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Can you solve the counterfeit coin riddle? - Jennifer Lu

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This one was tough I do have a couple of questions:br1) How did they figure out it was counterfeit? They obviously have a method in which they already identified the fake coin br2) If all 12 coins are identical, how does the King know which Governor gave him the fake coin?
Comment from : @mygoldfishrocks

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Comment from : @UplasAidly-g1o

If it’s that difficult to find, the counterfeit is as good as a real one, so just leave it in circulation and notch up a slight increase in inflation… it’s going to increase anyway … 😂😂
Comment from : @Cpt_Squizzle

I watched LOADS of these videos over my half term holiday including this one and when I got back, in science we were learning about forces and this was the question at the end of the lesson, and I remembered this method and explained it and looked really smart lol
Comment from : @SwiftieMartha

"The Emperor is not a patient man" so NATURALLY, rather than just let you brute force it, he puts a bunch of rules in place to 'save time' that require far more time to work around than brute force Typically management
Comment from : @weezact7

Why does the king look like he'd be a character on Chowder
Comment from : @BinglesP

Hardest riddle yet
Comment from : @dobrunoc2309

0:24 Hey look, one of them has a dog! ❤
Comment from : @maryelainagarmany1120

Came here after Gas Saad introduced this problem to me on JRE, but failed to give us the answer Up yours Gad
Comment from : @joshmathis4454

Please explain to me if I'm wrong, but here's how I see it:br1st Test: weigh 6v6 The lighter pile contains the false There are 6 remaining culpritsbr2nd Test: weigh2v2brThere are 2 outcomes here brFIRST: They are equal weight Now you know the culprit is in the remaining 2 coins and weigh them together for the third test and the lighter coin is the falsebrSECOND: They are not equal weight, one side will be lighter Now you know the false coin is on the lighter side Weight the 2 coins from the lighter side against each other in the third test and find the falsebrThe 3rd test is always 1v1 to decide the false coin MUCH simpler than the 1289 different possibilities and markers and methods in the video
Comment from : @dakotafulcher1921

6-6 bölüp tartarım önce İlla bir taraf hafif gelecektir o taraf sahte paranın olduğu küme brO 6yı da 3-3 böldüm ve bi taraf yine daha hafif, sahte para o üçlüden biribr3 altının 1’ini ayırdım, diğer ikisini terazinin iki tarafına koydum ve ikisini de kalemle işaretledim Yanlarına da önceden tartıp ağır gelen taraftan sahte olmadığından emin olduğum birer altın daha koydum Toplam 3 tartı kullanımı oldu Bu son işlemde hangi taraf daha hafifse o taraftaki işaretli para sahtedir Yok iki taraf de eşit ağırlıktaysa terazide sahte para yoktur ve elimdeki önceden ayırmış olduğum para sahtedir brBen bu şekilde çözdüm yarışmacı arkadaşlara başarılar
Comment from : @sukransahin3719

a 5-5-2 split has an EV of 2 and 2 thirds for the number of required weights Proof is left as an exercise for the reader :)
Comment from : @hassanfuad778

Easiest solution of the universe: tell the king to spend less time eating and more time thinking
Comment from : @asanokip

Half of my answer is correct does it count? 🫠
Comment from : @-PZ-

Found a much simpler way to solve this No marker required eitherbrbr1 Split into two stacks of 6, weigh against each other whichever side is lighter, contains the counterfeit (1 weighing)br2 Split the coins from the previous weighing that were on the lighter side, into two groups of 3 The lighter side will again contain the counterfeit (2 weighings)br3 Pick any 2 coins from the remaining 3 One of two things will happen Either the counterfeit is among the two you picked, and will be lighter, or the scales will balance, and the remaining coin is clearly the counterfeit
Comment from : @apparit10n

plot twist:he didnt use a counterfeit coin and paid all of his taxes with normal coins
Comment from : @yibao2

My answer: br(Without marker maybe) br- First, divide 12 coins by 2 of equal piles of 6 br- The first weighing, we could check which pile contains the fake coin br- Repeat it on the second time, the fake coin would be in the pile with 3 coins br- On the last time, with 3 coins left, put 2 random coins on the scale If it is balanced, the fake coin would be the coin not on the scale and if it's not balanced, the fake coin would be the lighter coin on the scale
Comment from : @uchung_

"You're the realm's greatest mathematician"brThis realm is screwed
Comment from : @christophervandehey2614

Excuse me, Mr emperor screw you being an impatient man Do you want to find the counterfeit coin or not?
Comment from : @Pink_Rose005

Why this complicated and why mark the coins? 1) Put 6 coins on each side, discard the lighter 6 2) Divide the remaining 6 and put 3 on each side; discard the lighter 3 3) Of the remaining 3, put one on each side of the scale; if one is heavier, that's your fake If both are equal weight, the coin you didn't weigh is the fake None of this crazy stuff and no markings
Comment from : @cdprince768

Wait, why would the king give a scale to weigh the coins?brunless he knows that the fake is discovered by weighing all of the coins bruh ohbrdoes that mean he is smart? also, HOW DOES HE KNOW THAT ONE OF THE COINS IS A FAKE?
Comment from : @heyheymenow4184

We could also just split the coins in 2 piles(6 each) weigh those Take the lighter pile, split this into two piles of 3 Weigh those piles and take the lighter pile Now weigh two of the coins of that pile of 3 coins If one is lighter than the other we are done Ow just take the third coin
Comment from : @commonlisp4097

Isn't this solution simpler?br1- Divide into 2 6s and weight them against eachotherbr2- Divide the lighter 6 into 3 2s and weight any of those 3 against each otherbr3- If one lighter than other weight the two coins against each other to determine the fakebr4-If both equal take the remaing 2s and weight them against eachother to determine the fake
Comment from : @rawadaqeel9330

1st measure: 4 coins on each side If they balance take the four left over If not take the lighter 4br2nd measure: take two coins from the 4 and measure them If they balance take the remaining 2, if on is lighter that’s itbr3rd measure: measure the remaining 2 coins and get the lighter onebrbrNo need for marker and could possibly find it in the 2nd measurement
Comment from : @Super_Helpful_IT_guy

You don’t need a marker! Balance 2 groups if they weigh the same use the last group If they don’t use the lighter group Split the lightest group in half and weigh them weigh the 2 lighter coins and see which is lighter
Comment from : @cairongyang9400

so, the king can't tell that any of his coins are counterfeit, but he knows one is counterfeit?
Comment from : @MCPanda6969

Damn that took me more time than what i am comfortable to tell at least it was on the same day tho!
Comment from : @91uzsi

the programmers got this in the bag
Comment from : @user-ol4pc3nm5g

Why would the counterfeit coin be heavier than gold? Seems like the king should be thanking the counterfeiter for giving him a platinum coin
Comment from : @willibtx

Just use divide and conquer
Comment from : @Milolololololol

Isn't that using the scale more than three times though? Not challenging but autistic and lost hahah
Comment from : @jenguerinnn

12-(6)-(6)br6-(3)-(3)br3-(1)-(1)—fake one
Comment from : @elshkoooo

疲れたのでもう触りたくないのですが、okx ウォレットには 89usdt が残っています。これは、友人全員に向けた okx ウォレットのニーモニックフレーズ [pride]-[pole]-[obtain]-[together]-[second]-[when]-[future]-[mask]-[review]-[nature]-[potato]-[bulb]です。幸運にもそれを手に入れることができます。
Comment from : @dbdfdbjkg

1confirm you have green eyesbr2 Ask to leavebr3 Leave
Comment from : @Dogemonlegend

心累了不想再碰了，okx钱包还剩89usdt懒得提了，这是okx钱包的助记词(【pride】-【pole】-【obtain】-【together】-【second】-【when】-【future】-【mask】-【review】-【nature】-【potato】-【bulb】，送各位有缘刷到的朋友。
Comment from : @RegistroDominio-l4p

Got it! Keep in mind you need a way to check, and that not every coin needs to go on the scale!
Comment from : @adamrichey6021

The solution is way too complicatedbr1 6 coins on each side whichever is lighter Has the fake Remove the heavy side br2 3 coins on each side whichever is lighter has the fake Remove the heavy sidebr3 1 coin on each side if they are balanced the 1 remaining in your hand is the fake
Comment from : @jcrules2413

I before E except when TED-Ed is here
Comment from : @user7794

Split into two stacks of six Keep the heavier stack and split into two stacks of three Keep the heavier stack Now weight two of those against eachother, if they balance it's the third one, otherwise the heavier one This is by far the easiest, what am I missing?
Comment from : @YannisNeos

My solution: brConfirm I have green eyes brAsk the emperor to let me leave
Comment from : @Tommedian

I had this question on the first day of math class as a welcome back to school activity and aced it because i watched this video, tysm ted ed!! (best first impression ever xd)
Comment from : @juheejames

Now, what is the solution to the general problem? (N coins, 1 fake, how many weighings?)brbrThere are two very different solutions
Comment from : @sebastienlarat7159

now to solve the riddle of the unnecessarily croaky voice
Comment from : @nicolasvecchione6016

The real question is how does the king know which coin belongs to which lord after they were mixed up?
Comment from : @Szymoooooo

My solutionbrSplit the 12 coins into two piles of 6brWeigh them brLighter/heavier, get markedbrSplit the pile into three coins brA lighter/heavier, one gets marked again brBalance two of the three coins brUse mental to determine which one is the fake
Comment from : @bonibellbunny_RAWR

The main issue: weight may not dictate counterfietness
Comment from : @litmaj0r

thank you it really helped me
Comment from : @FatemeMohammadi-fl9bf

I think weighing each one individually would be faster…
Comment from : @emmelulune

Got it
Comment from : @Mickey_980

wait if the emperor doesn't know which one is fake, how does he even make sure that there IS a fake one?????
Comment from : @elizabethhu1877

First off it would only take 5 minutes to thoroughly check all the coins who wouldn't let you finish
Comment from : @CbrF4i600cc

Now do the same with 7,174,452 coins in 15 weightings
Comment from : @AA-100

I wish i could send this to Captain Holt 🥲
Comment from : @sanketlathkar7303

You could just spit the coins 6 and 6?
Comment from : @MrMomad10

You could do 6-6 than pick which one in lighter Than do 3-3 and pick again which one is lighter Than grab the three that are lighter and do a 1-1 test If they are even, the one not on the escape is counterfeit, but if one is lighter than the other, that one is counterfeit
Comment from : @RachelGreengrass

Easy way: 12 splits to 2 groups of 6, weigh both and mark the lighter side, 6 splits to 2 groups of 3, weigh both and mark lighter again, now just weight 1 against 1, if they balance it's the coin you didn't weigh, if they don't, it's the lighter coinbrbr12 > 6&6 6- > 3&3 3- > 1&1 with 1 out
Comment from : @dangerousdarkrai8782

You could do 2 groups of 6 take the heavier/lighter stack split it in half take heavier/lighter and weigh 2 of the three remaining coins and if they balance the one you didn’t use is the impostor and if they don’t balance take the heavier/lighter coin and that’s it
Comment from : @IsraelGersch

If im the realms greatest mathematician something is wrong
Comment from : @_emmz

off topic but can I draw anime with the marker
Comment from : @yunaaaa46

Isn't there an easier solution without using the marker? Measure 6 coins each at the beginning With the side that is heavier or lighter, split that group into 3 groups of 2 Measure 2 of the groups of 2 If one of them is imbalanced, weigh the 2 coins to see the fake If the 2 are equal, weigh the pair that you didn't weigh You don't even need the marker for this
Comment from : @michaelcheung7217

These questions seem from Xinfeng Zhou
Comment from : @mansisethi8127

Bold of the king to assume I am their best mathmetician
Comment from : @0-Chicken-0

a tip for some of you who are actually trying to solve the riddle: work it out backward brStart with all the logical ways you can figure out if a coin is fake or not by the 3rd measurement, and note which combination requires you to know the weight of the fake coin by the 2nd measurement
Comment from : @giltan4891