Can you solve the counterfeit coin riddle? – Jennifer Lu

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.

100 thoughts to “Can you solve the counterfeit coin riddle? – Jennifer Lu”

1. Tan Yao Yang says:

Nothing like a MARKER in ancient times

2. Balaji Varadharajan says:

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

3. ArchtansterPG says:

> emperor is an impatient man
> lets you write out all of the possibilities on the board

4. Kabir Singh says:

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

5. Querty Beighteen says:

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.

6. NAKUL chorey says:

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

7. juckingstar209 says:

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

8. YUR1CH4N 19 says:

“You are the realm’s greatest mathematician.”

bish i almost failed math when i was grade 7

9. Madi G says:

Instructions unclear

My cat is in my washing machine

10. gerardo says:

This la the same riddle as the on from the islanders un Brooklyn 99?

11. Colm Kelliher says:

Three words:

Sniff the marker

12. Steffi Bernice Sarrosa says:

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

13. Jonathan Do says:

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

14. A A says:

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.

15. Some Guy says:

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

16. Jon Parkes says:

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…?

17. J W says:

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

18. Folioscope cartoons says:

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.

19. NODSCELOT T says:

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.

20. LeonardoDiSnaprio says:

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

21. Roger Alcaraz says:

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.

22. angie btw says:

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

23. Alessandro Pinto says:

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?

24. Familia Mancilla-Marquez says:

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

25. SnoozeCruise says:

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.

26. Hal Bassett says:

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.

27. Freckled Angel says:

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

28. Soundwave says:

There's a better answer

29. Esmae Redwana says:

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

30. Ashstar Lit says:

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.

31. Robert Hamilton says:

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

32. kantapon nitisermwong says:

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?)

33. Bennitti howlett says:

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

34. achsadness j says:

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

35. Nguyen Max says:

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

36. 김예찬 says:

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

37. Dominic Quadrini says:

Captain Holt needs to watch this video.

38. Catherine Wright says:

This took me five months to solve

39. Riceball and Lettuce says:

Me:just lock me up at least free fooddddddd

40. RyanHung says:

Another math question, ain’t I right?

41. Little tato says:

why don’t you just throw a magnet and if the coins that doesn’t stick, it is fake boom easy solve

42. 枕頭神 says:

We 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!!😂😂

43. System 3 says:

Or, establish a federal reserve and just print.

44. Lucas Possatti says:

"Can you solve the counterfeit coin riddle?"
Me: "Well… Back to the dungeon…"

45. Angel_playz 18 says:

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

46. joe says:

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

47. Kolby Saurer says:

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?

48. WeezyUnknownWeezy Ssj9k says:

You lost me at coin

49. Piero Rio says:

"Just the coins, a balance scale, and a marker."
Table: Am I a joke to you?

50. Yanny Kojongian says:

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

51. CatGirl The Little Retard says:

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.

52. R. Obias says:

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

53. Dhruva V says:

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

54. Raz Menashe says:

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?

55. The one who sucks says:

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!

56. Gao Industries says:

Marker ink has no weight apparently

57. shubham gupta says:

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.

58. Shreerang Vaidya says:

Then came cryptocurrency.

59. Андрей Чернышев says:

It also possible for 13 coins as well.

60. K Kothari says:

Just do this l = coin edit: __ = scale

A. 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

61. K Kothari says:

Please press read more it’s so much more simple

62. SANE SSS says:

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

63. candyfloss184 says:

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.

64. Ira Ballal says:

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

65. Rohan Vig says:

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

66. XxDarkNightmarexX 999 says:

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

67. Hyped Nation says:

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.

68. Elizabeth Martin says:

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!

69. Joseph M says:

Split the coins: 4 4 2 2 1 1

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

70. Hamster Main Here :3 says:

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

71. Zahrah Azhar says:

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…

72. art_lover says:

No on actually solves this right?

73. Egzon Sadriu says:

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!?

74. Kellvey Teruh Hinokuma says:

no

75. TMJ says:

Step 1: Find random coin
Step 2: Say it's fake
Step 3: Done

76. Pavan RVinod says:

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

78. Random Chicken says:

these guys got 11 coins

79. Kevin H says:

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.

80. Ricardo Zuniga says:

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

81. awesoman says:

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)

82. 10million subs no video challenge says:

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.

83. Colin Smith says:

This was easy

84. futurefox128 says:

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

85. Daniel Farbowitz says:

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

86. ITSmeSHUJI says:

The emperor looks like that guy from Chowder.

87. 이희정 says:

What about the weight of the markings?

88. miredrinn says:

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?

89. TheCracker Master says:

I thought I was a sock

90. Swatilekha Dutta says:

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.

91. FrostyRPGT says:

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

92. Bertie Pryke says:

Captain holt wants to know your location

93. Einstein Lol says:

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

94. Techspeed Gaming says:

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

95. Ajay Pandit says:

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…

96. Rowan says:

This is the Brooklyn 99 riddle with the islanders

97. Airwolf 276 says:

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.

98. Justin Shen says:

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

99. Tyler Liebert says:

Can’t we just… ya know

Pick it up?

100. brent tiu says:

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