101 Coins Puzzle - One coin is fake - Is it heavier or lighter ?

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Title	:	101 Coins Puzzle - One coin is fake - Is it heavier or lighter ?
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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 , brweight it with 1 of 100 coin and get the answer, brcase 2:
Comment from : anonymous yogi

very good video
Comment from : Victor Kamat

Unable to understand the concept about the remaining two coins weight calculation
Comment from : Sai Kishore

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
Comment from : agytjax

We can also solve it by taking 50 50 coins
Comment from : Saurabh Kumar Singh

There is a lot of possible solutions in this puzzle example to split 101 coins into 50,50,1brFirst 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 brbrSecond 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
Comment from : Tharani M

If i have 12 coins, how many weights would I need at least?
Comment from : Alex

Nice question and nice explanation
Comment from : Sagar Modi

Why only "bro" in all the comments? Gender bias
Comment from : Steven

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
Comment from : Shinchan

Hii
Comment from : sonia yadav

I solved it using a different approach:brWe make groups like, 50, 50, 1brNow we compare the piles of 50, 50:brCase1: 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 lighterbrCase2: 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
Comment from : Danyal Shamsi

I did with 34+34+33 coins Call them a,b,c respectivelybrbrWeigh a and bbr*If same, compare with any 33 of a or b with cbrbr*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 lighterbrbrIncase if u took lower n separate them, u can conclude lower one is having fake n it is heavier
Comment from : Psychøツ

Good video!
Comment from : Megha Aggarwal

What if 1 coin out of 2 coins we taken out is fake,we need 3 trials??
Comment from : Justus Ece

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
Comment from : sayed-sarfaraz

Amazing bro
Comment from : super sonic

5111lojes:compare 1020×4 br 1021 > imbalance or balance
Comment from : 고유환

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
Comment from : மணிகண்டன் ஆறுமுகம்

What to dowhen the fake coin presentin the last set of two coins
Comment from : VISHAL S P

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?
Comment from : A A

I took as 34,34 & 33
Comment from : SomBeri

But what if the two coins set aside by us contains the fake one 🤨🤔
Comment from : G_3094_Nikhil Bhardwaj

Nice
Comment from : anurag verma

You are awesome bro (ammar) how can I contact you
Comment from : Vasu Rateriya

Hi ammar bro I am big fan of your presentation skill How can I contact you to get suggestions from you
Comment from : sra1 kumar

Verrrrry slow 🤦🤦🤦
Comment from : Keerti Sharma

The solution in this video is wrongbrHow if on 1st weighing set 1 and set 2 equalbrAnd on set 2nd weighing, they are also the same The fake coin is on the remaining 2 coin rhat ledt out
Comment from : FARONA FARRIONA

What happened to the last 2 coins?brWhat 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
Comment from : Rajat Pratap Singh Bisht

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
Comment from : Alip Nandi

What if those extra 2 coins had that fake coin?? 🤔
Comment from : Panikiran Bharadwaj

Ammar can I ask you what is your job ?
Comment from : luka eric

I have an idea- make groups of 50 50 coins and break last coin
Comment from : moumita majumdar

This solution is tricky , try to understand brAnd my grammar is weak so grammatical mistakes are therebrbrFirst : like others divide the coins in set 50,50 and 1brbrWeigh the 50 and 50brEveryone knows ,if they balance equal brbrIf they weigh unequal brWe come to know the third set of only 1 coin is real brbrNow we just put that coin on the lighter set of 50 coins brbrIf weight balance show no change means the coin is heavy and is in the heavier set of 50 coins brbrAnd if lighter set of 50 coins comes down, means coin is lighter and it is now in the set of 51 coins brbrThis is my approach 🙄
Comment from : sunil kumar

I have third method and iam going to write it in comment soon
Comment from : sunil kumar

I dont want to explain fullbrBut i solved this by split coins like 50 + 50 +1 And u can assume the rest
Comment from : devendra parmar

Nice puzzles but i just want to know from where do u get soo many interesting puzzlesgood work though
Comment from : Ram Prasad

What if fake coin was in the set4 of (2)?
Comment from : Shriram Bhat

Thank you bro Its asked me at interview time
Comment from : ranjitha gunasekaran

Interesting for job interviews IT brGrowing to Greatness
Comment from : mihir tari

Set1-50 coinsbrSet2-50coinsbrSet3-1coin
Comment from : Vaibhav Sharma

Your videos are really amazingbrI am a big fan of you
Comment from : Nanda peela

It is more easy if we take 50,50 and 1
Comment from : uma nagaralli

How about that last 2 coins ? brWhat if the fake coins is on the last 2 coinsbrbr1 , 2 and 3 group coins have equal weight
Comment from : Andi Dermawan

It was so easybrbrThat I got it in 3 seconds
Comment from : uma nagaralli

Puzzel 3:brbrसात चोरों ने मिलकर हीरे चुराए वो हीरों को लेकर जंगल में भाग गए, रात वोही काटने लगे दो चोरो ने सोचा की जब सब सो रहे हैं तो वो सारा हीरा आपस में बराबर बाँट कर भाग जाए लेकिन उन्होंने देखा की बराबर बांटने पर एक हीरा ज्यादा बच जा रहा हैं उन्होंने तीसरे चोर को जगाया सोचा की तीन में बराबर बाँटले लेकिन उन्होंने देखा की बराबर बांटने पर फिर से एक हीरा बच गया , उन्होंने चौथे चोर को उठाया फिर पांचवे छठे हर बार एक हीरा बच जाता जब उन्होंने सातवे चोर को जगाया तब जाके सबमे बराबर बराबर बंट पाया उन्होंने minimum कितने हीरे चुराये थे
Comment from : Dhiraj Govindvira

Puzzel 2:brbrआपके 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 ?brEasy right
Comment from : Dhiraj Govindvira

Looks like you forget that,brThere's 50 coins in each side in second condition
Comment from : Yuvrajsinh Rana

Another solution,👇👇👇brDivide in 50coin, 50 coin and 1brbrCompare 50 with 50 {br1st case they equal:brIf they equal means the last one coin is fake compare one genuine coin with last remaining fake coin and you get the resultbrbr2nd case they unequal:brIf they does not equal that means fake coin is in either 1st 50 coin bunch or in 2nd 50 bunchbrTake 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 heavybr}
Comment from : sachin baranwal

I still don't get itbrHow to make sure the EXACTLY one fake coinbrIt's still confusing
Comment from : Monica Nirbito

You didn't tell about the remaining 2 coins in case2
Comment from : Himanshu Agrawal

Very good explanationbrbrWill definitely try for the other approach
Comment from : mohammed Zubair

👍
Comment from : Arghya Sahoo

Thanku for uploading such type of coin puzzle which generally asked during interview
Comment from : prateek jain

Puzzel 1 : brbrYou have been given a list of words out of these words one is the secret word: brAIM brDUE brMOD brOATbrTIEbr 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?
Comment from : Dhiraj Govindvira