HARD Logic Puzzle - The Seemingly Impossible Counterfeit Coin Problem

An evil warden holds you prisoner, but offers you a chance to earn your freedom. You are given 101 coins, of which 51 are genuine and 50 are counterfeit. Each genuine coin is identical. And each counterfeit coin is identical to a genuine coin, except that it differs in weight by exactly 1 gram (all are lighter or all are heavier, only the warden knows). The warden gives you a randomly selected coin from the 101 coins. You have to guess whether it is genuine or counterfeit. If you are incorrect, you are imprisoned forever. If you are correct, you are set free. You are given a weighing balance that displays the difference in weight between its left and right pans (for example, if the left pan has 8.3 grams and right pan has 10.3 grams, the display would show -2 grams). You can weigh any of the 101 coins, but you only get to use the weighing balance once before you have to guess. What is your best strategy to identify your coin and escape the prison? Watch the video for a solution.
My blog post for this video
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Source
Putnam-style practice from Olena Bormashenko
www.ma.utexas.edu/users/olena...
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[@kennkong61]

I'm beginning to sense a theme here. Apparently, if you like creating logic problems, you develop a desire to imprison illogical people for life.

[@user-kl8vj2qy9w]

😂😂😂😂😂

[@ArnavJainprofile]

There's actually logic behind this as well since illogical people won't be able to use logic to get out

[@miguelheat]

😂😂👏👏👍

[@JohnMoseley]

It's the nightmare AI future.

[@charlesbromberick4247]

That´s like saying the Democrats created Covid so that the (anti-science, anti-vax) Republicans would kill each other off - I don´t believe the Democrats are that smart.

[@artweaver6963]

It never ceases to amaze me how complex math proofs can get when expressing the simplest logical concepts! I almost instantly realized the solution. But the proof gave me a headache!

[@Buphido]

The proof honestly doesn’t have to be that complicated. Here’s an alternate logical proof: After splitting the 100 remaining coins into two heaps of 50, assume the left coins are all genuine. Case 1: Weight is 50 or -50, then the right coins are all counterfeit, then your coin is genuine Case 2: Weight is 49 or -49, then the right coins are all counterfeit with one genuine, then your coin is counterfeit So for all left coins genuine, even number implies genuine and odd number implies counterfeit. Now assume you move a random coin from each scale to the other. Let w be the initial weight displayed. If both coins are of the same type, w will not change, if they are of differing types, one side will grow lighter by 1 and the other heavier by one for a total difference of 2. Hence after any number of swaps, w and the new weight w' can only ever have a difference of -100

[@ziyad1809]

This is an extremely helpful video! Just the other day, my brother in law was kidnapped and was only let free after he successfully solved this riddle (by guessing).

[@sceylo9573]

And I was doing the math like a dumbass... I talked with your brother outside the prison ever since we teamed up on the crime we were arrested for. He told me he was smarter than me and just did the math quicker. wp to your brother I guess

[@yourcurtainsareugly]

I always love it when these videos contain the words "let's go through a proof" rather than "viral on Facebook".

[@sylvaeon860]

+1 The "viral on Facebook" ones are like 2 + 2*2 = ???

[@vicr123]

Which is undoubtedly 6, correct? How does one argue? :)

[@WhovianMinecrafter]

I know, they should at least do us the benefit of making it a function, so f(a,b)==???. That way they could make the function something weird.

[@geryon]

One can always argue. There are three question marks implying the answer must be three digits. So the answer is actually 110 which is 6 in binary.

[@vicr123]

geryon Well... 6(sub 10) = 110(sub 2) so you could _technically_ replace 6 with 110(sub 2) :)

[@unclemunch]

The formulas in the middle of the video likely perplexed a number of people. He is how to figure it, without knowing a lot about equations. You don't have to think too long to realize you have to weigh 50 against 50. Any less cannot show anything, with just one weighing. Set aside the coin you chose and place 50 coins on each side. If you chose a real coin, there will be 50 counterfeit coins. Regardless of how they are distributed, 50/0, 49/1, 48/2, etc., the scale will always show an even number, since it displays in one gram units the difference in counterfeit coins on the two sides . If you chose a counterfeit coin, there will be 49 left. Any way they might be distributed, 49/0, 48/1, 47/2, etc., the scale has to show an odd number. There is nothing wrong with using math, and I did understand the equations, but you can get the same answer by just thinking it through.

[@sonalidasgupta3562]

...now if you only write up the distribution of genuine coins too.....it would become more wordy and ...for me...more complex than the equations...but..to each his own.

[@rantlord8373]

Nice one. Much more logical. Thanks

[@johnmneu]

You don't even need to think that far, just consider the case that you are given a genuine coin and anything that isn't 0, 2, 4, 6... difference on the scale (that is even diff) has to be counterfeit by default. Note the special case of zero difference on the scale when you have a genuine coin and you split the remaining 100 exactly on the scale: 25/25 on one pan and 25/25 on the other.

[@howtungchong174]

Just put the coins in a bag and use it to whack the warden.

[@RedsBoneStuff]

It doesn't matter.

[@clieding]

Ah! The Alexander the Great solution to the Gordian Knot.

[@kylerivera3470]

Where are you getting the bag though?

[@typingcat]

@@kylerivera3470 I thought of the exact same solution. You can use one of your your socks.

[@ymj5161]

Then you will meet the warden in prison because of corruption within no long time.

[@engi98]

Phew, I'm so glad I know this, now I know I'll be safe next time the evil warden finds me

[@colinbrash]

Not sure if you’ve been keeping up but the evil warden has many more devious puzzles, unfortunately...

[@user-rw9no4vt7e]

@@colinbrash As the evil warden I can confirm

[@jeffreycanfield1939]

No, I can't solve it. Thanks for asking, though.

[@Qermaq]

Looks like we're cellmates forever, man. I couldn't either.

[@xereeto]

Look on the bright side, you still have a 50% chance at freedom...

[@vidicate3963]

+xereeto Is it 50%? _stirring the pot, mwahaha_

[@blackbacon08]

Is it me or does his name at 0:00 sound like "Fresh Tall Licker?

[@mickgough3574]

MistaTwoJeffreyTwenty Yaay a

[@thartwig]

the directions need to be a little clearer, upon trying too solve I assumed I would only be able to weigh one other coin since it does not specify. "You can weigh any of the 101 coins..." should say any NUMBER of the 101 coins.

[@gimmydicrosta9817]

well...it clearly says you can use the scale only once.Doing what you say means using it multiple times

[@chrisdeep8417]

By the same logic could argue that he had to specify "any one of the 101 coins..." in order for your assumption to be true. It is a bit ambiguous but there is a clear hint in the brackets as it mentioned -2. This means at least 4 coins were weighed which is more than one per side.

[@sustainablelife1st]

You are correct, the way it is worded implies you may weigh any (one) of the coins, one time. A math genius may have written the puzzle, but he clearly failed grammar.

[@sustainablelife1st]

@@gimmydicrosta9817 No, the way the puzzle is worded means that you may weigh any one coin, one time only. You are extrapolating.

[@rybiryj]

@@sustainablelife1st It literally says you can weigh any of the 101 coins, not "any one coin" as you wrote

[@superWTF64lol]

I remember a puzzle like this in a Professor Layton game. It has a very clever solution like this one, and is one of my favourite puzzles.

[@davidjames9935]

I would tell the wardon yes, the coin you gave me is either genuine or counterfeit. My answer of yes should be correct 100% of the time. Ahhh, sweet freedom!

[@jialiu805]

David James except you've answered the question in the wrong context which makes your answer incorrect.

[@davidjames9935]

Wrong context (my ass).

[@fejfo6559]

He answered the question the warden asked him so his answer was correct.

[@vkulanthaivel]

David James A BADA*S EQUATION?

[@SlyMaelstrom]

You'll be etching that on the wall of your cell for the rest of your life.

[@markosp1816]

i have only one question..Why did i become a prisoner?

[@Trias805]

Maybe aliens placed you in their zoo

[@markosp1816]

Trias00 then why they let me out? maube they dont want to have inteligent life in their zoos

[@Trias805]

My name is Jeff They lied about letting you out. Next thing, they're gonna test their meds on you. Or just make you do tricks for their amusement.

[@markosp1816]

AstoundingPilot -SW- are you the meme police?

[@markosp1816]

AstoundingPilot -SW- oh god i see now...

[@PhizetaHD]

Fuck it. I'd guess genuine, just cuz of the higher chance.

[@dariobarisic3502]

What if evil warden guessed you'd do that so instead he gave you counterfeit coin?

[@PhizetaHD]

The question said he gave me one at random.

[@dariobarisic3502]

***** fair enough

[@zerothefaceless4888]

so you'd bet your freedom on 51/101 chance?

[@smacinnes15]

Forget about whatever happens to be going viral, do more like these!

[@Pumbear]

Well since the guy apparently wrote REAL and FAKE on the coin I'd geuss I'd say whatever is on my coin.

[@yurenchu]

He wrote REAL on one side of each coin, and FAKE on the other side. "Each counterfeit coin is *identical* to a genuine coin, except that it differs in weight".

[@R3lay0]

@@yurenchu I'm quite sure hasn't the power to decide to change a coin design

[@kylerivera3470]

What if he wrote "I give up, kill me now" on your coin?

[@anujmishra9077]

One of the greatest puzzle I've ever solved... It's too beautiful... The person who framed it must be very genius... Thank mindyourdecision for uploading such videos... your channel is one of my favorites...

[@christianfunintuscany1147]

I am sure Presh will have a nice picture to explain the solution, I can just tell what was in my mind: my strategy would be to devide the 101 coins into three groups: group A = 50 coins, group B = 50 coins and group C = 1 coin. Let R means “real” and F means “fake”. By weighting A and B it is possible to tell if C is real or fake. There are two cases: 1) A and B = 50R + 50F C = 1R 2) A and B = 51R + 49F C = 1F In case 1) when I weight A and B I can measure a difference in weight that is always even: for example on the left pan I can have 4 R and 46 F while on the right pan 46 R and 4 F measuring a difference of 42 grams. All the other combinations give an even difference, because C = 1 R and a equal number of R and F coins are left in the A and B groups. On the other hand in case 2) the difference in weight that I measure is always odd. So by measuring A and B whatever the coins are distrubuted I can deduce if C is real or fake.

[@jsignuy]

Another simpler explanation. Take 50 on each side. Suppose there're 25 counterfeit on each side, it's trivial that the difference is zero. Then move one to the other side, the whole subtraction of the panes moves by -2 (or +2). If we keep doing this (moving by one coin) we show that in any possible distribution of the initial 50/50 setting the parity remains the same.

[@Stevearnell1]

I'm sure you know you can only do ONE weighing?.. so what you're saying is that these are all the possible outcomes assuming you picked a GENUINE coin - always an EVEN difference in the weight between the 2 pans containing 50 coins each. You need to state the converse - an UNEVEN difference would always mean you picked a FAKE coin, making your decision simple...as proved in the video. This was very clearly explained by Ricardo Ricky above.

[@goldenwarrior1186]

@@Stevearnell1 Switch the genuine coin in ur hand with a counterfeit from the scales. Now there will be an odd number of counterfeits instead, and the parity always remains the same

[@theroxenes]

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.

[@jumpman8282]

Here's how I did: Let 𝐖 represent the weight of a genuine coin. If the coin we're guessing on is genuine, the rest of the coins are made up of 50 genuine coins and 50 counterfeit, and their weight would add up to 100𝐖±50. If the coin is counterfeit, the other 100 coins will weigh 100𝐖±49. Now, we divide these 100 coins into two piles, so that each pile holds 50 coins. If there are 𝐍 counterfeit coins in the first pile, then the other pile holds either 50-𝐍 or 49-𝐍 counterfeit coins, depending on whether the coin we're guessing on is genuine or counterfeit. The weight of the first pile is 50𝐖±𝐍, while the second pile would weigh either 50𝐖±(50-𝐍) or 50𝐖±(49-𝐍). This means that if we put the first pile on the left pan of the weighing balance and the other pile on the right pan, the display will show either 1) 50𝐖±𝐍-(50𝐖±(50-𝐍)) = 50𝐖 - 50W ± 𝐍 ∓ 50 ± 𝐍 = ±(2𝐍-50), or 2) 50𝐖±𝐍-(50𝐖±(49-𝐍)) = 50𝐖 - 50W ± 𝐍 ∓ 49 ± 𝐍 = ±(2𝐍-49). Since 𝐍 is an integer, (2𝐍-50) is always even, while (2𝐍-49) is always odd. So, if the weighing balance displays an even number, then we know that the coin we're guessing on is genuine, but if it's an odd number, then we know the coin is counterfeit.

[@sundaramdevakottai9619]

I solved it differently, same as what David Paven has mentioned below. The natural thought that occurred to me was to put all the remaining 100 coins in one pan and the one on hand in the other. If you then work through the possibilities, you come to this conclusion: Divide the pan’s reading by 99; if your reminder is 49 or 50, you have a genuine coin; if the reminder is 48 or 51, you have a fake. This seems simpler (to me). Why isn’t this mentioned as another solution?

[@DAndre98]

I did something very similar just now, and it looks like makes perfect sense. Let's say for the sake of ease the real and fake coins weigh either 1g or 2g. There's only four possible scenarios, which boil down to your two options. If you've put a real coin in the left side (of which the chances of doing so are marginally higher), the right would then have 50 real and 50 fake coins, weighing 150g. If a real coin weighs 1g, the display would read -149g; if 2, then -148g. If you've put a fake one, the display would either read -150g (1g fake coins), or -147g (2g fake coins), given you'd have 49 fake coins and 51 real coins in the right side. This works for any coin weights within a gram of each other, the lowest value scenario on the scale means you were given a lighter fake coin, the highest means a heavier, and in-between means you got a genuine coin. It doesn't appear to violate any of the rules, and seems to be way more intuitive.

[@quasarcia]

I solved it exactly same as you, and I also think this solution is worth mentioning (especially for two reasons: first the fact that it's not necessary weighing 50 coins on each side, as stated by some others here AND even more important to me: with this method, we're not only able to tell whether we got a genuine or a counterfeit coin, but we can tell from this method also the weight g of a genuione coin and whether the counterfeit ones are heavier or lighter...)

[@frederf3227]

Fun tip: Since the genuine coin mass can be anything try setting it to 0g if the counterfeit is heavier or 1g if the counterfeit is lighter. This only applies if there are the same number of coins on each pan.

[@tp2hu]

Your solution works even if each counterfeit coin can be either heavier or lighter by 1g, so I don't see why the "all are lighter or all are heavier" should be in the puzzle description.

[@BigDBrian]

(paused at 1:22) Weighing the given coin itself against another coin will not give you any information, as there are limited options that don't give any clues: the scale outputs 0: the two coins are either both genuine, or both fake. the scale outputs 1 or -1: you have two different coins but you don't know whether the real one weighs 1 more or 1 less. Oh! instead of weighing the coin you are given, perhaps you can weigh all the others? So 50 on the left, 50 on the right. It doesn't matter which ones, just pick at random. Let's say the real coins weigh x and the fake ones x+1 (we'll do x-1 separately if necessary) if you grabbed a real coin, there would be an equal amount of fake and reals left. the total weight would be 100x + 50. To see how this would be divided let's take a smaller example of 4 reals and 4 fakes. I've made a small table of all possibilities in notepad, but it's inconvenient in the comments box. However from the results it becomes easy to conclude that if you have n reals, and n fakes, then the number the scale will output is even if n is even, and odd if n is odd. You can figure this for yourself. 50 is an even number so **If the coin is genuine, the scale will tell you an even number**. The outcomes are symmetrical in a way that makes it obvious that it wouldn't matter if the fake coins weighed x+1 or x-1. Now let's say you grabbed a fake. The total weight would be 100x + 51(or 100x + 49). You don't even have to check with smaller numbers, actually. You have an odd total. if you divide an odd number in two, the difference must also be odd! So, the scale will output an odd number(positive or negative) Now the answer is trivial. If the scale outputs an *even* number, you will say the coin you didn't weigh is genuine. If the scale outputs an *odd* number, you will say the coin you didn't weigh is fake. And thus you will be set free, if the evil warden keeps his word that is.

[@Kendots]

mrBorkD thats my zombie, christina!

[@kevinderoo3880]

Bruh... here's an EASY explanation: Let a genuine coin weigh "a" and a counterfeit coin weigh "b". If a is even, it can be represented as 2*k. b is either 2*k+1 or 2*k-1, which represents an odd integer. If a is odd, this will be the other way round, so b will be even. CASE 1: - If you were given a qenuine coin, you would have 49 genuine coins and 51 counterfeit coins on the balance. - The balance would always have different amounts of genuine and counterfeit coins on each pan. For example: 24 genuine coins + 26 counterfeit coins [vs] 25 genuine coins + 25 counterfeit coins. No matter how you distribute the 100 coins, in this case you will always have even amounts of genuine and counterfeit coins on one pan, and odd amounts on the other pan. - Left pan - Right pan will give you: odd * a + odd * b - If a is even, b is odd or vice-versa, and using the rules like odd*odd=odd and odd+even=odd, we can deduce this sum will be odd. - Therefore the difference in weight of these two pans will always be odd and you can now deduce that you were given a genuine coin. CASE 2: - If you were given a counterfeit coin, you would have 50 genuine coins and 50 counterfeit coins on the balance. - Example of distribution: 24 genuine coins + 26 counterfeit coins [vs] 26 genuine coins + 24 counterfeit coins. No matter how you distribute the 100 coins, in this case Left pan - Right pan will give you: even * a + even * b - If a is even, b is odd or vice-versa, and using the rules like even*odd=even and even+even=even, we can deduce this sum will be even. - Therefore the difference in weight of these two pans will always be even and you can now deduce that you were given a counterfeit coin. Please like if you understood and if it made the problem clearer. Thanks :)

[@mikefochtman7164]

I solved it a bit differently. Divide the remaining coins into two piles and weigh them. First, if I assume the remaining coins are 50/50, then it's possible the balance reads zero if they just happen to be evenly distributed. Then I realized that if they are not evenly distributed, the scale would indicate an even number. This is because you subtract one counterfeit coin from one side and placing it on the other. Shifting one gram across will make the scale increase/decrease by 2. So zero or any even number indicates 50/50 in the remaining coins. Therefore, if the scale indicates an odd number, the remaining piles are 51/49. So now I know what I have in my hand.

[@GretgorPooper]

It was such an eureka moment when I finally figured it out xD

[@Kris89BCA]

Nice one, solved it however it shoudn't be said - "you can weigh any of the 101 coins" but "you can use the balance only once"; otherwise it says that you can weigh only one coin not whatever number of coins you want.

[@quantum-atom]

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 information. The 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!

[@codediporpal]

Use the scale only once... hmm, that scale look pointy and dangerous. Attack the evil warden with the scale? Did I get it right?

[@sceylo9573]

How about putting your coin alone on the right pan and all the other 100 coins on the left. 1.) If you happen to have a genuine coin the left pan is on a mix of 50 genuine and 50 counterfeit coins. 2.) If your coin is a counterfeit the mix on the left is 51 genuine to 49 counterfeit. Let's make an easy example and set the weight of an genuine coin to 2g. 1.) 100 coins will always be heavier than your single coin. 2x50 + 1x50 = 150g 2x50 + 3x50 = 250g The weight balance will be -148g or -248g -> If your last digit is an 8 you have a genuine coin. 2.) 51x2 + 49x1 = 151g 51x2 + 49x3 = 249g The weight balance will be -150g or -247g. -> If your last digit is not an 8 the warden gave you a counterfeit coin.

[@puliverius]

What if the genuine coin weights 3 grams? Then you can get 247 gram difference and you will be imprisoned forever :).

[@francoism2232]

Very good puzzle. My solution: put 50 coins in the left pan of the balance, 50 coins in the right pan of the balance, and keep the 101th coin in your hand: * If the weight difference is odd, then the coin in your hand is counterfeit. * If the weight difference is even, then the coin in your hand is genuine.

[@kenhaley4]

Much simpler: Put all the coins (except the selected coin) on one side of the scale, leaving the other side empty. The scale now shows the total weight of the remaining coins. If it's even the selected coin is good, otherwise it's counterfeit. Not only easier to do, but easier to prove correct.

[@gofer9156]

You don't know the weight of a single coin, so weighing all the coins could produce any number.

[@NikolayZabrodin]

@@gofer9156 wrong. The total weight of the rest of the coins will always be odd if selected coin is fake, and always be even is selected coin is real.

[@NikolayZabrodin]

Proof Let's say: W - total weight of the rest of coins (100) R - real coin (51) F - fake coin (50) E - even number O - odd number 1. If selected coin is Fake: W=51R+49F=50(R+F)+(R-F) 50(R+F)=E; R-F=+-1=+-O E+-O=O 2. If selected coin is Real: W=50R+50F=50(R+F)=E

[@gofer9156]

@@NikolayZabrodin , no they won't because you don't know the weight of a single coin. You only know the weight difference between a fake and a real coin. For example, suppose a single real coin weighs 0.5 ounces; then weighing 100 real coins gives the reading 50 ounces, contradicting your assertion.

[@sonalidasgupta3562]

You are assuming weight of each coin is INTEGER.....what if its say 1.23456 grams....?? Total would not be even or odd

[@tylerkane1238]

This puzzle reminds me of the Cyberchase episode where in one scenario, the kids had to find the right key out of eight by using the scale and find which key is the lighter to them all. The coin scenario given here is similar to the problem given above.

[@vitriolicAmaranth]

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" The Statistician: "Guess that it's genuine because odds are that it is" The 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." The Engineer: "Flip the coin to decide if it's real or counterfeit"

[@user-kd4mi8xb7p]

"The warden gives you a randomly selected coin from the 101 coins." I assumed that we would only have one coin and a scale... Then MindYourDecisions says, "... you end up weighing all of the other coins!" What part of using the other coins were in the question?

[@GueVonez]

El Blanco yeah i thought you had to choose just one of the pile of coins... waisted 20mins thinking it through

[@aday07271996]

El Blanco The last sentence did mention of weighing ANY of the 101 coins. Basically, any amount of coins on the balance is acceptable, but mentioned that the weight can be used only once.

[@user-kd4mi8xb7p]

***** I paused to read the question. Probably missed him saying it

[@ZeroSleap]

This kind of puzzle seems to require some cunning thinking XD You had to outsmart the warden and wiegh all the OTHER coins at once

[@marylee936]

it says "you are given 101 coins". read more carefully next time.

[@beepbadapup8523]

So proud I finally got one of these riddles :)

[@jaglinuxmint]

I almost came close and gave up! Thanks for making me work my gray matter

[@sorsocksfake]

- the only way you can determine it, is by knowing whether the other stack has 50 or 49 fakes. - To get a definite answer, you'd need to weigh all 100 remaining coins. - Assume we weigh them 50-50 (because intuition). - The weight of each pan will be 50x, +/-y, where y is the number of false coins on it (depending whether they're heavier or lighter). Since both pans have 50x, that cancels out. - The remainder is either 50 or 49 fake coins, which are either 1g lighter or heavier. No matter how they're distributed among the pans, one will give an even number (if your coin is real) and the other gives an odd number (if your coin is fake). (for maths: it's either 50-2y, with y being an integer positive or negative, vs 49-2y)

[@hozelda]

Alternate but riskier solution. If you can set one of the scales and look at the value quickly before the other scale is set, then put the selected coin on a scale and then all the other 100 on the other scale. By doing the algebra, the four cases amount to the following: Call x the weight you see when the coin is placed. Then the final value after both scales are loaded will be one of the following, 99x+51 (yours=fake,lighter) 99x-51 (yours=fake,heavier) 99x+50 (yours=real,lighter) 99x-50 (yours=real,heavier) You will know x (the positive of the value you quickly saw on the scale) and from it can calculate all those 4 cases. The final value will match one of them. If it matches either of the first two cases, then your coin is counterfeit. If it matches either of the bottom two, it's a genuine coin.

[@DavidPAVEN]

Hi, I just fund this problem (thanks to KZbin recommandation) and I solve it in another way. I call w the weight of a Genuine coin. I put my coin in right pan I put all the remaining coins in left pan. The balance show a number D. With a few simple maths, I find that : If we have a genuine coin, D = 99*w +/- 50 : D = 99*w + 50 or D = 99*(w-1) + 49 If we have a fake coin, D = 99*w +/- 48 : D = 99*w + 48 or D = 99*(w-1) + 51 I make the euclidean division of D by 99 (D = 99*q + r) and I get the answer : - If r is 50 or 49 : the coin is a genuine one. - If r is 48 or 51 : the coin is a fake one Note that we can find the value of w at the same time : if r is 50 or 48 then w = q , else w = q + 1

[@sundaramdevakottai9619]

This is exactly how I solved it too. Why isn’t this mentioned as another solution? The natural thought is to put all the remaining coins in one pan and the coin in your hand on the other. If you work out the possibilities, you end up with this answer: If you divide the pan reading by 99, you will get the reminder as either 49 or 50 if you have a genuine coin; you will get 48 or 51 if you have a fake. This seems simpler than what Presh says.

[@4dzxk332]

Oh wow.... what a lovely prison... back to reality .

[@M1cr0p47]

What if the weighing scales is imperial? It's going to be really awkward to do the conversion when you don't know the conversion factor for ounces to grams

[@xxJing]

I figured it like this. Distribute the remaining 100 coins 50 and 50 on the scale and assume each side is as equal as possible. That means if it’s possible both sides weigh the same and display a difference of 0 then they will, if not then they will display the smallest possible difference. Assume you have the genuine coin. This means that enough coins exist to distribute them such that the difference in weight displayed is 0. Now try replacing a heavier coin with a lighter coin between sides. You will notice that one side looses a net gram while the other gains one. This means that every time you unbalance the scale by switching coins of different weights from each side, it changes by 2 grams, you can assume you can do this until all coins of a specific type are on one side, which will yield a difference of 50 grams. Therefore, if the difference is an even number between 2 grams and 50 grams (the maximum difference for a genuine coin route) then you have a genuine coin. If it’s not, you don’t.

[@yvesbrunet5199]

I'm using logic instead of math to prove the solution: I'll name the genuine coins G and the fake coins F. There are 50 coins randomly distributed on each side. I'll separate those into 3 groups: The first group is the number of genuine coins that are on both sides: xG. The second group is the number of fake coins that are on both sides: yF. The third group is the number of mismatch coins where one side has zG coins and the other side has zF coins. Side A Side B group1 xG xG group2 yF yF group3 zG zF On both side of the balance, x + y + z = 50. Groups 1 and 2 will not affect the balance reading because they even themselves out. Group 3 is the one that determine the reading. The sign of that result will be determined by whether plate A is on the right or left side of the scale and whether the genuine coin is heavier of lighter than the fake one. Either way, the absolute value of the result will be the value of z. The total number of genuine coins: G = z + 2x. The total number of fake coins: F = z + 2y. If the coin I have to guess is a genuine coin then there are 50 genuine and 50 fake coins on the balance. Since 50 is an even number and 2x and 2y can only be even numbers, z has to be even. If the coin I have is a fake coin then there are 51 genuine and 49 fake coins on the balance. Both are odd numbers and since 2x and 2y can again only be even numbers, z has to be odd.

[@davidb5205]

Ok, before looking at the solution. I'm guessing that because the difference in weight is one gram exactly, the solution has something to do with even and odd numbers. (Even-Even) = Even, (Odd-Odd) = Even, (Even - Odd) = Odd, (Odd - Even) = Odd. If I take a genuine coin then I am left with 50 times the weight of each, which means that both total weights will be even and thus the difference will be even. If I take a counterfeit coin, then I am left with 51 and 49 which means that one total weight will be odd and one will be even, which means the difference will be odd. Although, this logic is kinda flawed, that's my best guess.

[@wschmrdr]

It's easiest way to explain this.

[@JSSTyger]

What a truly sadistic warden...giving inmates complex math questions and weighing balances.

[@insidetrip101]

Holy shit, I actually figured out one of the ones that don't involve elementary arithmetic and geometry--and in a reasonable amount of time! Spending time on your channel is definitely improving my critical thinking--or maybe I just got lucky? who knows! It was great to actually succeed completely for once!

[@pragadeeshkannan4938]

I can't believe I solved it. I just experimented a lot of times then i got the trick

[@JurijFedorov]

I can't even solve what is worse. Suffering through this math or staying in his dungeon.

[@AkaomeNanashi]

An year ago coworker gave me this problem, and it took me like 15 min to solve. It's very interesting problem and really easy if you try to think. I give this from time to time to other people around me and the success rate is pretty much good. And the variation i know is with 10 bags of coins and 1 of the bags is with fake ones which are slightly different in weight. And you can take what ever amount of coins from whatever bag, but you can measure the weight of the sample only once and the question is how you say which bag is with fake money.

[@nlb137]

Based on dicking around with a spreadsheet for a bit; split the coins into two piles of fifty each and weigh both against eachother. if you have a genuine coin, the difference will be an even number between -50 and 50, and if you have a false coin, the difference will be an odd number between -49 and 49.

[@Ahmedyaser7]

I can't describe the solution with mathematical terms, but weighing all the coins at 50 to 50 division will give us a specific number ranging from minus fifty "-50" to fifty "50" which is going to interpret the distribution of coins on both sides of the scale (which is a bonus info!) in addition to finding the solution. This could be done by doing a chart of all the possible scenarios starting with all genuine coins placed at the right side

[@mauriceshipsey]

If a coin has an integer or simple decimal (non recurring) weight, then it is possible, with ONE weighing, to tell •The weight of a good coin •Whether counterfeits are heavier or lighter •what kind of coin you have !!! Put your coin in one pan, and the other 100 in the other pan. If all coins were normal, the readout on the scales would be divisible by 99, or 9 for ease.(1 coin versus 100, difference is 99w) But an added difference is introduced by the counterfeit weights.. either +50,-50,+48or-48. None of these differences is divisible by 9. (So, eg, 99w+48 is not divisible by 9) Now see which operations changes the weighing scale readout into a number divisible by 9 If you have to subtract 50 then your coin is good, counterfeits are 1g heavy, and a good coin weighs (readout less 50)/99 If you have to add 50 then your coin is good, counterfeits are 1g light, and a good coin weighs (readout plus 50)/99 If you have to add 48 then your coin is bad, counterfeits are 1g light and a good coin weighs (readout less 48)/99 If you have to subtract 48 then your coin is bad, counterfeits are 1g light and a good coin 5weighs (readout plus 48)/99

[@djyotta]

A much easier way... Consider the two cases: case G - you have a genuine coin. then there are 50 genuine coins, and 50 counterfeit coins remaining. if x genuine coins is in the left pan, then x counterfeit coins is in the right pan. let y be the number of counterfeit coins in the left pan. we have x + y = 50 and the following equation let g be the weight of genuine coin and c be the weight of counterfeit coin x *g + y *c - x*c - y*g = x*g -y *g - (x*c - y*c) = (x-y)(g-c) = (x-y) * (+-1) = D now for all x, y such that x+y = 50, we have x-y is even, so D is even case C - you have a counterfeit coin. then there are 51 genuine coins, and 49 counterfeit coins remaining. if x genuine coins and y counterfeit coins are in the left pan, then we have x+y = 50 therefore, as above, 51 -x = y+1 genuine coins are in the right pan similarly, 49-y = x-1 counterfeit coins are in the right pan we have the following equation x*g + y*c - x*c + c -y*g -g= x*g -y*g - (x*c +y*c) +c-g = (x-y)*(+-1) + (+-1) = D now for all x,y such that x+y = 50, we have x-y is even, so D is odd

[@Timmers97]

you could also put 1 coin left and 100 coins right. then you calculate what the weight difference would be depending on what coin is on the left part. if the coin is 49, then on the right you'd have 51x50 + 49x49, the difference being -4902. if you got a coin weighing 51 left, then on the right you'd have 51x50 + 49x51, the difference being -4998. and if you got a coin exactly 50 in weight on the left, depending on what the counterfeit coins weigh, you'd either have 50x50 + 50x51, or 50x50 + 50x49 on the right.

[@pvzpokra8602]

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.

[@luisalejandrohernandezmaya254]

I think is not necesary that ALL fake coins weight 1 gram heavier or lighter. We can assume that each fake coin weight randomly ±1 gram and still works.

[@jessstuart7495]

The use of "integer" was a little confusing to me at first, until I remembered that the difference in weight (c-w)=±1g.

[@HostileInsight]

Who else doesn't actually take the time to solve these, but instead just waits for the explanation?

[@12thtime]

There is a far easier way to solve this, with a much easier to understand calculation. Put the 100 coins you didn't pick on one side of the scale and nothing on the other. If you picked a counterfeit con, there will be 51 real coins of weight W and 49 counterfeit coins of weight W+1 on the scale. 51W + 49(W+1) = 100W + 49 grams. Since there's nothing on the other side, the scale will show the total weight, which in this case has to be an odd number. If you picked a real coin, there will be 50W + 50(W+1) =100W + 50 grams on the scale, which will be even.

[@GraemeMcRae]

In this puzzle, *all* the counterfeit coins are lighter by 1 gram, or *all* counterfeit coins are heavier by 1 gram. A seemingly harder puzzle would be if *some* of the counterfeit coins are lighter, and *some* of them are heavier, by 1 gram. In this case, the procedure would be the same, and the result, if even would mean the prisoner is holding a genuine coin, and the result, if odd, would mean the prisoner is holding a counterfeit coin. So this seemingly harder puzzle would have the same solution, and the prisoner would be set free.

[@legendhero-eu1lc]

Thank you for the video! All of you friends are super awesome!

[@rodrigoaceves9020]

the warden is kidnapping me under false pretenses, possessing counterfeit coins, and blackmailing me. I'd like to speak to a lawyer

[@KyleMart]

Solved it. I think a good way to think about this is to assume arbitrary values to the coins and the ending result. Pretend real coins are 3 grams and fake coins are 2 grams. Let's say 50 on one side happen to be real, this would be 50*3 = 150. On the other scale you have 50 fake. This would be 50*2=100. The difference of these numbers is 50. If you swap any two coins the difference will be 2 greater, 2 lesser, and eventually if you swap a coin of the same type, no change at all. Therefore evens mean real. If you hold the counterfeit coin it will follow the same pattern except be one off making it always odd.

[@ayub5030]

This video was very enticing, I nearly got the answer and I guessed. *Luck is on my side*

[@biegeldog]

I solved this through induction. I started with the two cases, one where I had a genuine coin, and one where I had a counterfiet coin. I also gave a weight for the genuine coin at 10g and the counterfeit at 9g just so I could do some easier math. I assumed that I would have to weigh the remaining 100 coins. Let's say I was super lucky and one pan ended up with 50 real coins. This would give them a weight of 500g. Case 1: The other pan has 50 fake coins weighing 450g. Difference of 50g. If I swap a coin between the pans, they will now have a weight of 499 and 451g respectively. Difference of 48. By swapping a fake coin and a genuine coin, I will always increase one pan by 1 and decrease the other by 1 making the difference always change by 2. So if I have a genuine coin, the difference will always be even no matter how many real or fake coins are on either pan. Ok, so case 2, I have a fake coin. Like before, I am really lucky and have 50 real coins in one pan with a weight of 500g. The other pan has 49 fake coins and 1 real one with a weight of 451. Difference of 49. Swapping a real and fake coins between the pans will always change the difference in weight by 2 so when you have a fake coin, the difference is always odd. Even difference, real coin. Odd difference, fake coin.

[@eskhaphey2873]

True, but there is also no point in weighing two separate piles, just weigh all the remaining coins and you get an identical solution.

[@matthewc.7301]

You made your proof so much harder than it had to be. If you just considered that you either end up weighing 50 of each coin or 51 genuine and 49 fakes, this turns simple. In the case of the first, all whole numbers multiplied by even numbers(50) give a result of an even number, thus an even number as the difference means that you are holding a genuine coin. But since you are using consecutive numbers for the weights of the coins, one of them is odd and the other even. Multiplying these by odd numbers(49 and 51) results in one odd number and one even number, since odd x odd = odd, but even x odd = even. Thus the weight difference when holding a fake coin will be an odd number.

[@Logan-dk8of]

There is a simpler way. If you put all the 100 coins on one side you get the total weight. Average that. Then you put your coin on the other side (technically one use) and subtract that weight from the other total weight (using all absolute values), and that will give you your coins weight. If your coin is exactly .5 grams different (assuming the one gram difference was exact) in weight than the average then you know you have a counterfeit. If it differs from exactly .5 grams from the average (even by .01) then you have a real coin. This is because if you have equal of both coins then your average of them should be exactly in the middle of both their weights, which leaves one left over coin which means it has to be a fake.

[@oldestries]

ill just walk away and out from the prison. When my freedom is only depended on those 101 coins authority, ill just let the warden have them all thanks. Even better if them actually are gold coins, genuine or fake he could still find the use of them. Thanks warden. I c u in millions of millions years.

[@assalane]

you could also do: w ->weight of genuine X ->number of genuine in left balance 50-X ->number of counterfeit in left balance then simplify those 2 expression which represent the value read on the scale: Xw+(50 - X)(w +/- 1) - [(50 - X)w+(50 - (50 - X))(w +/- 1)]= - /+ 50 +/- 2x (genuine, result is even) Xw+(50 - X)(w +/- 1) - [(51 - X)w+(49 - (50 - X))(w +/- 1)]= - /+ 51 +/- 2x (counterfeit, result is odd)

[@blanmorrison1794]

I was getting ready to dislike the video, but then I realized it was lighter OR heavier, which is different than the usual one. Nice, I'm stumped!

[@jeffiek]

Hah! Silly warden. He's not going to imprison *_me_* forever. I'm going to die of old age first.

[@jarvismckenzie6843]

Of all of the puzzles I've looked at, I've successfully solved 3 - all the rest were close but not quite or left me dumbstruck! By now I'm comfortable with feeling inadequate - let me get really comfy.. ;-)

[@idontwantanusername]

I resolved it in a slightly different way, but still based on the idea of evaluating the even/odd result of the weight difference between left and right. If I take the 100 remaining coins and I put 99 of them on one side (left) and one on the other side, right, then there are 8 possible scenarios (the combination of my coin being real/fake, the coin on the right side being real/fake and the weight of the real coin being more or less than a fake coin). Then I'll solve the problem by exhausting of all the possible cases (I'd often go for a "brute force" approach if number of cases is limited, even if it's not the most elegant solution... I'm lazy! :P) Notation: I'll call "L" the weight on the left side and "R" the weight of the right side. A) If a real coin weights more than a fake one, then weight of a real coin = x + 1 and weight of a fake coin = x If my coin is real, then there are 50 real coins and 50 fake coins, so the total weight is 50(x+1) + 50x and we can distribute it in these 2 ways: A1) fake on the right: L = 50(x+1) + 49x = 99x + 50 ; R = x -> weight difference L-R = 99x + 50 - x = 98x + 50 -> even + even = even A2) real on the right: L = 49(x+1) + 50x = 99x + 49 ; R = x + 1 -> weight difference L-R = 99x + 49 - x - 1 = 98x + 48 -> even + even = even If my coin is fake, then there are 51 real coins and 49 fake coins, so the total weight is 51(x+1) + 49x and we can distribute it in these 2 ways: A3) fake on the right: L = 51(x+1) + 48x = 99x + 51 ; R = x -> weight difference L-R = 99x + 51 - x = 98x + 51 -> even + odd = odd A4) real on the right: L = 50(x+1) + 49x = 99x + 50 ; R = x + 1 -> weight difference L-R = 99x + 50 -x - 1 = 98x + 49 -> even + odd = odd B) If a real weights less than a fake one, then weight of a real coin = x - 1 and weight of a fake coin = x If my coin is real, then there are 50 real coins and 50 fake coins, so the total weight is 50(x-1) + 50x and we can distribute it in these 2 ways: B1) fake on the right: L = 50(x-1) + 50x = 99x - 50 ; R = x -> weight difference L-R = 99x - 50 - x -> 98x - 50 -> even - even = even B2) real on the right: L = 49(x-1) + 50x = 99x - 49 ; R = x - 1 -> weight difference L-R = 99x - 49 - x + 1 -> 98 x - 48 -> even - even = even If my coin is fake, then there are 51 real coins and 49 fake coins, so the total weight is 51(x-1) + 49x and we can distribute it in these 2 ways: B3) fake on the right: L = 51(x-1) + 48x = 99x - 51, R = x -> weight difference L-R = 99x - 51 - x = 98x - 51 -> even - odd = odd B4) real on the right: L = 50(x-1) + 49x = 99x - 50, R = x - 1 -> weight difference L-R = 99x - 50 - x + 1 = 98x - 49 -> even - odd = odd We can notice that, when my coin is real (A1, A2, B1, B2) the weight difference is even, while it's odd when my coin is fake (A3, A4, B3, B4); hence I can say if my coin is true or fake with just one measurement. Probably there are others solutions, but I guess they probably revolve around this scheme of reasoning: - realising that we can only work on properties of a single measurement/number, so even/odd is a likely choice. - finding a grouping strategy for the coins so we can express the weight difference as an unknown weight multiplied by an even number (so that quantity is always even) plus an even/odd quantity, which we can use to determine if we are in a real/fake coin case.

[@rasraster]

Whew - I got released from the prison.

[@ArunVutnoori]

If the prisoner is that smart , he would never be caught or if caught ,he would have somehow escaped the prison even before the warden had a question for him. Hahaha

[@bg6b7bft]

I'd put that coin down, and weigh the remaining 50 coins vs 50 coins. If the balance is off by an even amount, then all 50 fake coins are on the scale. If its off by an odd amount, then you're holding a fake coin.

[@sunnyilya7876]

I did this assuming every genuine/counterfeit pair had a different weight, that was a mess, but you can still get 100% if the probability of two genuine coins adding up to a whole number is 0. If the 2 genuine coins left over in the 49 pair scenario add up to an odd number, then you essentially have a false even. Just my initial thoughts, correct me if I'm wrong.

[@girlmore4484]

I read it as "you can have a weight difference before you know which coin you need to tell" and found no way

[@iTs-JP-916]

my issue your not specifying if all the counterfits are in one bag and the real are in another. or are all 100 other coins are all mixed together?

[@eddietheeclectus230]

Yeh the setup to this logic puzzle never tells the viewer that they can interact with all the other coins. This problem is impossible to solve ahead of time if you assume you can't touch the other coins, which is sort of implied by the fact that the setup to the puzzle doesn't explicitly state that you CAN interact with the other coins.

[@hummesse]

An induction proof would have been nicer imho. Put 50 in each and observe some integer difference. Swapping a genuine for a counterfeit makes the number change by two, keeping the evenness/oddness. Swapping two identical coins does not change anything. Now find a single example of the last coin being genuine and the scales showing an even number. Then find an example that the last coin being counterfeit and the sclaes showing an odd number. Choose the most trivial examples to make it easy. Now you are done.

[@cgamin6921]

at first I was thinking weigh all the coins (separately) against 1 and if it was lighter or heavier (depending on which coin you picked) put them into one pile and the even ones into another and whichever has more you pick that one but then I saw the step where you can only use the scale once. (basically I'm saying I don't know)

[@Games-mw1wd]

I don't think you can describe a "best strategy" question as "seemingly impossible." There is always a best strategy you can find, even if it's not very good.

[@ddebenedictis]

True, and besides, after hearing the problem, I knew it was possible, it never seemed impossible.

[@kreia187]

this is not actually a 'best strategy' question (even though he worded it that way). There is a 100% surefire method of getting it correct

[@SmileyEmoji42]

I guessed that the solution must involve a 50/50 weighing. I then imagined taking a good coin and weighing the most unbalanced possible 50/50 split giving a ballance of 50d where d is the difference in weight betwen good and counterfeight. I then considered what would happen if I swapped 1 coin (48d), then another (46d) and immediately saw that the difference must always be even. A quick check of the reverse case (49d, 47d....) and I had the solution. It seems that quite a lot of these puzzles depend on looking for an odd/even trick.

[@lipak45]

I think we need to add that the weight of the coins are integers otherwise the balance will show an non-integer difference from which we can't determine the parity.

[@yurenchu]

The weight of each single coin doesn't need to be an integer. The point is that we put an equal number of coins on both sides of the scales.

[@coconutjuice7777777]

You are correct. Everyone missed that.

[@elSethro]

The solution I came up with is much more convoluted and involves cheating. When you load a scale, you're not expected to load both sides simultaneously, right? So I put all 100 'other' coins on one side, and sneak a quick peek at the result (hopefully the warden doesn't notice) before putting the coin I was given on the other side and noting that result, too. I have managed to effectively cheat a second use of the scale. This assumes the scale produces the reading instantly, and you don't have to load both sides and then push a button or something. The first reading will be some number Y1, and is the difference between the 100 'other' coins and the empty balance. If the weight of a light coin is X, then there are 3 possibilities: 1. If I was given a real coin, then Y1 = 100X + 50, regardless of which type of coin is heavier. 2. If I was given a fake coin, and real coins are heavier, then Y1 = 100X + 51 3. If I was given a fake coin, and real coins are lighter, then Y1 = 100X + 49 The second reading will be some number Y2, and is the difference between the 100 coins and the coin I was given. One of 4 scenarios must be true: 1. If I was given a real coin, and real coins are heavier, then Y2 = 99X + 49 2. If I was given a real coin, and real coins are lighter, then Y2 = 99X + 50 3. If I was given a fake coin, and real coins are heavier, then Y2 = 99X + 51 4. If I was given a fake coin, and real coins are lighter, then Y2 = 99X + 48 Now, for each of the 4 possible scenarios, solve both equations for X and set them equal to each other. Then compare the 2 readings from the scale and determine which one of the following is true: 1. If (Y2 - 49)/99 = (Y1 - 50)/100, then I was given a real coin, and real coins are heavier 2. If (Y2 - 50)/99 = (Y1 - 50)/100, then I was given a real coin, and real coins are lighter 3. If (Y2 - 51)/99 = (Y1 - 51)/100, then I was given a fake coin, and real coins are heavier 4. If (Y2 - 48)/99 = (Y1 - 49)/100, then I was given a fake coin, and real coins are lighter

[@skifast_takechances]

Yess I solved this after thinking for a mere 30 seconds or so! I was initially so stumped but then aha! I thought of the fact that there is a difference in the weight of the coins.

[@ADDmonkey55]

So to explain it more simply, weigh 50 coins on the left and 50 on the right. If all coins were genuine, then the difference would be 0, but of course there's gonna be either 49 or 50 fakes. If the left and right side both happen to hold the same # of fakes, then the difference will again be 0. But most likely one side will have more fakes than the other side. Doesn't matter which side. If the left side has 2 more fakes than the right side, than the difference is 2 (or - 2, doesn't matter). So the number produced on the scale tells us how much more fake coins are on one of the sides of the scale. This number will always be even if there are 50 fakes, and will always be odd if there are 49 fakes. Thus that tells us if we have the genuine coin (50 fakes) or the fake coin (49 fakes).

[@TheDashACorner]

When I read this question I thought that all genuine coins have different weights. But each genuine coin (except one) got a counterfeit sibling with +-1g.

[@bojandimovski1504]

Thanks for giving me an excellent idea, Presh. Now I am going to kidnap someone and become an evil warden.

[@joelamond9536]

My line of thinking: (before watching the solution) It seems there are only two possible strategies to proceed with: you could put your coin on the weighing scale with another coin or you could put two other coins on the weighing scale. Let's have a look at the former. Since you cannot tell whether the counterfeit coins are heavier or lighter you cannot make any deductions from the exact measurement on the weighing scales, only whether the coins are the same weight or a different weight. If the coins are the same weight, then they have to be the same coin. In this case it seems they are more likely to be genuine coins. The probability that the first coin you are given is genuine is 51/101 and the probability that the second coin you are given is genuine is 50/100 or 1/2. The chances these both happen are 51/101 x 1/2 = 51/202. The probability of them both being counterfeit coins is smaller, 50/101 x 49/100 = 49/202. 51/202 > 49/202. So, it seems that if they are the same coin it is more likely they are genuine. The second case is when the coins weigh differently. In this case there must be one genuine coin and one counterfeit coin. The odds work out like this: 51/101 x 1/2 = 51/202. Or if your coin is counterfeit: 50/101 x 51/100 = 51/101. So the events are equally likely. However, your particular coin is more likely to be genuine as 51/101 > 50/101. So you should also expect to have a genuine coin. The other strategy seems strange. But I'll take a look at it anyway. If the two coins you weigh are the same then they are more likely to be genuine coins. Let's suppose you have a genuine coin. In this case there are 50 genuine coins are 50 counterfeit coins left to pick. So the probability of them both being genuine is 1/2 x 49/99 = 49/198. So is the probability of them both being counterfeit, as there are an even number of genuine and counterfeit coins. But what if you have a counterfeit coin? In this case the chance of picking two genuine coins is higher than picking two counterfeit coins as there are fewer counterfeit coins. As you have no way of telling whether you have a genuine coin or counterfeit coin for certain it is logical to assume that you have a genuine coin. If the two coins you weigh are different then, as I said before, you have one genuine coin and one counterfeit coin. It is still more likely that you have a genuine coin, with similar reasoning as above. So it seems logical to assume you have a genuine coin. EDIT: All that time wasted ... Only now do I see that you could weigh more than one coin on each scale.

[@zelandakhniteblade5436]

To those writing how obvious it is that you need to weigh 50 vs 50, you clearly have not understood what is going on here. The solution presented here is just one from a family of possible solutions. As long as you weigh all 100 remaining coins, it does not matter which constellation you choose, only what the total weight is. Take for example the extreme case of weighing 100 vs 0. In this case we either have 50x + 50(x+1) or 50x + 50(x-1) for a genuine coin, or 51x + 49(x+1) or 51x + 49(x-1) for a counterfeit. It is trivial to see that the former is always divisible by 50 while the latter is always odd. This solution is actually "better" than that of the video in 2 ways, firstly because if the scales show an even number not divisible by 50 we can conclude that the warden is lying to us, and secondly because throwing all of the coins in a single bucket is much faster than counting out 50 vs 50. More importantly imho, this solution also is much clearer for non-mathematicians to understand the point rather than having to go into more esoteric mathematical notation that will just cause such viewers to turn off. This is in reality not a hard logic puzzle at all; it is merely presented as such by showing a solution that obfuscates rather than explains the reason why this works.

[@EliSitbon]

X can be any weight (might not be an integer). If X=0.01 than 51X+49(x+1)=100X+49=1+49=50 ---> You will not be able to tell which type of coin you have. The solution must cancel the weight factor from the equation

[@LeeleeThe9]

The video's question is missing the most important sentence for solving the problem: Each genuine coin is identical. It says it in the descriptions problem but does not in the video.

[@totalme302]

The proof is bizzare, but the puzzle at once seemed hard, I solved it easily when I gave it a little thought.

[@ville-mattiamonsen2092]

The description says "Each counterfeit coin is identical to a genuine coin, except..." This 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.

[@GraveUypo]

or you could guess at random. 50ish% chance is pretty good if you're lazy. that warden is actually quite generous.

[@ojasrox]

yup i got it! I used the same method!

[@arielfuxman8868]

What does it mean I only get to use the weighing balance once?