Title | : | Can you solve the counterfeit coin riddle? - Jennifer Lu |
Lasting | : | 4.35 |
Date of publication | : | |
Views | : | 8,9 jt |
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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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I appreciate you sharing that! I have a quick off-topic question: I have USDT in my bybit wallet, and I possess the 12-word passphrase: host hair point assault nominee walk plate focus idea analyst metal liberty What is the process to transfer my USDT to Binance? Comment from : @UplasAidly-g1o |
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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 |
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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 |
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"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 |
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Why does the king look like he'd be a character on Chowder Comment from : @BinglesP |
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Hardest riddle yet Comment from : @dobrunoc2309 |
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0:24 Hey look, one of them has a dog! ❤ Comment from : @maryelainagarmany1120 |
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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 |
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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 |
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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 |
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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 |
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Easiest solution of the universe: tell the king to spend less time eating and more time thinking Comment from : @asanokip |
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Half of my answer is correct does it count? 🫠 Comment from : @-PZ- |
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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 |
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plot twist:he didnt use a counterfeit coin and paid all of his taxes with normal coins Comment from : @yibao2 |
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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_ |
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"You're the realm's greatest mathematician"brThis realm is screwed Comment from : @christophervandehey2614 |
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Excuse me, Mr emperor screw you being an impatient man Do you want to find the counterfeit coin or not? Comment from : @Pink_Rose005 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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so, the king can't tell that any of his coins are counterfeit, but he knows one is counterfeit? Comment from : @MCPanda6969 |
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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 |
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the programmers got this in the bag Comment from : @user-ol4pc3nm5g |
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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 |
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Just use divide and conquer Comment from : @Milolololololol |
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Isn't that using the scale more than three times though? Not challenging but autistic and lost hahah Comment from : @jenguerinnn |
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12-(6)-(6)br6-(3)-(3)br3-(1)-(1)—fake one Comment from : @elshkoooo |
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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 |
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1confirm you have green eyesbr2 Ask to leavebr3 Leave Comment from : @Dogemonlegend |
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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 |
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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 |
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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 |
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I before E except when TED-Ed is here Comment from : @user7794 |
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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 |
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My solution: brConfirm I have green eyes brAsk the emperor to let me leave Comment from : @Tommedian |
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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 |
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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 |
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now to solve the riddle of the unnecessarily croaky voice Comment from : @nicolasvecchione6016 |
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The real question is how does the king know which coin belongs to which lord after they were mixed up? Comment from : @Szymoooooo |
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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 |
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The main issue: weight may not dictate counterfietness Comment from : @litmaj0r |
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thank you it really helped me Comment from : @FatemeMohammadi-fl9bf |
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I think weighing each one individually would be faster… Comment from : @emmelulune |
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Got it Comment from : @Mickey_980 |
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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 |
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First off it would only take 5 minutes to thoroughly check all the coins who wouldn't let you finish Comment from : @CbrF4i600cc |
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Now do the same with 7,174,452 coins in 15 weightings Comment from : @AA-100 |
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I wish i could send this to Captain Holt 🥲 Comment from : @sanketlathkar7303 |
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You could just spit the coins 6 and 6? Comment from : @MrMomad10 |
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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 |
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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 |
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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 |
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If im the realms greatest mathematician something is wrong Comment from : @_emmz |
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off topic but can I draw anime with the marker Comment from : @yunaaaa46 |
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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 |
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These questions seem from Xinfeng Zhou Comment from : @mansisethi8127 |
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Bold of the king to assume I am their best mathmetician Comment from : @0-Chicken-0 |
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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 |
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I just had a test with a question very similar to this Thanks TedEd for saving me 😅 Comment from : @polaquinha |
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0:45🥬🍐 Comment from : @As_Sulay |
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hopefully it's not permanent marker Comment from : @AnLamHousehold |
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my 6th grade teacher asked us this Comment from : @uncool254 |
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I’m only watching this cuz of Brooklyn 99 Comment from : @poopfartpoopy |
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