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HARD Logic Puzzle - The Seemingly Impossible Counterfeit Coin Problem

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can't you just put all of the 100 coins on 1 plate?
Comment from : @nmrs4342

Yup, I'm glad I figured that out
Comment from : @prabaharanaece

If you have a real coin, the rest are 50 real & 50 fake If you have a fake coin, the rest are 51 real & 49 fake In each case if you weigh any 50 vs the other 50, in the first case, the difference must be an even number of grams, and in the second case an odd number of grams So when you weigh the other 50 vs 50 if the scale reads an even number of grams, state you have a genuine coin, otherwise state you have a counterfeit
Comment from : @patsk8872

And what if the counterfeit coin is 1/49 gramm lighter/heavier than the real one?
Comment from : @belagoblyos7822

here's a variation: What if the warden also wants to know if the counterfeit coin is lighter or heavier?
Comment from : @larryb9198

The Lateral Thinker: "Test the coins by hand, if it feels like it weighs the same as 50 other coins it's genuine and if it feels like it weighs the same as 49 it's counterfeit"brThe Statistician: "Guess that it's genuine because odds are that it is"brThe Logician: "Weigh a random 50 of the other coins against the other random 50 If the difference is even or 0 you have a genuine coin, and if it's odd you have a counterfeit coin"brbrThe Engineer: "Flip the coin to decide if it's real or counterfeit"
Comment from : @vitriolicAmaranth

Use it multiple times but replace the coins on one side at a time so you can argue that it's like the ship of Theseus so you've only used it one time
Comment from : @pvzpokra8602

I was thinking if they were the same, as in the same material, you can just see how many are bigger than the other ones
Comment from : @themessageman2760

You forgot to mention that all genuine coins have to the same weight, I first thought genuine and fake coins came in pairs with one genuine coin without a fake coin as a pair
Comment from : @jemandanderes7075

I did it you people!
Comment from : @MoldingMatty

Why can't we just put all 100 coins on 1 side of the weighing balance? If the number is divisible by 50 , we were given a genuine coin Otherwise,a fake one
Comment from : @killmonger4532

but you only know 1 of 51 coins now
Comment from : @kales901

I have a similar one : suppose you have 12 coins and you want to find the counterfeit in 3 tries the properties of the counterfeit is exactly the same you can draw on the coins and this will not change their properties thanks to ted ed for the riddle
Comment from : @Guys-s5v

just guess
Comment from : @smoceany9478

What if some of the fake coins are heavier by 1, and some a lighter by 1? The way the problem is stated- such situation is possible, but the solution doesn’t cover it
Comment from : @AlBert-ti3nc

The proof is bizzare, but the puzzle at once seemed hard, I solved it easily when I gave it a little thought
Comment from : @totalme302

So it wouldn’t matter if you had 20 counterfeit on the left and 30 on the right with genuine coins mixed on both?
Comment from : @hemp6473

This is not a hard problem, I solved it But it is to be noted that I have solved similar types of problems before
Comment from : @anibeto7

I tried the first method, but I am not a mathematical genius, so I got stuck But I think I found out an easier methodbrbrThere is 51 genuine coins, and 50 counterfeit coins Place 52 coins on a side, and there will be at least one counterfeit coin So place your coin on the other side If it's counterfeit, the weight difference will be 51 grams If it's not, so it's a genuine coin
Comment from : @johnassisdelima3193

The question is misleading The warden only gives you a single coin It never tells you you even have the option to weigh all of the others
Comment from : @critik_l

I think it also works and it's easier if you put all the coins (except yours) in the left pan, with the same results: Even: genuine Odd: counterfeit
Comment from : @Leankin

Simply remove one coin and put the other 100 on the balance, if the sum is even, the coin is genuine, if it is odd, the coin is fake
Comment from : @sergiosereno1489

Hello can anyone tellme if I solved this correctly ? brI did not understand the proof, I'm no good with math I also used odd /even to decide , but I put all 100 coins on one side of the scale and nothing on the other
Comment from : @dariogerardi4587

I just weighed all 100 coins in one pan If the difference is even I have a real coin If the difference is odd I have a fake coin
Comment from : @Tmwyl

good solution but there is a third possibility when the scale is balanced, and in this case the coin given is genuine
Comment from : @bushfires1760

I don't get it You don't know if you put the same number of light/ heaver coins in each pan, without an even split it won't work out
Comment from : @earlmcgill5867

Interestingly, I think this also works even if the counterfeit coins are a mix of heavier and lighter The weighing process is essentially a parity check
Comment from : @theroxenes

i thought to place all the coin (except for the one that the warden gave me) in one side of the balance and see if the difference is an odd or even number
Comment from : @tommaso1476

Excellent question! To tell whether your coin is genuine or counterfeit, you need at least one bit of information; Knowing whether the scale reads odd or even is exactly one bit of informationbrThe difference between two integers and the sum of them have the same parity, and therefore by putting 50 coins on the left and 50 on the right side of the scale, the reading should tell you the parity of these 100 coins' total weight, and therefore tell you whether the one coin in question is genuine or counterfeit Brilliant solution!
Comment from : @quantum-atom

Couldn´t you just weigh the coin you want to know against all other coins and get to a result, because a counterfeit coin would result in more comma digits as a genuine one?
Comment from : @Gatschknoedel

no way don't know weight of "real" coin what if coins were mixed? false proof
Comment from : @mudfloodwasarmageddon

Logic or mathematics?
Comment from : @alphabetsoup7918

Your solution seems flawed It's relying on putting the real coins in one pan and the fake coins in the other?brIf you can tell the difference between them already the there was no need to weigh them brYou would just pick one of the 51 and discard the other 50
Comment from : @rd9337

yeah seemed pretty straight forwardbryou either end up with an odd or even number of fake coins and their weith is the only relevant variable
Comment from : @TJMiton

"Evil" warden? Almost 50/50 chance to escape prison without using Math
Comment from : @aFruitDealer

Balance Scale Nonesene?brI got caught up because I didn't realize it was said that the Scale would tell me the DIFFERENCE in weight I had assumed it would only tell me Balance vs Not Balanced, but not the DifferencebrI don't believe the hypothetical scale in the question can physically exist Balance scales are used by comparing a Test Object vs a Calibrated Weight if level they are equal, it not level they are either heavier or lighter, but it CANNOT tell you the exact difference in weightbrSince the scale in the question couldn't exist, I couldn't imagine the hypothetical usage of it
Comment from : @adriansue8955

Or use Occam's razor Put 50 in each pan If it balances the coin is genuine, if not, it isn't Don't make a big project out of a simple question
Comment from : @iggybeauchance7406

It isn´t that tough
Comment from : @charlesbromberick4247

You don't even have to split the coins 50/50 You can place all of the hundred remaining coins on the left side of the scale and you'll still get even when you're given a genuine coin and odd when you're given a counterfeit coin
Comment from : @realnazarene5379

I won the Squid Game because I knew this
Comment from : @thatonechick7569

Appearently they were an avarage 102g difference and you screwd up by a rounding issue - life sucks
Comment from : @christian9540

Easy easy
Comment from : @dweeds

I didn't go as far as to figuring out the solution, but I figured the solution was from the 50/50 or 51/49 imbalance in the remaining heap of coins
Comment from : @ivanscottw

blogic/b puzzle with a bmathematical scheme/b brAight
Comment from : @TramvayValera

The description says "Each counterfeit coin is identical to a genuine coin, except" brThis means each individual fake coin can weigh anything and that there are 50 real coins that are counter-parts for those fake coins There is also one wild real coin that can weigh anything So the mass of the 51 real coins is not 51x and the mass of the 50 fake coins isn't 50x±1 I'm still staring at the puzzle description and trying to figure it out Atm it seems very much impossible but I bet I'm missing something very obvious
Comment from : @ville-mattiamonsen2092

Surely this is made too complicated - if your coin is genuine the scales will balance if it isnt they wont
Comment from : @MarcDebenham

I got this one!
Comment from : @RJ-Isaac-TSOML

Well, there’s gonna be a lot more crime in our communities now, just as soon as this information spreads throughout all our prisons, and I’m sure it will, soon I hope you’re happy, you jerk It was very irresponsible of you to post this video, and you should be ashamed of yourself This needs to be removed, immediately, and you need to be punished, to the fullest extent of the law You deserve to go to prison for this, to be honest Unfortunately, we all know that would be pointless, though, of course
Comment from : @chriswebster24

If I were that evil warden, I would grab at random one of those math problems (like this one) and offer them to my prisoners
Comment from : @usptact

If the real coins weigh an even number of grams apiece, the fake coins will weigh an odd number, and the total weight will be an even number If the real coins weigh an odd number of grams apiece, the fake coins will weigh an even number, and the total weight will be oddbrThat's about as far as I got
Comment from : @safrprojects

😌😌😌😌💙💙😌
Comment from : @mintudas5724

50 of the other coins on one side, 50 on another In case yours is a genuine coin there will be the same amount of each type If 25 fake are on both the difference is 0, switch one genuine with one fake and you have 24 to 26, switch two and you have 23 to 27 basically the difference is always an even number because moving a coin changes it by two (Or to but it differently since the sum of fake coins is even splitting them in two mean either both are even or both are odds) Similarly if yours is a fake the difference between fake coins will be an odd number Same goes for the difference in gram
Comment from : @user-zj9rr6yc4u

Why can't I just put all of the coins except mine on one pan, and it'll be 1050 if my coin is genuine?
Comment from : @pythagorasmathematik9585

if there is tons of if in the question avoid it That doesnt make sense at all
Comment from : @keremyaka1

i can hear clock in background
Comment from : @mad4hat

Can you please clarify? 101 coins How does it work when you have 25 of each coin Fake and real on each side of the scale Even weight You still don't know if the coin in your hand is real or fake
Comment from : @bryblaze

Are the coins mixed or are they separated?
Comment from : @behrensf84

Just reduce it to three coins - the rest are all good forget any more than 3
Comment from : @edstefaniuk5822

given what life has become now, I'd probably go with 50/50 guess
Comment from : @ilet3049

I think that that evil is you in ur previous birth That evil is too genius like u
Comment from : @babluji8893

About half way, I guess
Comment from : @christopherreese4546

I, pole 3 out, to solve it
Comment from : @christopherreese4546

The easier solution is to hold the subject coin and compare 50 coins in each pan If the result divided by 50 is an integer the subject coin is genuine, if fractional the coin is counterfeit
Comment from : @SirMarkey

I know, it is beside the point, but scales (at least the ones in the video) do not work that way You will get the same reading, if you double the weight on each side That is why you normally only use the information, which side is heavier, but not how much It only works, if you have a fixed (known) weight on one side
Comment from : @Highlander07