In a group of 80 coins,exactly one is counterfeit and weighs less than others. You are provided to weigh the coins. The minimum number of weighing required to determine counterfeit coin is?

• The weighing method is,
  Make the three groups as 27, 27 and 26
  Compare two 27 groups, if they are equal, the wrong coin is in the third group.
  If the first 27 group is heavier, the wrong coin is in the secong 27 group.
  If the first 27 group is lighter , the wrong coin is in the first 27 group
  
  After the first attempt, we could figure out, the group of 27 or 26 coins , where the wrong coin is.
  
  Now, the selected 27 group should be made as 9, 9 and 9 groups(Make 9,9 and 8 if 26 is selected)
  
  Of the three 9 groups, compare the first 9 group and the second 9 group, if they are the same, the wrong coin is in the third group.
  If they are not the same, we can select the corresponding group of 9 coins, where the wrong coin is
  
  So, after the second usage of the balance, we can finalise, the 9 coins group.
  
  In the third usage, make them groups of three each and we can find the three coin group by using the balance for the third time.
  For the fourth time, we are left with three coins.
  Compare one each in each side of the balance, if they are the same, we know the third one is the wrong coin.If they are not the same , we know which one of them is the wrong coin.
  Answer is 4 times.
• 10 years agoHelpfull: Yes(27) No(2)
• This question follows a pattern,
  Upto 3^n coins, we can find out the wrong one ( the heavier or the lighter coin as mentioned in the question) , by using the balance for (n) times.
  
  For any number up to 3, we have to use the balance for once
  For any number up to 9, we have to use the balance for 2 times.
  For any number up to 81, we have to use the balance for 3 times
  
  For any number up to 81, we have to use the balance for 4 times
  For any number up to 243, we have to use the balance for five times
  
  So, the answer is 4 times
  
  Note: The pattern will be different if we do not know the coin as lighter or heavier
• 10 years agoHelpfull: Yes(10) No(0)
• 6 turns....
  1. 80 (divide in two parts 40 40)=take less one
  2. 40(less) (divide into 2 parts 20 20)= take less one
  3. 20(less) (divide into 2 parts 10 10)= take less one
  4. 10(less) (divide into 2 parts 5 5)= take less one
  5. 5(less) (divide into 2 parts 2 2)= equal then 5th one left is defected one
  inequal then take less one
  6. 2(less) (weigh 1 and 1) can detect defeted one
  
  thus ans =6
• 10 years agoHelpfull: Yes(9) No(6)
• 14 time weighing required
  bcoz first we divides coin in 40+40 n weight them n find which 40 coins has that coin
  now these 40 divided into 20+20
  then in 10+10
  then in 5+5
  now we add a coin n divide 3+3
  now again add one coin n 2+2
  n finally we got that coin
• 10 years agoHelpfull: Yes(0) No(5)
• 14 times , becouse any thin 80
• 10 years agoHelpfull: Yes(0) No(3)