Here is a proof of why his method works:there are 10 H coins on table 1. Now we pick any 10 coins from table 1 and place those on table 2. Suppose we picked x no of H coins and 10-x no of T coins and put then in table 2. So now we have 10-x no of H coins in table 1. Now if we flip the coins in table 2 then no of T coins will become no of H coins=10-x H coins. So we have 10-x H coins in both the tables. Awesome puzzle and solution!!
I first saw this many years ago. It was a fun 'magic' trick, however it was done with 7 coins and the coins were covered with a cloth. You were to reach in and make the switch 'sight unseen'! It's an impressive trick. Problem was, I lost the trick somehow, forgot about it and could never find it again. It didn't really have a name, and searching for 'coin tricks' brought up way too many to search for. Eventually gave up looking and I could not figure out how it was done. I was unhappy with myself because of it. Then here you are with essentially that same trick! Many, Many Thanks! subscribed :)
@@kishlayamourya3141 puzzle is "Only the total no. Of coins with heads facing needs to be the same". Then how can you just take randomly pick and flip the coins and place them on table? This is not correct. It's just impossible. Coz there must be equal no.of coins facing heads on both the tables. 😐😐😐😐😐
@@HALLvsTON there is equal no. of heads facing up even if you pick any 10 random coins. You can try any combination. It will always work. Try yourself if you're not getting it.
@@HALLvsTON look think that u picked 10 random coins in which 3 are head up (means 7 heads remaining on first table) then flip and put on other table there'll be 7 heads that is equal to first table heads up.... Try for other combination you'll get same result
I know this answer is not good but..... I think it is also a correct answer On the table which have 35 coins.... Throw all the coins to the floor... Now both the tables have equal number of heads that is 0. 😀😂
In the 2 coins case is there no guaranty one coin is with head face and another one is tail face so when we pick up one coin and flip them that coin will be unequal number
Hey Deepraj.... if you have two coins (one head-up, one tails-up), flip anyone while placing on second table... and you will either have one heads-up on both tables... or zero head-up on both tables... in any case they are equal.
I think you've missed the part where he states "consider exactly 1 coin has heads facing up". This is an important part to understand the logic used in solving the problem. You could also state the same thing (meaning no guarantee about how many coins are heads up) about his 5 coin example where exactly 2 are heads up.
Dude you've misunderstood. In each case we will know the exact no. Of heads initially present. Hence if we have two coins and they say no head is present then it is certain that both are tales. And if they state 1 H is present then further procedure is same.
HEY so sorry to bother but do you, or anyone else here, know the original study where they put "two" teams in a dark room with a puzzle and a loud buzzer and the scientists in a different room and are buzzing them to get their reaction to see if some people are more aggressive than others? its a psychological study! PLZ HELP IM WRITING A PAPER DUE TONIGHT!!! thank you so much!
But here we have a darker room and we can't feel the texture of the coins then how will we know whether the coin is head faced or tail faced after moving.
To cover the case of 0 heads up coins, transfer half of the remaining coins to table 2 and flip them (In this case 25/2 = 12). Take 12 coins on table 1 and flip them too.
The condition is only the number of heads are need to be equal and the number of total coins on each table does not have to be equal so if there are no heads on table 1 you don't even have to move a single coin to table 2 cuz technically there are 0 heads on table 2 also .... But I appreciate your thinking ❤️
Alt. solution 1: Put all the coins on their side (edge) on either table, now you'll have exactly zero coins with the heads side up on each table. Just make sure they stay on the tables if that's a requirement. Alt. solution 2: Move the tables together so that their edges touch each other. Now stack all the coins on top of each other and place this single stack of coins on this border so that it sits on both tables at the same time. Now each coin with the heads side up will contribute to the coin heads count on both tables. (Of course, this requires the tables to be of the same height.)
You move 10 coins. Then the first table will have x amount facing up and the second table will have 10-x facing up. Flipping all 10 on the second table will make 10-(10-x)=x facing up. I directly solved like this instead of starting with smaller numbers.
Please make video is this puzzle named Chaos in the bus: There is a bus with 100 labeled seats (labeled from 1 to 100). There are 100 persons standing in a queue. Persons are also labelled from 1 to 100. People board on the bus in sequence from 1 to n. The rule is, if person ‘i’ boards the bus, he checks if seat ‘i’ is empty. If it is empty, he sits there, else he randomly picks an empty seat and sit there. Given that 1st person picks seat randomly, find the probability that 100th person sits on his place i.e. 100th seat.
Supposedly you move 25 coins from table1 to table 2 and coincidentally all the coins were heads facing down in table 1. So, on moving them to table 2 they become heads facing up. So table 2 has 25 coins heads facing up and table 1 has 10 coins with heads facing up. So your solution fails in this case. Can you justify or give a better solution???
Suppose the no. of heads up coins on table 1 was 'x'. So you have to move only 'x' random coins and flip them. You don't have to move more or less than 'x'. If you move more or less, then the solution will fail. The solution to this puzzle is that whatever number of heads up coins is given in the question (here 10) you will move that many coins and flip them.
My method ( just flip coins in a way , 1st coin should be flipped 2nd should be same , 3rd should be flipped , 4th would be same , and so on until 20 coins)
take 10 coins and flip them over.let's say that you got 3 "heads up"the first time,the number of remaining=10-3=7.when flipped:7 heads and 3 tails to match 7silver.we can prove this using Algebra: s1(pile 1)=20-n g2=20-n \flipped: \g2=s2 \s1=s2=20-n pile 2
Hey Tuan... thanks for sharing the approach(try to generalize for x number of coins instead of 20.)... I really enjoyed understanding the equations as the title for variables was not mentioned.
I've a doubt if I flip only tail coins or above 10 number of tail coins on the other table then the number of head faced coins on the two tables are not same
My doubt is, u said it is dark.So how can u know where those 10 H coins are in T1? If u take any random coins as to be flipped, then why cant i say all the 10 were tail facing up. How this puzzule is sloved without touching clarifications for my(kind of) questions???,.
How do we know that we are moving coins that have head or tail on them to the second table, as there is no light in the room at all. We can't tell which coin, wether head or tail , we are moving to second table.
I thought something is wrong with me that I am not able to solve these puzzles but now I am relaxed. Guys are not able to understand the solution 😂 You don't need to know, you just need to move 10.coins to other table and flip them.
Trick question, humans and robotic drones both emit some type of light so neither I nor a robot drone acting in my behalf are in the room. But if I assume no viable light and I have no goggles for the purpose... My guess is you want me to flip them before placing them on the tables but that would not make an even distribution.
Sir how to solve the puzzle......If there are two coins on the table with heads facing up on both the coin and then if one shifts any one coin from one table and puts it on another table by flipping it then on ist table there will be one head and on 2nd table there will be one coin with tails facing up....and no of coin with with heads facing up will not be equal?
first I thought I won't b able to solve it... may b it's a probability /pnc question ... but than just trying it, I got the solution 😄😄😄 feeling like genius 😅😅 though I know it's not a big deal ... anyways I loved this riddle ♥
if u don't feel the texture of the coins, then how would know us to flip coins make it an equal number of heads. anyhow, we can't see them bcz it was a dark room.
Noone says all other coins are tails. They may be on the side or in superposition. Noone says, that all coins have one head and one tail side. Anyway solution: move all coins out from table. 0 is equal to 0.
I guess in theory it works, if the conditions are perfect, but realistically you could grab all tails and if you only flipped two, you'd still have 8 heads up coins on the other table.
But how would you get the exact number of head facing coins which you r picking from table 1 as you can't see them and can't feel the surface??? In this way your answer is not solving this puzzle ....
Here is a proof of why his method works:there are 10 H coins on table 1. Now we pick any 10 coins from table 1 and place those on table 2. Suppose we picked x no of H coins and 10-x no of T coins and put then in table 2. So now we have 10-x no of H coins in table 1. Now if we flip the coins in table 2 then no of T coins will become no of H coins=10-x H coins. So we have 10-x H coins in both the tables. Awesome puzzle and solution!!
I have a doubt y can't he just throw all the coins down so that both table have 0 coins facing head as well 0 on each😂😂😂
Here already written :you can move the coins to 2nd table
But it is nice thought
I was going to comment that
you can get selected even after giving this answer! they will appreciate.
🤣🤣🤣🤣
I first saw this many years ago. It was a fun 'magic' trick, however it was done with 7 coins and the coins were covered with a cloth. You were to reach in and make the switch 'sight unseen'! It's an impressive trick.
Problem was, I lost the trick somehow, forgot about it and could never find it again. It didn't really have a name, and searching for 'coin tricks' brought up way too many to search for. Eventually gave up looking and I could not figure out how it was done. I was unhappy with myself because of it.
Then here you are with essentially that same trick! Many, Many Thanks! subscribed :)
Welcome to the channel :)
Wow man I am addicted to your channel...keep up bro,,
Highly appreciated :) that really gives inspiration to make more and more videos.
You are making us logically yours
Of course; please don’t quit making videos; I am addicted to them too and I like your explanation; which inspires me to become more and more logical…
If I can't see or feel texture of coins how am I able to know that the coins I am placing on second table is head or tail
U dont need to know that....just take any 10 random coins and flip them and put them on 2nd table....
@@kishlayamourya3141 puzzle is "Only the total no. Of coins with heads facing needs to be the same". Then how can you just take randomly pick and flip the coins and place them on table? This is not correct. It's just impossible. Coz there must be equal no.of coins facing heads on both the tables. 😐😐😐😐😐
@@HALLvsTON there is equal no. of heads facing up even if you pick any 10 random coins. You can try any combination. It will always work. Try yourself if you're not getting it.
@@HALLvsTON look think that u picked 10 random coins in which 3 are head up (means 7 heads remaining on first table) then flip and put on other table there'll be 7 heads that is equal to first table heads up.... Try for other combination you'll get same result
Try it at home 😄
Super explanation sir...
I know this answer is not good but..... I think it is also a correct answer
On the table which have 35 coins.... Throw all the coins to the floor...
Now both the tables have equal number of heads that is 0. 😀😂
Nice 😂😂
He has asked to move coins to other table.
Gr8....U r Genius....I'm a student of physics n interested in Maths....u R simply Awesome...Hatssoff...!!!
In the 2 coins case is there no guaranty one coin is with head face and another one is tail face so when we pick up one coin and flip them that coin will be unequal number
Hey Deepraj.... if you have two coins (one head-up, one tails-up), flip anyone while placing on second table... and you will either have one heads-up on both tables... or zero head-up on both tables... in any case they are equal.
My que is if both the coins are with same face then we pick up any one and flip them then they are unequal
LOGICALLY YOURS His logic is right... What will be the guarantee that one of the two will be head a ND other will be tail?
I think you've missed the part where he states "consider exactly 1 coin has heads facing up". This is an important part to understand the logic used in solving the problem. You could also state the same thing (meaning no guarantee about how many coins are heads up) about his 5 coin example where exactly 2 are heads up.
Dude you've misunderstood. In each case we will know the exact no. Of heads initially present.
Hence if we have two coins and they say no head is present then it is certain that both are tales. And if they state 1 H is present then further procedure is same.
This was an incredible riddle! I came close to the answer but didn't think to flip the coins each time you move them. Really awesome solution!
I got this question in an interview !! now seeing it here makes me waana watch all videos of your channel!!!
HEY so sorry to bother but do you, or anyone else here, know the original study where they put "two" teams in a dark room with a puzzle and a loud buzzer and the scientists in a different room and are buzzing them to get their reaction to see if some people are more aggressive than others? its a psychological study! PLZ HELP IM WRITING A PAPER DUE TONIGHT!!! thank you so much!
Again awesome!!!!!!
Great video
Woww...
I exactly solved the answer with the same approach as urs😋😁😁
But here we have a darker room and we can't feel the texture of the coins then how will we know whether the coin is head faced or tail faced after moving.
Yes
To cover the case of 0 heads up coins, transfer half of the remaining coins to table 2 and flip them (In this case 25/2 = 12). Take 12 coins on table 1 and flip them too.
The condition is only the number of heads are need to be equal and the number of total coins on each table does not have to be equal so if there are no heads on table 1 you don't even have to move a single coin to table 2 cuz technically there are 0 heads on table 2 also .... But I appreciate your thinking ❤️
really nice logic......you are just a professional in solving puzzles :) ......
Logically Yours :D :D :D
Alt. solution 1: Put all the coins on their side (edge) on either table, now you'll have exactly zero coins with the heads side up on each table. Just make sure they stay on the tables if that's a requirement.
Alt. solution 2: Move the tables together so that their edges touch each other. Now stack all the coins on top of each other and place this single stack of coins on this border so that it sits on both tables at the same time. Now each coin with the heads side up will contribute to the coin heads count on both tables. (Of course, this requires the tables to be of the same height.)
simply awesome
One suggestion - You can divide your videos in categories levels . e,g Easy, medium, hard, expert.
You move 10 coins. Then the first table will have x amount facing up and the second table will have 10-x facing up. Flipping all 10 on the second table will make 10-(10-x)=x facing up. I directly solved like this instead of starting with smaller numbers.
Please make video is this puzzle named Chaos in the bus:
There is a bus with 100 labeled seats (labeled from 1 to 100). There are 100 persons standing in a queue. Persons are also labelled from 1 to 100.
People board on the bus in sequence from 1 to n. The rule is, if person ‘i’ boards the bus, he checks if seat ‘i’ is empty. If it is empty, he sits there, else he randomly picks an empty seat and sit there. Given that 1st person picks seat randomly, find the probability that 100th person sits on his place i.e. 100th seat.
oh no!!!! really it's Awesome!!! you are a genius
Man this is awesome.......first....I di not even understand the questiin properly...but the answer was quite simple!
Thanks Amit
Your work is highly appreciated , Keep up the good work!!!!
jakas ...
keep it up ..
Hats off ..I thought it impossible. Untill I came to know about the answer
Nice one..
yaar laate kaha se ho....jo v ho mzza aa jata ....i subscribed only 4 channels on you tube,and Ur channel beating all.
will explained thanks
Nice puzzles sir.
Super questions man
Keep on doing man...u r something different level...
Thanks Vinith for your kind appreciation : )
Awesome 💥
Supposedly you move 25 coins from table1 to table 2 and coincidentally all the coins were heads facing down in table 1. So, on moving them to table 2 they become heads facing up. So table 2 has 25 coins heads facing up and table 1 has 10 coins with heads facing up. So your solution fails in this case. Can you justify or give a better solution???
Suppose the no. of heads up coins on table 1 was 'x'. So you have to move only 'x' random coins and flip them. You don't have to move more or less than 'x'. If you move more or less, then the solution will fail.
The solution to this puzzle is that whatever number of heads up coins is given in the question (here 10) you will move that many coins and flip them.
ITS WONDERFUL LOGICAL VIDEO BRO.
nice puzzle... i like it much.
I got it ✌🏻...
Wow.... I didn't know this. Thanks a lot
Amazing sir
Thanka for sharing!! Really challenging..
This question is of Pre RMO or RMO level
I loved the question
Thanks!!
Superbbbbbbb
Very nice channel. Keep up the good work...
My method ( just flip coins in a way , 1st coin should be flipped 2nd should be same , 3rd should be flipped , 4th would be same , and so on until 20 coins)
Really... Great brother!!!!
take 10 coins and flip them over.let's say that you got 3 "heads up"the first time,the number of remaining=10-3=7.when
flipped:7 heads and 3 tails to match 7silver.we can prove this using Algebra:
s1(pile 1)=20-n
g2=20-n
\flipped:
\g2=s2
\s1=s2=20-n
pile 2
Hey Tuan... thanks for sharing the approach(try to generalize for x number of coins instead of 20.)... I really enjoyed understanding the equations as the title for variables was not mentioned.
Marvelous bro.. I like it..😱😵
Thanks Danish
Your videos are awesome bro. Keep up the good work and thank you so much.Waiting for more videos😊
or just knock all coins off first table and now you have same number of heads up coins on both tables, aka Zero
It's awesome
I've a doubt if I flip only tail coins or above 10 number of tail coins on the other table then the number of head faced coins on the two tables are not same
My doubt is, u said it is dark.So how can u know where those 10 H coins are in T1? If u take any random coins as to be flipped, then why cant i say all the 10 were tail facing up. How this puzzule is sloved without touching clarifications for my(kind of) questions???,.
I'm loving it bro keep it up
It's was really very very interesting and tricky
Superb Man...Best Puzzle...From where u get such inspirations???
Do Answer :)
Thank u Samaksh :) I have been solving puzzles since childhood, but I feel explaining the puzzles is even more fun :)
Mindbolwn
this is awsome
How would know in darkness whether it is facing heads or tails while flipping
😂😂😂
How do we know that we are moving coins that have head or tail on them to the second table, as there is no light in the room at all.
We can't tell which coin, wether head or tail , we are moving to second table.
I thought something is wrong with me that I am not able to solve these puzzles but now I am relaxed. Guys are not able to understand the solution 😂
You don't need to know, you just need to move 10.coins to other table and flip them.
Damn that was a tricky one
Same thing can be done with face up and face down cards
beautiful
Trick question, humans and robotic drones both emit some type of light so neither I nor a robot drone acting in my behalf are in the room.
But if I assume no viable light and I have no goggles for the purpose...
My guess is you want me to flip them before placing them on the tables but that would not make an even distribution.
Oh, I can just throw them all off both tables, duh!
,1st of all in that dark how could we know which 10 coins are facing heads up in order to equalise the heads on 2 tables
Sir how to solve the puzzle......If there are two coins on the table with heads facing up on both the coin and then if one shifts any one coin from one table and puts it on another table by flipping it then on ist table there will be one head and on 2nd table there will be one coin with tails facing up....and no of coin with with heads facing up will not be equal?
Hey can u tell me how we came to know that how many coins of head and how many coins of tells while moving the coins from one table to another table.?
You don't need to know. You just need to pick 10 coins and flip them on other table.
I solved this😀😀😀😀
But how would we know which coin is faced up or down as we also cannot feel the texture of the coin
You don't need to know that. Just move 10 coins and flip them. It will work with any combination.
Blew my mind
The instructions says that you cannot see the coins and also can't feel the texture...? 😕
Sir u r great... what's ur age and where do u live
Thanks Ayush for the appreciation :) I am 30.. from Hyd
Didnt get it..its dark so how do v knw which coins r v fliping???
What if i just put all the coins on the second table without flipping them
I'm gonna subscribe ur channel
It was easy btw .😍
first I thought I won't b able to solve it... may b it's a probability /pnc question ... but than just trying it, I got the solution 😄😄😄 feeling like genius 😅😅 though I know it's not a big deal ... anyways I loved this riddle ♥
Perfect :)
This guy is Crazy.
if u don't feel the texture of the coins, then how would know us to flip coins make it an equal number of heads. anyhow, we can't see them bcz it was a dark room.
Pick 10 coins from from first table and keep it reverse on another table
I saw this kind of puzzle using playing cards....
Cool
And if odd no. of coins are facing up
Then??
The same idea will work.
Noone says all other coins are tails.
They may be on the side or in superposition.
Noone says, that all coins have one head and one tail side.
Anyway solution: move all coins out from table. 0 is equal to 0.
tysm
Very Enjoyable Game
I guess in theory it works, if the conditions are perfect, but realistically you could grab all tails and if you only flipped two, you'd still have 8 heads up coins on the other table.
Wow...
I ll show this as a magic,lol😎
ty bhai
Got it. Yahh
What if the coins are of even number
Hey Arya... any number of coins (even or odd) and any number of heads-up coins (even or odd) can be solved with this way.... give a try :)
How can u see in dark
Consider 3 coins heads facing up how to make heads equal on both sides..
But how would you get the exact number of head facing coins which you r picking from table 1 as you can't see them and can't feel the surface???
In this way your answer is not solving this puzzle ....
@podho dolan thanks bro !!!
Ohh u can know how and which one to flip fool stop Ur nonsense answers
In your 2nd example what if the man select 3 coin out of which 2heads and 1 tail ... Then it will become inequal
Always missing crucial informations... Like you have the right to flip the coins and the heads up coins do not need to be the same from the beginning
I solved It ,am I genius?
But In a dark room how can we know which coin is head and which is tail
What if majority are heads up like 18,19, .... In 35 ,does I need to follow the same concept
It works ok its the matter of flipping and complimenting each time
no at that time it wont works .... it will work only if the condition to be satisfied is "no of head facing down should be equal"