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i will use quantem magnetism to make the heavier side go down, leaving all coins to be in the same side, and then count the amount of coins andthen split it equally
Actually if you toss the coin the gold side should fall on the bottom, as gold is denser than silver. Do that with every coin a few times just to be sure then split them in any pattern, then you get 0 silver sides up on both sides
most of them yes. this one tho is more about understanding a simple concept. all the math was only them proving that concept, the math isnt needed to solve the puzzle
@@janjeikobu5163 no, so u basically take any number of cards, and flip any number of them over, then you blindfold yourself or something and take the exact number of the the cards you flipped over from your pile. (it could even be all of them)
It happens that I watched Michael's video before this so, here's how it is done with a deck of cards : ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-aqKkwTKWWUQ.html
For the final riddle, you have to have an odd # of gold coins to be able to flip them to all gold. Including the #'s 1 and 3. If any of the 3 coin piles start with an odd # of silver coins, it wont be possible. This includes starting with 3 coins facing silver up and 1 coin facing silver up. This means that 4 out of the 8 3-coin piles are able to be turned to all gold solutions
@@richardl6751gss is shown as ssg and sgg is shown as ggs to denote the fact that the pairs are the same point in the riddle because of parity symmetry, for anyone who is flipping and jumping between the 8 combinations shown there. Thus confusion is avoided.
Essentially, since silver coins can only be removed in pairs (flipping one of each becomes...one of each), there needs to be an even amount of silver coins showing.
this one was one of few riddles i got everything correct in. It was a dead giveaway when he said there was a surprisingly easy method of getting it right.
This is the first puzzle I could do in the entire series. I tend to overthink problems, so when he said there is a surprising solution it was actually a huge help.
Actually, because pure gold used in old coins is soft, you can bite both sides of a coin and the one with a dent is the gold side, so there IS a way to determine which side is facing up. According to the riddle, that's cheating, but if you're stuck in an underground dungeon in total darkness cheating is the best option 😁.
Technically, it doesn't say you only have one chance. You could flip one coin into a new pile, then flip a second coin into the new pile, etc., and when the number of coins in the new pile is the same as the original number of silver-side-up coins, you'll win.
It is 2:13am and I'm watching these on a bit of a binge. I usually try to solve them, but I haven't been until this one. I can't believe my half-awake solution was right! I was so pleased with my reasoning and was praying it was sound
Even if that worked reliably, you now have no idea whether that coin was silver or gold to begin with, and with hundreds of coins in the pile it's not plausible to spin every single one of them.
@@kaliyanelson6106 Not really, since you can flip any coin you want, you can still win. You can start with 21 silver side up coins and have 0 or 100 silver side up coins in each pile. Doing this in the dark is an issue though.
Why don't you just bring one or more of the following: 1: Flashlight 2: Matchbox 3: Cellphone 4: Literally anything that can create light Edit: Wow, I make a list of things people should have while exploring and it gets more then 150 likes. Not that I'm complaining.
There is a way I have found out, but a little more difficult. Stand the coins up on its side. Since gold is more dense than silver, it’s side is more massive, assuming that the silver and gold go right down the middle of the coins in equal volumes. That means when you put the coin up on its side, it will always land gold side down and silver side up.
Algebra is used to prove what you should be able to tell intuitively. Advanced or difficult math is not actually required to solve the riddle. If you make a new pile of coins (Pile B), the original pile (Pile A) will have one less silver coin for every silver coin you took. So if 1 of the coins you take is silver, Pile A now has 19 silver, and if 3 of the coins you took were silver, Pile A now has 17. This is true no matter how many coins you take, and it's actually pretty obvious. Subtracting numbers from 20 may be a math operation, but it's not exactly a riddle. If you form Pile B from 20 coins, you're also guaranteeing that the number of gold coins in Pile B is equal to the number of silver coins in Pile A. This isn't a complicated math problem either. The number of silver coins in Pile A is 20 minus the silver coins you took, and the number of gold coins in Pile B is 20 minus the silver coins you took, because you took 20 coins and all the non-silver ones in pile B are gold. Once you determine these two numbers have to be equal, all that you need to do is flip Pile B. This makes all the previously-gold coins in Pile B silver, and you don't need to worry about any of the previously-silver ones, because the same motion turned them all gold.
For the bonus riddle: If I am not incorrect, any pile with two silver coins can become all gold. If the two are adjacent you simply flip them and leave the initially gold coin alone. If they are on opposite sides of a gold coin then you flip each of them over once, the gold coin is then flipped twice since flipping coins on either side flips the middle coin and this remains gold, this leaves you with three gold coins.
Before I watch the video: Gold is heavier than silver. You put every coin on it’s side and gently spin it. Whichever side it lands on is gold. Then you just sort them.
For the bonus riddle, I figure that if there's an even number of silver-side-up coins in the arrangement, then it is possible to make them all gold-side-up, whereas if there's an odd number then it's not possible. This is because each flip you make always affects exactly two coins, and adding 2 to or subtracting 2 from an odd number will always leave you with another odd number but 0 (the number of silver-side-ups you want) is an even number.
That's correct, but a better reasoning is as follows Assume the number of face up silver coins to be x At any operation, you're either 1. Adding 2 to the number of face up silver coins (flip 2 golds) 2. Subtracting 2 to the number of face up silver coind (flip 2 silvers) 3. No change (flip 1 silver 1 gold) Thus you'll always change the number of face up silver by 2, and no matter move you do, you can't change odd into even just by adding even numbers, so that proves it's impossible if the number of face up silver is not odd. Then you'll only need to prove it's possible for all odd configuration.
Bonus riddle solution: as long as the top has 1 gold-facing coin, you can always make it into 3 gold. Never with 0 or 2. This is because of parity. If we start with 3 silver, flipping 2 will make it into 2 gold. Flipping the same coins sets us back to all silver, so if we flip 1 gold and 1 silver, we still have 2 gold coins face up. Every flip movement has the capability to either +2, -2, or +0 to the value of gold coins facing upwards (depending on which coins you flip). This means that even-facing trios will forever stay even, and same thing with the odd-facing piles. Knowing how to solve Rubik’s cubes, surprisingly, helped me to solve this, since certain puzzles, like 4x4 and square-1 have their own parity cases, since doing any turn on the cube changes the parity of all edges and corners on the cube. The solved state has 0 swaps of edges and 0 of corners, and doing a turn adds 3 swaps to corners and edges at the same time, which works similarly to this riddle, except every turn (or flip) changes the parity, rather than keeping it the same.
That doesn't affect the silver coind (odd/even) since we dont know the original total number of coins and thus will not affect the solution XD good try
Answer for last riddle: The ones with 1 or 3 gold coins, since you always move 2 coins. An even number + another even number, will always be an even number, therefore, you can never make 3 gold coins, no matter what you do, you can only get 0 or 2.
Or biting the coin and if what ever the the top or bottom coin Is soft then it Is gold if it hard the it silver. This way I can I can divide equally with silver side up and gold side down.
Gold and silver have a distinctive sound. You can use the ping test to decide which one is silver and which one is gold. Silver will usually sound like a C and gold will usually sound like an A. This is usually the case no matter the size of the coin, unless the actual silver/gold content is super low. Even then, a 35% war nickel has a distinctive sound over a copper-nickel nickel.
@@finchfamily7268 perfect pitch does help, but you don’t need it. You can do it just fine using relative pitch if you have a sample sound and you compare it to the one in question.
Bonus solution: I don’t know if this is exactly the logic to it, but it’s the best way I can think of to describe it. The piles with an odd number of gold-facing coins (or even number of silver) can be converted to all gold, the others can’t. Because you flip two coins each time, there are 3 possibilities for a given flip: - If you flip two gold, they become two silver (-2 gold) - If you flip two silver, they become two gold (+2 gold) - If you flip one of each (in either order), the ratio stays the same (+0 gold) And because of the math principle that “odd + even = odd” and “even + even = even”, any trio with 0 or 2 gold at the start cannot be made into 3 gold. So in essence, exactly half of these scenarios can be made into all gold (the others can be made into all silver since they begin as opposites).
For the bonus puzzle, you can tell whether you can flip all the coins in a trio to gold side up if two of the coins are silver side up. Otherwise, if there are one or three coins silver side up, you can't make them all gold side up.
well the bonus riddle is simple... any pile with an odd Nimber of gold up coins can be made into an all gold pile while piles with an even number of gold up coins can't be made into all gold up since there are an odd no. of coins in each pile
Piling has made this problem quite difficult. It is quite easy to understand if instead of piles, we take individual coins lying on the table. Then just take 20 coins and flip them over.
For the bonus riddle, all the arrangements with a odd number of gold coins up can be made all gold and all the arrangements with an even number of gold coins facing up cannot. This is because whenever you swap, you either swap two gold and subtract the number of gold coins up the same, one gold and one silver where the number of gold coins up stays the same, or two silver where the number of gold coins goes up by two, all even differences. Since the final result must be 3, an odd number, you have to start with 3 or 1 gold coins up.
For the sake of others who also don't want to Google it: it is a term in the cryptocurrency world which originated from a misspelling of the word "hold" meaning to hold on to the coins in hope that they will appreciate in value.
man i got asked this question in a final round interview for a consulting group and i had no idea what to say and even after watching this i still have no idea what's going on
Step 1: Count them all Step 2: Organise them into two equal piles Not one rule states that the silver coins must be facing upwards at the end. Due to this rule not being there, every coin can be classed as a silver coin. As long as you have two equal piles, you know they both contain an equal number of silver coins.
@@crk.s_kermo i would say to just test your luck and put one of the sides to test if it was gold or silver (there is not anything against failing, just flip the coin) BUT if it is always silver it becomes unsolvable
I knew this from a magic trick using 10 or 20 cards, lets say 10. 5 would be face up and a spectator would shuffle without flipping cards, give it to you without looking, put it behind your back, count out 5 and flip that, now both decks have the same amount of face up and down
The bonus riddle was easy (well, I just happened to tackle the same problem before). In odd numbered arrangement like this, you can never change which side has odd number of occurences in the sequence. Two differently sided coins will always just exchange their sides, two of the same will just change the count by two, so the odd sum will remain odd. So the only option is to see which side is shown odd number of times, either just on one coin or already on all three, then flip accordingly. - for one that has the odd one on the edge, flip the other two, - for the one that has the odd one in the middle, flip any pair you want, then follow the instructions in the above line
