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.
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.
Same reasoning... of course in the first case tge second weight is to determine if the 101st coin is heavier or lighter. It is enough to replace one coin from one of the two 50-set with the 101st coin, or compare one of the true coins with the 101st coin
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 }
Here we even not divide in 25-25 then also we get the answer. I will talk about the case where when we take 50-50 coins in each pan and the weight is not balanced. Then take lighter pan and replace any one coin from lighter pan to remaining one coin . Then if weights are balanced then coin which we replace is fake and it is lighter. And if weights are not balanced even then fake coin is in other pan and it is heavier than usual coin
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.
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.
Complete solution for 12 coins (I suppose so) Select 1,2,3,4 and weigh it with 5,6,7,8. If scales balance then coins 1-8 are real and u weigh 1,2,3 with 9,10,11 If scales balance again then 12 is fake and u weigh 1 and 12 to find out. If 9,10,11 is heavier than 1,2,3 weigh 9 and 10. If scales balance again then 11 is fake and is heavier. If they dont balance then the fake coin is the one thats heavier on the scales. Same goes for when 9,10,11 is lighter. Suppose scales don't balance at the start and assume 1,2,3,4 is lighter than 5,6,7,8. Weigh 1,2,5 with 3,4,6. If scales balance then either 7 or 8 is fake and is heavier, and weigh 7 with 8 to determine. If 1,2,5 is lighter than 3,4,6 again then either 1 or 2 is lighter or 6 is heavier. Weigh 1 with 2. If they balance 6 is heavier, otherwise the lighter coin is fake. If 1,2,5 is heavier than 3,4,6 means either 3 or 4 is lighter or 5 is heavier. Weigh 3 with 4. If they balance 5 is heavier, otherwise the lighter coin is fake. Same goes for the opposite outcome. Lol I'm not even gonna bother trying to solve the 36 coins and 4 tries version of this problem.
I think the more appropriate formula for this is where *k* is the number of times u can use the scales, the max coins where this is solvable in every possible outcome is 0.5( 3^k - 3), by that I mean it is possible to do 39 coins with 4 weighings
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?
Pal I really like your puzzle videos, your selection of probems and your presentation. I think you are the best I have found for these kind of channels. Thank yououou
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.
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.
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?
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..
@@anseilerIf we take 50 coins one side & 50 coins another side in the classical balance, Case 1: if they show equal remaining one coin is fake & we can replace remaining coin with one side 50 coins then Sub case (a): if replaced side 50 coins lesser weight then fake coin is less weight. Sub case (b): if replaced side 50 coins showing more weight then fake coin is more weight. Case 2: if two sets of 50 coins in the balance showing un- equal weight then... Sub case (a): take lesser weight 50 coins & devide them into 25 & 25 then weight them... If they show equal, the defect coin is more weight. If they showing un- equal the fake coin is less weight.
If fake coin will be in last 2 then set 1 and 2 will weigh equal. Which means all these 66 coins are genuine. Now take any 35coins from these 60 and weigh it with remaining 35 coins.
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
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
Split 101 coins in 3 sets as 30,30,41 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 🤔
1. We know only one coin is fake.(given in ques.) 2. we do not have to identify the fake coin. Instead, we have to tell if the fake coin is heavier or lighter than a genuine coin.
Let's say the LEFT 50 coins are lighter than the RIGHT 50 coins. You split LEFT into two 25 sets and weigh them: 1. The resulting 25-25 split is balanced, so the fake coin is in RIGHT. Since RIGHT was heavier than LEFT, the fake coin is heavier. 2. The resulting 25-25 split is unbalanced, so the fake coin is in LEFT. Since LEFT was lighter than RIGHT, the fake coin is lighter. An analog can be made for the case were LEFT is heavier than RIGHT.
@@Tomix4kand if the two groups of 50 are balanced, we know the extra coin is fake, and we can measure the one fake coin against any one of the real coins
@@nathen9085 You can't know which 50 set holds the fake coin, so you are most likely to end up weighing a normal coin with another normal coin, gaining no information.
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 🙄
No.... not valid... consider each coin weighs 1 gm... and fake coin weighs 1.5 gm(it's heavier)... so the 50 coins pan that goes down should have fake coin...In your second case you place the last genuine coin in the lighter pan... now the weight on 50 coins pan would be 50.5 gm.... and on the 51 coins pan would be 51 gm... So the lighter pan comes down, but that does not mean the fake coin is lighter . Please reply if you got my point.
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..
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
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:
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
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.
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
Split the coins into 2 piles of 50 and weigh them. If they are the same weight then the coin left over is the fake one 😜 BUT .. more likely .. if they are different weights .. take the heavier pile of 50 and split it into 2 piles of 25 .. and weigh them. If they are the same weight then the fake coin is in the other pile of 50 .. and is lighter. If they are different weights then the fake coin is in the heavier pile .. and is heavier.
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 😆.
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
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.
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
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!
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.
Wait wait... pls elaborate 2nd case... for example.. if left pan of 50 goes down, the fake coin can be heavy.. but the fake coin might also be lighter (in the right pan).
@@LOGICALLYYOURS ok sry I just didn't elaborate it. Take heavy 50 bunch of coin , divide it with 25 coins each. Weigh it. If its equal , dn fake coin is light or vice versa.
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
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.
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
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.
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
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...
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
Puzzel 2: आपके पास 8 सिक्के हैं जो देखने में एक दम एक समान हैं . आपको पता चलता हैं की उसमे से एक सिक्का नकली हैं , और नकली सिक्का बाकी सब सिक्को से जरा सा वज़नी है. अगर आपके पास एक बैलेंस तराजू है (पर कोई बाट नहीं) आप सिर्फ 2 बार ही तौल सकते हैं तो नकली सिक्के को कैसे पहचानेंगे ? Easy right.
Bhai sabhi siko ko 3, 3 ,2 main baat lenge fir ek side 3 rakhenge or dusri side 3 agar barab hue to iska matlab in 6 siko main koi bhi nakli ni hai or baki ke do siko ko taul kr pta kr lenge kon sa bhari sika nikli hai Or agar 3, 3 sike jo suru main taule the vo barabar nhi nikalte to jo 3 sabse bhari hai unme se 2 sike lenge or unhe taulenge agar barabar nikle to samjhna baki ek bacha sika nakli hai or agar barabar ni nikal to jo bhari hoga vahi nakli hai
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.
Yep, this was my solution as well.
Yes bro
Same here
I got so confused when he said the other solution because I got this one as well
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.
well done bro
Same solution
Good one bro..
Same solution 😋
Same reasoning... of course in the first case tge second weight is to determine if the 101st coin is heavier or lighter. It is enough to replace one coin from one of the two 50-set with the 101st coin, or compare one of the true coins with the 101st coin
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
}
well done bro
Great solution
I thought the same bro well done
Here we even not divide in 25-25 then also we get the answer.
I will talk about the case where when we take 50-50 coins in each pan and the weight is not balanced.
Then take lighter pan and replace any one coin from lighter pan to remaining one coin .
Then if weights are balanced then coin which we replace is fake and it is lighter.
And if weights are not balanced even then fake coin is in other pan and it is heavier than usual coin
This is how I solved it.
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.
Perfect !! this is the expected answer.
I too thought of the same answer
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.
Complete solution for 12 coins (I suppose so)
Select 1,2,3,4 and weigh it with 5,6,7,8. If scales balance then coins 1-8 are real and u weigh 1,2,3 with 9,10,11 If scales balance again then 12 is fake and u weigh 1 and 12 to find out.
If 9,10,11 is heavier than 1,2,3 weigh 9 and 10. If scales balance again then 11 is fake and is heavier. If they dont balance then the fake coin is the one thats heavier on the scales. Same goes for when 9,10,11 is lighter.
Suppose scales don't balance at the start and assume 1,2,3,4 is lighter than 5,6,7,8.
Weigh 1,2,5 with 3,4,6. If scales balance then either 7 or 8 is fake and is heavier, and weigh 7 with 8 to determine.
If 1,2,5 is lighter than 3,4,6 again then either 1 or 2 is lighter or 6 is heavier. Weigh 1 with 2. If they balance 6 is heavier, otherwise the lighter coin is fake.
If 1,2,5 is heavier than 3,4,6 means either 3 or 4 is lighter or 5 is heavier. Weigh 3 with 4. If they balance 5 is heavier, otherwise the lighter coin is fake.
Same goes for the opposite outcome.
Lol I'm not even gonna bother trying to solve the 36 coins and 4 tries version of this problem.
I think the more appropriate formula for this is where *k* is the number of times u can use the scales, the max coins where this is solvable in every possible outcome is 0.5( 3^k - 3), by that I mean it is possible to do 39 coins with 4 weighings
@@AA-100 Can you give a procedure for the case of 39 coins not using the balance for more than 4 times ? Should one start comparing 12 coins ?
Avijit Panja. Should start with weighing coins 1-13 with 14-26
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?
Pal I really like your puzzle videos, your selection of probems and your presentation. I think you are the best I have found for these kind of channels. Thank yououou
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.
Nice puzzles. ..but i just want to know from where do u get soo many interesting puzzles.....good work though
Thanku for uploading such type of coin puzzle which generally asked during interview.....
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.
*101st
Very good explanation..
Will definitely try for the other approach
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
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?
I think the code word is TIE
Due for the U
@@lucazara9137 No
@@too_ql Right
@@too_ql no because if he tells the code word has a D then you can't know if the code word has 1 or 2 vowels
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..
if remaining coin is defect, you can‘t tell wheater it is lighter or heavier.
@@anseilerIf we take 50 coins one side & 50 coins another side in the classical balance,
Case 1: if they show equal remaining one coin is fake & we can replace remaining coin with one side 50 coins then
Sub case (a): if replaced side 50 coins lesser weight then fake coin is less weight.
Sub case (b): if replaced side 50 coins showing more weight then fake coin is more weight.
Case 2: if two sets of 50 coins in the balance showing un- equal weight then...
Sub case (a): take lesser weight 50 coins & devide them into 25 & 25 then weight them...
If they show equal, the defect coin is more weight.
If they showing un- equal the fake coin is less weight.
@@Goury30 I do understand the correct solution. I was answering and referring to the suggested version ii) of chevulanaresh6001.
I'm pleased with myself that many comments are mentioning the 50/50 method. That was my immediate idea.
Interesting for job interviews.... IT
Growing to Greatness.....
How about that last 2 coins ?
What if the fake coins is on the last 2 coins...
1 , 2 and 3 group coins have equal weight...
If fake coin will be in last 2 then set 1 and 2 will weigh equal.
Which means all these 66 coins are genuine.
Now take any 35coins from these 60 and weigh it with remaining 35 coins.
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
@@parveengoel9051
I really hope that he explained it this way :)
Hi ammar bro.. I am big fan of your presentation skill.. How can I contact you to get suggestions from you..
What if 1 coin out of 2 coins we taken out is fake,we need 3 trials..??
no because you will compare all 35 remaining coins with 35 coins from set 1 and 2 (they are all of same weight)
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
Your videos are really amazing
I am a big fan of you
Thanks Nanda :)
Split 101 coins in 3 sets as 30,30,41
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 🤔
I still don't get it
How to make sure the EXACTLY one fake coin
It's still confusing
1. We know only one coin is fake.(given in ques.)
2. we do not have to identify the fake coin. Instead, we have to tell if the fake coin is heavier or lighter than a genuine coin.
It is more easy if we take 50,50 and 1
you lose if the 50s are unbalanced
Let's say the LEFT 50 coins are lighter than the RIGHT 50 coins. You split LEFT into two 25 sets and weigh them:
1. The resulting 25-25 split is balanced, so the fake coin is in RIGHT. Since RIGHT was heavier than LEFT, the fake coin is heavier.
2. The resulting 25-25 split is unbalanced, so the fake coin is in LEFT. Since LEFT was lighter than RIGHT, the fake coin is lighter.
An analog can be made for the case were LEFT is heavier than RIGHT.
@@Tomix4kand if the two groups of 50 are balanced, we know the extra coin is fake, and we can measure the one fake coin against any one of the real coins
@@Tomix4kbut is it not easier if you take one coin from the unbalanced 50/50 and you measure it with the 1 coin you left at the beggining?
@@nathen9085 You can't know which 50 set holds the fake coin, so you are most likely to end up weighing a normal coin with another normal coin, gaining no information.
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 🙄
No.... not valid... consider each coin weighs 1 gm... and fake coin weighs 1.5 gm(it's heavier)... so the 50 coins pan that goes down should have fake coin...In your second case you place the last genuine coin in the lighter pan... now the weight on 50 coins pan would be 50.5 gm.... and on the 51 coins pan would be 51 gm... So the lighter pan comes down, but that does not mean the fake coin is lighter .
Please reply if you got my point.
Ooh ,thanks to correct me
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..
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
Nice question and nice explanation
Unable to understand the concept about the remaining two coins weight calculation.
I dont want to explain full.
But i solved this by split coins like 50 + 50 +1 . And u can assume the rest
how u find heavier or not heavier?
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:
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
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.
Good video!
We can also go with 50,50 ,1
compare 50 and 50...... if they are equal
Then exchange the 1 coin from any 50 coin set
very good video
You are awesome bro (ammar) how can I contact you
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
Split the coins into 2 piles of 50 and weigh them.
If they are the same weight then the coin left over is the fake one 😜
BUT .. more likely .. if they are different weights .. take the heavier pile of 50
and split it into 2 piles of 25 .. and weigh them.
If they are the same weight then the fake coin is in the other pile of 50 .. and is lighter.
If they are different weights then the fake coin is in the heavier pile .. and is heavier.
Who is losing the weight for face reveal??
Almost there :)
Find the coin with altered perimeter. Solved. Fake coins are always lighter. Else there is no motive to create a fake coin.
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 😆.
It was so easy....
That I got it in 3 seconds
Thank you bro. Its asked me at interview time.
In which interview can u tell?
What.... to do...when the fake coin present....in the last set of two coins
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
Amazing bro
Bro we can also put 50 50 in each case and one coin in any one case so we can find fake coin lighter or heavier
51 coins r more heavily than 50 coins for sure
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.
If i have 12 coins, how many weights would I need at least?
2
What if those extra 2 coins had that fake coin...?? 🤔
Watch again
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
Second method is we can split them into 50/50/1
Then also we will be able to find the fake coin weight
5111lojes:compare 1020×4
1021 > imbalance or balance
I have third method and iam going to write it in comment soon
Nice
You didn't tell about the remaining 2 coins in case2.
not needed
Yes what happen if the fake coins is one of the last two coin...
@@andidermawan6233 He told in the video, see carefully.
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
I know more than 10 ways to solve this puzzle.
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!
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.
Wait wait... pls elaborate 2nd case... for example.. if left pan of 50 goes down, the fake coin can be heavy.. but the fake coin might also be lighter (in the right pan).
@@LOGICALLYYOURS ok sry I just didn't elaborate it. Take heavy 50 bunch of coin , divide it with 25 coins each. Weigh it. If its equal , dn fake coin is light or vice versa.
@@alipnandi9964 Perfect... :)
@@LOGICALLYYOURS thank you sir, Big fan of your channel. Keep sharing us such amazing puzzles. It's very healthy workout for brain.
What if fake coin was in the set4 of (2)?
That's why we made a set of remaining 35 coins R(35)... that has those 2 coins in it.
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
Set1-50 coins
Set2-50coins
Set3-1coin
50~50 +1 : if 50=50 then 1 is fake .now 51 real 50real+1. Like this we can predict.
I think this solution is wrong. If you want to know why then reply to me.
I have an idea- make groups of 50 50 coins and break last coin
I took as 34,34 & 33
Ammar can I ask you what is your job ?
😂😂😂😂
👍
Verrrrry slow 🤦🤦🤦
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.
But what if the two coins set aside by us contains the fake one 🤨🤔
In that case first comparison of 33 and 33 coin will be equal. So take any 35 from first 66 coins and 35 remaining and compare. If remaining 35 weigh more than fake coin has more weight and vice versa.
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.
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
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...
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
That's exactly what I thought, how is he going to determine that
Watch the video again. Even i thought the same thing so i rewatched. He's clearly used all the 101 coins. Watch it carefully.
Looks like you forget that,
There's 50 coins in each side in second condition .
We can also solve it by taking 50 50 coins
Hii
Why only "bro" in all the comments? Gender bias.
First one to comment
Puzzel 2:
आपके पास 8 सिक्के हैं जो देखने में एक दम एक समान हैं . आपको पता चलता हैं की उसमे से एक सिक्का नकली हैं , और नकली सिक्का बाकी सब सिक्को से जरा सा वज़नी है. अगर आपके पास एक बैलेंस तराजू है (पर कोई बाट नहीं) आप सिर्फ 2 बार ही तौल सकते हैं तो नकली सिक्के को कैसे पहचानेंगे ?
Easy right.
Bhai sabhi siko ko 3, 3 ,2 main baat lenge fir ek side 3 rakhenge or dusri side 3 agar barab hue to iska matlab in 6 siko main koi bhi nakli ni hai or baki ke do siko ko taul kr pta kr lenge kon sa bhari sika nikli hai
Or agar 3, 3 sike jo suru main taule the vo barabar nhi nikalte to jo 3 sabse bhari hai unme se 2 sike lenge or unhe taulenge agar barabar nikle to samjhna baki ek bacha sika nakli hai or agar barabar ni nikal to jo bhari hoga vahi nakli hai
@@sachinbaranwal6987 Right.
See my other puzzel comment.