101 Coins Puzzle - One coin is fake - Is it heavier or lighter?
video description
Fist compare 30 coins with 30 coins There is 2 possiblities
i) They will be equal
ii) They will not be equal
Lets see first condition
i) If They are equal in weight then these 60 coins are real.
Then compare remaining 41 coins with any 41 coins out of 60 real coins.
If 41 real coins are weighing more than 41 fake coins then fake coin must be lighter and vice versa
Now second condition
ii) If They are not equal
fake coin must be in these 60 coins.
but either of 30 coins must be real coins.
At the same time remaining 41 coins are real.
Now replace 30 coins which is weighing more by remaining 30 coins out of 41 real coins
After replacement if both are came to eqaul then fake coin must be heavier.
If this causes no change then fake coin must be lighter.
I think I cracked this one correctly.
I think so -
Date: 2023-11-15
Related videos
Comments and reviews: 29
ABHISEK
I do have a more easy approach.
First, we should divide the coins in 3 grups - A - 50 coins, B - 50 coins, C - 1coin.
1. Now, we have to balence A and B. If they are balenced, we can easily make out that coin C is fake and exchange any 1 coin from A or B with the fake coin and we can find out weather it is heavier or lighter.
2. If, A and B are not balenced, then we have to divide either A or B in two grups of 25 coins each and balence them. Example - Suppose, A is heavier than B. So, divide A into 2 groups- of 25 coins - A1 and A2. Now, is we see A1 and A2 balences each other then the fake coin is in B and so it is lighter or if we see A1 and A2 dosen't balences each other, then A has a fake coin which is heavier.
reply
I do have a more easy approach.
First, we should divide the coins in 3 grups - A - 50 coins, B - 50 coins, C - 1coin.
1. Now, we have to balence A and B. If they are balenced, we can easily make out that coin C is fake and exchange any 1 coin from A or B with the fake coin and we can find out weather it is heavier or lighter.
2. If, A and B are not balenced, then we have to divide either A or B in two grups of 25 coins each and balence them. Example - Suppose, A is heavier than B. So, divide A into 2 groups- of 25 coins - A1 and A2. Now, is we see A1 and A2 balences each other then the fake coin is in B and so it is lighter or if we see A1 and A2 dosen't balences each other, then A has a fake coin which is heavier.
reply
Danyal
I solved it using a different approach:
We make groups like, 50, 50, 1
Now we compare the piles of 50, 50:
Case1: They are equal in weight. Thus, the last coin is the counterfeit coin, and we can compare it against any one of the other genuine coins in the second round, thus letting us know if the counterfeit coin is heavier or lighter.
Case2: One of the piles of 50 is heavier. This would mean one of them contains the counterfeit coin. Let's divide the heavier pile into 25, 25, and weight them against one another. If they are equal, the other pile of 50 had the counterfeit coin and it was lighter. If one of the piles is heavier, it contains the counterfeit coin, and the counterfeit coin is heavier than the rest.
reply
I solved it using a different approach:
We make groups like, 50, 50, 1
Now we compare the piles of 50, 50:
Case1: They are equal in weight. Thus, the last coin is the counterfeit coin, and we can compare it against any one of the other genuine coins in the second round, thus letting us know if the counterfeit coin is heavier or lighter.
Case2: One of the piles of 50 is heavier. This would mean one of them contains the counterfeit coin. Let's divide the heavier pile into 25, 25, and weight them against one another. If they are equal, the other pile of 50 had the counterfeit coin and it was lighter. If one of the piles is heavier, it contains the counterfeit coin, and the counterfeit coin is heavier than the rest.
reply
Jaya
Take 50, 50 and 1. Weigh 50 and 50 first. If it is equal, no problem. Weigh the remaining 1 coin with any of the 100 and say whether the last fake is lighter or heavier. At the first attempt if both are not equal then take one of the 50. Say lighter 50 and divide equally 25, 25 and weigh. If equal then the heavier in the first attempt has fake coin and the fake is heavier. In the second attempt if both differs then the fake coin is in the lighter 50 and the fake coin is lighter. Same way if you divide the heavier 50 for the same attempt. If equal first lighter 50 has fake and fake coin is lighter. If differs then the heavier 50 has fake and the fake coin is heavier.
reply
Take 50, 50 and 1. Weigh 50 and 50 first. If it is equal, no problem. Weigh the remaining 1 coin with any of the 100 and say whether the last fake is lighter or heavier. At the first attempt if both are not equal then take one of the 50. Say lighter 50 and divide equally 25, 25 and weigh. If equal then the heavier in the first attempt has fake coin and the fake is heavier. In the second attempt if both differs then the fake coin is in the lighter 50 and the fake coin is lighter. Same way if you divide the heavier 50 for the same attempt. If equal first lighter 50 has fake and fake coin is lighter. If differs then the heavier 50 has fake and the fake coin is heavier.
reply
sunil
This solution is tricky, try to understand
And my grammar is weak so grammatical mistakes are there
First: like others divide the coins in set 50, 50 and 1
Weigh the 50 and 50
Everyone knows, if they balance equal.
If they weigh unequal
We come to know the third set of only 1 coin is real
Now we just put that coin on the lighter set of 50 coins
If weight balance show no change means the coin is heavy and is in the heavier set of 50 coins
And if lighter set of 50 coins comes down, means coin is lighter and it is now in the set of 51 coins
This is my approach -
reply
This solution is tricky, try to understand
And my grammar is weak so grammatical mistakes are there
First: like others divide the coins in set 50, 50 and 1
Weigh the 50 and 50
Everyone knows, if they balance equal.
If they weigh unequal
We come to know the third set of only 1 coin is real
Now we just put that coin on the lighter set of 50 coins
If weight balance show no change means the coin is heavy and is in the heavier set of 50 coins
And if lighter set of 50 coins comes down, means coin is lighter and it is now in the set of 51 coins
This is my approach -
reply
Paulo
Wanna face tough challenges?
1) -12 similar coins-, 1 of them different in weight from the others. Use the balance at most -3 times- to identify the odd one, and wether it is lighter or heavier than the others.
2) -36 similar coins-, 1 of them different in weight from the others. Use the balance at most -4 times- to identify the odd one, and wether it is lighter or heavier than the others.
In general, -12 x 3-k- coins (k = 0, 1, 2,, and you may use the balance -k + 3- times, always identifying wether the odd one is lighter or heavier than the others.
reply
Wanna face tough challenges?
1) -12 similar coins-, 1 of them different in weight from the others. Use the balance at most -3 times- to identify the odd one, and wether it is lighter or heavier than the others.
2) -36 similar coins-, 1 of them different in weight from the others. Use the balance at most -4 times- to identify the odd one, and wether it is lighter or heavier than the others.
In general, -12 x 3-k- coins (k = 0, 1, 2,, and you may use the balance -k + 3- times, always identifying wether the odd one is lighter or heavier than the others.
reply
Tharani
There is a lot of possible solutions in this puzzle example to split 101 coins into 50, 50, 1
First test two 50 to 50 is there is equal another one coin is fake so test one fake coin and one geniune coin we can get the solution
Second step if there is not equal two set of 50 coin, again split the one heavier 50 coins into 25, 25 to test it now the test will equal the fake coin is in the another set of 50 coins and it was lighter incase second test is not equal the fake coin is heavier and is there in heavier side of 25 coins
reply
There is a lot of possible solutions in this puzzle example to split 101 coins into 50, 50, 1
First test two 50 to 50 is there is equal another one coin is fake so test one fake coin and one geniune coin we can get the solution
Second step if there is not equal two set of 50 coin, again split the one heavier 50 coins into 25, 25 to test it now the test will equal the fake coin is in the another set of 50 coins and it was lighter incase second test is not equal the fake coin is heavier and is there in heavier side of 25 coins
reply
sachin
Another solution, ---
Divide in 50coin, 50 coin and 1
Compare 50 with 50 -
1st case they equal:
If they equal means the last one coin is fake compare one genuine coin with last remaining fake coin and you get the result
2nd case they unequal:
If they does not equal that means fake coin is in either 1st 50 coin bunch or in 2nd 50 bunch.
Take heavy 50 bunch coin than divide them in 25 and 25 if they equal means fake coin is light in weight and if 25 and 25 coin doesnot equal means the fake coin is heavy
-
reply
Another solution, ---
Divide in 50coin, 50 coin and 1
Compare 50 with 50 -
1st case they equal:
If they equal means the last one coin is fake compare one genuine coin with last remaining fake coin and you get the result
2nd case they unequal:
If they does not equal that means fake coin is in either 1st 50 coin bunch or in 2nd 50 bunch.
Take heavy 50 bunch coin than divide them in 25 and 25 if they equal means fake coin is light in weight and if 25 and 25 coin doesnot equal means the fake coin is heavy
-
reply
adarsh
SECOND METHOD
We split coins into set of 50. We weigh these two sets. If the weight of both the sets is equal then the one coin left is fake.
If the weight of 2 sets of 50 coins is unequal then we pick any one set(say lighter set. We break the set again in 2 set of 25 coins. We weigh the coins. If the weight is equal then the coin was heavier(fake coin was in the heavier set which we discarded. But if two set of 25 coins are unequal in weight then the fake coin is lighter because we picked the lighter set of 25 coins.
reply
SECOND METHOD
We split coins into set of 50. We weigh these two sets. If the weight of both the sets is equal then the one coin left is fake.
If the weight of 2 sets of 50 coins is unequal then we pick any one set(say lighter set. We break the set again in 2 set of 25 coins. We weigh the coins. If the weight is equal then the coin was heavier(fake coin was in the heavier set which we discarded. But if two set of 25 coins are unequal in weight then the fake coin is lighter because we picked the lighter set of 25 coins.
reply
shubham
How about we split it like 50, 50, 1. Compare 50-50first if they are similar then the last coin is fake and by comparing it with any coin from last two sets of coin we will find if its fake coin is light or heavy. 2 condition. If first 50 50 don't balance then take the light set and devide it in two sets of 25 if it doesn't balance then it contains fake coin and scenes it was lighter in first comparison we can say the fake coin ways less than original one. We can do same thing with heavy set.
reply
How about we split it like 50, 50, 1. Compare 50-50first if they are similar then the last coin is fake and by comparing it with any coin from last two sets of coin we will find if its fake coin is light or heavy. 2 condition. If first 50 50 don't balance then take the light set and devide it in two sets of 25 if it doesn't balance then it contains fake coin and scenes it was lighter in first comparison we can say the fake coin ways less than original one. We can do same thing with heavy set.
reply
CHEVULA
I have another way.
30, 30, 41 coins.
And one more way is.
25, 25, 50, 1 coin.
First weigh 25, 25 coins, if not equal, take one set 25coins and 25coins from 50 coins. We identify.
Second: if 25, and 25 coins are equal. mix these coins and put them one side and 50coins in other side. i) if 50 coins less wt, defect coin is less wt. If more wt, then defect coin is more wt.
ii) if equal remaining coin is defect.
reply
I have another way.
30, 30, 41 coins.
And one more way is.
25, 25, 50, 1 coin.
First weigh 25, 25 coins, if not equal, take one set 25coins and 25coins from 50 coins. We identify.
Second: if 25, and 25 coins are equal. mix these coins and put them one side and 50coins in other side. i) if 50 coins less wt, defect coin is less wt. If more wt, then defect coin is more wt.
ii) if equal remaining coin is defect.
reply
Psych--
I did with 34+34+33 coins. Call them a, b, c respectively
Weigh a and b
-If same, compare with any 33 of a or b with c
-If not same, take upper (a or b. Now call it u as it's upper) n separate into 17 each as u1 n u2. Now compare u1 n u2. If not same upper one is having fake n is lighter
Incase if u took lower n separate them, u can conclude lower one is having fake n it is heavier
reply
I did with 34+34+33 coins. Call them a, b, c respectively
Weigh a and b
-If same, compare with any 33 of a or b with c
-If not same, take upper (a or b. Now call it u as it's upper) n separate into 17 each as u1 n u2. Now compare u1 n u2. If not same upper one is having fake n is lighter
Incase if u took lower n separate them, u can conclude lower one is having fake n it is heavier
reply
Srinivasan
My answer before watching the video:
Put 50 coins each on either side of the scale. If one side is lighter, the counterfeit is lighter, if one side is heavier, the counterfeit is heavier. But if both sides weigh the same, then the last coin is counterfeit. Clear the scales, take that coin and any other coin. Weigh them against each other and we have our answer!
reply
My answer before watching the video:
Put 50 coins each on either side of the scale. If one side is lighter, the counterfeit is lighter, if one side is heavier, the counterfeit is heavier. But if both sides weigh the same, then the last coin is counterfeit. Clear the scales, take that coin and any other coin. Weigh them against each other and we have our answer!
reply
Vamshi
Let's assume two parts has 50 coins each. Weight them and take heavier part and divide into 25 each. In second weight comparison if they both are equal then fake coin is lighter in weight if not heavier. Just don't mind 101th coin incase if it fake then u will know it in first comparison that shows equal weight. And compare fake one with genuine one in next step.
reply
Let's assume two parts has 50 coins each. Weight them and take heavier part and divide into 25 each. In second weight comparison if they both are equal then fake coin is lighter in weight if not heavier. Just don't mind 101th coin incase if it fake then u will know it in first comparison that shows equal weight. And compare fake one with genuine one in next step.
reply
---------
Split the coins into 3 groups of 37. Weigh a random 2 groups. if weight is same, the fake is in the remaining group. If not equal, choose any of the previously weighed group and exchange it with the remaining group. if weight equal, the remaining group has fake. If not equal, using the previous weighing comparison, we can determine the fake
reply
Split the coins into 3 groups of 37. Weigh a random 2 groups. if weight is same, the fake is in the remaining group. If not equal, choose any of the previously weighed group and exchange it with the remaining group. if weight equal, the remaining group has fake. If not equal, using the previous weighing comparison, we can determine the fake
reply
Alip
Ok. If my sets of coin are 50, 50 & 1. If first case, two sets are equal, dn obviously 1 coin which is left is fake and we can weigh the 2nd term to determine its heavy or light. And if 2nd case. The fake coin is in first of two 50's set, dn we can say easily say by weight the fake coin light or heavy. So simple, why so much confusing.
reply
Ok. If my sets of coin are 50, 50 & 1. If first case, two sets are equal, dn obviously 1 coin which is left is fake and we can weigh the 2nd term to determine its heavy or light. And if 2nd case. The fake coin is in first of two 50's set, dn we can say easily say by weight the fake coin light or heavy. So simple, why so much confusing.
reply
Dhiraj
Puzzel 1:
You have been given a list of words out of these words one is the secret word:
AIM
DUE
MOD
OAT
TIE.
With the list in front of you, if I tell you any one character of the code word, you would be able to tell the number of vowels in the code word. Can you tell which is the code word?
reply
Puzzel 1:
You have been given a list of words out of these words one is the secret word:
AIM
DUE
MOD
OAT
TIE.
With the list in front of you, if I tell you any one character of the code word, you would be able to tell the number of vowels in the code word. Can you tell which is the code word?
reply
W---d
My approach
Make 3 sets of 50, 50 and 1
Compare 50=50
1)If equals then,
next turn compare 1 with any coin from 50 you can get.
2) If not equal then
Take heavier 50 make sets of 25, 25
Compare if equal coin is lighter and present in another 50
If not equal then coin is heavier
reply
My approach
Make 3 sets of 50, 50 and 1
Compare 50=50
1)If equals then,
next turn compare 1 with any coin from 50 you can get.
2) If not equal then
Take heavier 50 make sets of 25, 25
Compare if equal coin is lighter and present in another 50
If not equal then coin is heavier
reply
Jaish
It's so simple no need to be make it so difficult just
Divide the coins into 50 and 51 coins and then suppose 50 coins weigh 50g
Then 51 coins weigh 51. 5g means coin is heavier by. 5 g
And if weight of Set b is smaller than 1 g means it is lighter
So easy
-IF YOU AGREE GIVE A LIKE-
reply
It's so simple no need to be make it so difficult just
Divide the coins into 50 and 51 coins and then suppose 50 coins weigh 50g
Then 51 coins weigh 51. 5g means coin is heavier by. 5 g
And if weight of Set b is smaller than 1 g means it is lighter
So easy
-IF YOU AGREE GIVE A LIKE-
reply
Rajat
What happened to the last 2 coins?
What if the fake coin is in this set of 2 coins. Hence the balance will b equal in both the weighing cases. But since we have used up our both the chances of weighing. Not it will b impossible to determine the heaviness of fake coin.
reply
What happened to the last 2 coins?
What if the fake coin is in this set of 2 coins. Hence the balance will b equal in both the weighing cases. But since we have used up our both the chances of weighing. Not it will b impossible to determine the heaviness of fake coin.
reply
Kush
Divide the set of 100 coins into 50, 50, 1 and then weigh the 50-50 set if they are equal then the one coin which is left out is the fake one and then measure it with any one coin of the 50-50 set whatever the result came out is the final result
reply
Divide the set of 100 coins into 50, 50, 1 and then weigh the 50-50 set if they are equal then the one coin which is left out is the fake one and then measure it with any one coin of the 50-50 set whatever the result came out is the final result
reply
Nitin
Knowing that you gave the idea of 3-1=2, i. e equal numbered coins in balance( 2 tries, and which gave hint for most of them in comment section as 101-1= 50(again 2 tries enough. Yet you gave a different final answer which is not efficient -.
reply
Knowing that you gave the idea of 3-1=2, i. e equal numbered coins in balance( 2 tries, and which gave hint for most of them in comment section as 101-1= 50(again 2 tries enough. Yet you gave a different final answer which is not efficient -.
reply
Yohan
i have one more solution for this case. split them into 2 groups. take off one coin from the heavier groups, if the scale changes in weight then the fake coin is lighter. and if it doesn't change or it balance, the fake coin is heavier.
reply
i have one more solution for this case. split them into 2 groups. take off one coin from the heavier groups, if the scale changes in weight then the fake coin is lighter. and if it doesn't change or it balance, the fake coin is heavier.
reply
Saikat
There is not only one more way. there are many more ways. we just need to create 3 sets of coins. such that (number of coins in set 1)=(number of coins in set 2) - (number of coins in set 3) and total number of coins= 101.
reply
There is not only one more way. there are many more ways. we just need to create 3 sets of coins. such that (number of coins in set 1)=(number of coins in set 2) - (number of coins in set 3) and total number of coins= 101.
reply
anonymous
divide the coin into 50 and 51 case 1: obviously 51 is greater, pick one from greater, if weight balance are are equal, picked coin is faulty coin,
weight it with 1 of 100 coin and get the answer,
case 2:
reply
divide the coin into 50 and 51 case 1: obviously 51 is greater, pick one from greater, if weight balance are are equal, picked coin is faulty coin,
weight it with 1 of 100 coin and get the answer,
case 2:
reply
FARONA
The solution in this video is wrong.
How if on 1st weighing set 1 and set 2 equal.
And on set 2nd weighing, they are also the same. The fake coin is on the remaining 2 coin rhat ledt out.
reply
The solution in this video is wrong.
How if on 1st weighing set 1 and set 2 equal.
And on set 2nd weighing, they are also the same. The fake coin is on the remaining 2 coin rhat ledt out.
reply
sayed-sarfaraz
In second 2 you have used only 99 coins, what if its balance is equal after two time, you can't determine it between remaining 2 coins bcz there is no other chance left. So it is wrong
reply
In second 2 you have used only 99 coins, what if its balance is equal after two time, you can't determine it between remaining 2 coins bcz there is no other chance left. So it is wrong
reply
William
Another way: split it into two sets of 50 and a single coin. Divide the heaviest side into 2 sets of 25. You can then deduce if the counterfeight coin is lighter or hevier than the others.
reply
Another way: split it into two sets of 50 and a single coin. Divide the heaviest side into 2 sets of 25. You can then deduce if the counterfeight coin is lighter or hevier than the others.
reply
education
Given that you have 2 uses of the scales and you only needed to determine if the fake coin is lighter/heavier, wouldn't this be possible with infinite coins?
reply
Given that you have 2 uses of the scales and you only needed to determine if the fake coin is lighter/heavier, wouldn't this be possible with infinite coins?
reply
agytjax
The easiest solution is to split it into 50, 50 and 1 (as many have indicated below. The division into 3 sets is not every elegant
reply
The easiest solution is to split it into 50, 50 and 1 (as many have indicated below. The division into 3 sets is not every elegant
reply
Add a review, comment
Other channel videos
