in my treasury forever!

Marker ink has no weight apparently

You can do it without the marker, divide the coins in 2 groups of 6, the overall lighter group will contain the counterfeit, then divide 6 into group of 3 and again repeat the same process the one with counterfeit will be lighter, with 3 coins left, take any random 2 and weigh them if it balances then the third one is the counterfeit and if one is heavier than the other then the answer is obvious.

Then came cryptocurrency.

It also possible for 13 coins as well.

Just do this l = coin edit:

__= scaleA. B

llllll llllll

_________If A is lighter take it if B is lighter take it

Take your 6 and put it into 3 groups of 2

A. B. C

ll ll. ll – extra

______If A is lighter take it, if B is lighter take it, if A and B are even take C.

Take your two and place it on the scale.

A. B

l. l

__If A is lighter it is counterfeit, if B is lighter than its counterfeit

Please press read more it’s so much more simple

Split the 12 into six

Then do it for the lighter side and split it to 3 do the lighter side again except do 2 coins and 1 of 2 things will happen

1 one of the last to measured and one is lighter that's the counterfeit

2 they balance making it so the remaining one is the counterfeit

Too many combinations. The assumption that i ll be selecting the heavier coin in case in first measurement the balances are not equal…is actually a probability of 0.25. In short i wont be able to find in exact 3 attempts.

I'm pretty sure u guys used the scale more than three times

Wouldn’t u do 6v6 then take the lighter/heavier pile out then do 3v3 and take the lighter/heavier pile and do a 1v1 and if one of those are heavier/ lighter then u know and if they are both the same then the one u didn’t weigh is the imposter

Another way: balance 8 coins with 4 on each side. Let’s say they balanced. Put a mark on all of them that you just weighed (1 use of balance).Put the 4 other coins on with 2 on each side. One side will not balance. (2 uses of balance) put the 2 coins on the scale with 1 on each side. Whichever one you choose gives you a 50% chance of it getting right. That’s all I got. Can’t be precise but you have a chance

Split them into piles of 6. The one thats heavier or lighter is the one with the counterfeit. Then split that pile into 3 and 3. The heavier or lighter one has the imposter coin. Weigh tow of the last pile and if they balance out the last one is counterfeit and if they have different weights the heavier or lighter on is the counterfeit one.

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!

Split the coins: 4 4 2 2 1 1

I'm actually learning problem solving skills from the other riddles

Does this work? Do 6 by 6, R = real, – = possible, F = fake. Take that 6 and half it. Now it’s 3 left. Than put 2 coins in,

If it’s the same weight than the remains coin is fake. If it’s lighter than it’s the fake coin. It’s 3 moves on the scale. If I’m wrong let me know

I'm so confused. I divided it into 4 piles of 3s and even then you can figure it out. No? You weigh 2 of the piles. if they balance they are eliminated. If not they contain the coin and the other 2 piles are eliminated. Then you switch out one pile with the other. If they balance the the pile you removed has the coin. If it unbalanced, the pile you didnt remove has the coin. This will help you narrow it down to 3 coins either way. It will also help you figure out if the coin is heavier or lighter. Oh wait noooooooo.. if it balances you cant figure out the lighter or heavier thing.. sigh* ignore me…

No on actually solves this right?

Why can't we do it this way and without the marker!? And since the easiest explanation is the best one.

1. Divide 12 coins in to two groups and weight them,

2. Then the six lighter ones divide again in to two group of threes than weight them against each other,

3. At the three lighter ones you take two coins randomly and weight them and if they're equally the other unpicked coin is the fake one, otherwise the lighter coin is the fake one.

Am i missing something or is this the easiest way to do it!?

no

Step 1: Find random coin

Step 2: Say it's fake

Step 3: Done

Wait a second…

What if the marker’s ink or whatever was over?

I just did 6 on each side for the first time, get the lighter side, then weigh half on each side on the lighter 6, then get the lightest side, then get the heavier 3, and weigh 2, if they are equal, the other one is counterfeit, and the lighter one is in the scale, that ones the counter feit

these guys got 11 coins

Just split the coins in half, put one half on the left scale and the other half on the right scale. The scale will then become slightly unbalanced. Just keep taking out coins from both sides of the scale, until you can see the scale becoming in equilibrium. Then identify the coin you took out, which left the scale in equilibrium. That coin is the counterfeit one.

If it's balance–>unbalanced–>balanced there would be 2 coins left.

I can just split it into two piles of 6 and if one is lighter or heavier i'll split it into two piles of 3 and if one is lighter or heavier i'll put one coin aside and weight two coins, if they weight the same then the coin left aside is the fake and if they don't the fake one is the heavier or lighter one (my previous tries will determine if it's light or heavy)

Just do:

Weigh 6 coins on each side, pick the heavier/lighter

Weigh 3 coins on each side, pick the heavier/lighter

Weigh 1 on each, if there is one that’s heavier or lighter, it’s that one. Otherwise it’s the one that wasn’t weighed.

This was easy

This riddle is super sick. I even knew how to do it with 9 coins and only 2 weighings (you have to have the prior information if the fake one is heaver or lighter for that), which is basically a much easier version of this… and lots of smart people already struggle with that. But this takes it to a whole new level! Mind == blown

The emperor threw you in jail for criticizing him? Use your few moments of freedom to overthrow the government.

The emperor looks like that guy from Chowder.

What about the weight of the markings?

Guys i got a question. i have found the answer but i dont think i found by logical thinking but only think and trying many ways. i first tried to divide them into 6 coins then i realized it is useless, tried divididng them into two's and then realized i dont have enough oppurtunity to balance them. in the end, by trying like these i reached the answer. But i dont think i done the job well, i feel like i was a very slow computer(computers are like millions of babies thinking true or false just that). My question is; how i can improve my thinking/problem solving? Not just mindlessly trying evey way but asking the real/usefull question and logical thinking?

I thought I was a sock

Divide it into 6-6 first, select the lighter lot, divide it into 3-3 again, select the lighter lot, weigh two out of the three, if same then third one is fake, if not then whichever is lighter.

This is the exact same riddle Captain Holt asked his Detectives.

Captain holt wants to know your location

But then as a progressive Mathematician he still pointed the wrongdoings of the Emperor with its tax collection was sent to jail once again.

Plot twist, all 11 are counterfeit and 1 is real.

Theres one more way

Just divide the stack into 6 and 6

In the first try. The side which is heavier or lighter should be taken.

In step 2 divide the 6 into 3 and 3.

Again the side which is heavier or ligther should be taken . Now u remain with 3…

In step 3 ,measure any 2 coins ..

If they bith weigh equal,the 3rd is the fake, or anyone which weighs lighter or heavier should be taken… Please tell me if I am wrong…

This is the Brooklyn 99 riddle with the islanders

That was easy:

Put six on each side, then remove the heavier side.

Then put three on each side then remove the heavier side again.

Finally, put two coins on the table: if the scale is balanced, then the third coin not on the scale is fake. If one side of the scale is heavier then the lighter side of the scale has the fake coin.

Why would the emperor be okay with you marking the coins?

Can’t we just… ya know

Pick it up?

If all the coins look a like, and you find the fake coin, how will you identify the counterfeiter? All the coins would get mixed up thus randomizing the identity of the payor