There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?

• 3
• 14 years agoHelpfull: Yes(10) No(7)
• It requires a maximum of 4 weighings. Divide the 9 balls into sets of 3 each. Let them be set1, set2 and set3. Weigh set1 and set2 against each other. If they weigh equal, then set3 contains odd coin. Otherwise, weigh set1 against set3. If they weigh equal, set2 contains odd coin. If in both cases, they do not weigh equal, then set1 contains odd coin. Now, we have identified set with defective coin in maximum 2 weighings. Repeat this process on individual coins of the found set. This requires a max of 2 more weighings. Hence, the odd coin can be found in a maximum of 4 weighings at all times.
• 14 years agoHelpfull: Yes(8) No(10)
• 2
  take all coins into a group of 3
  on first iteration we can identify the group in which the odd coin resides
  then on second iteration we can find the coin
• 14 years agoHelpfull: Yes(8) No(9)
• Actually, it is not given whether the ball is overweight or underweight. It can be either. So, in first iteration, we cannot find identify the group containing the odd ball. We need two iterations for that. So, if there needs only 'n' iterations to find which ball is overweight (or underweight, and is given), then it requires 2n iterations for finding which ball is odd. here, n = log9/log3 = 2. So, 2n = 2*2 = 4
• 14 years agoHelpfull: Yes(4) No(5)
• wrong solve by TIJO this time
  divide 9 coins in 3 groups of 3 coins
  weight any two group two cases then 1= equal 2= unequal
  1-is its equal then = coin from 3rd group this 6 coins are equal
  2-unequal take the wrong wight (lesser 1 if coin light and heavy one is coin is heavy )
  take the wrong weight group
  and weight 1 coin against other then 2 casee
  1 = equal = the coin u left is odd one
  2 = is unequal u can find the odd weight
  
• 12 years agoHelpfull: Yes(4) No(6)
• One of these 8 coins is the odd one. Name the coins on
  heavier side of the scale as H1, H2, H3 and H4. Similarly,
  name the coins on the lighter side of the scale as L1, L2,
  L3 and L4. Either one of H's is heavier or one of L's is
  lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is
  one coin remaining in intial weighing.
  
  o If both are equal, one of L2, L3, L4 is lighter. Weigh L2
  against L3.
  
  ? If both are equal, L4 is the odd coin and is lighter.
  
  ? If L2 is light, L2 is the odd coin and is lighter.
  
  ? If L3 is light, L3 is the odd coin and is lighter.
  
  o If (H1, H2, L1) is heavier side on the scale, either H1 or
  H2 is heavier. Weight H1 against H2
  
  ? If both are equal, there is some error.
  
  ? If H1 is heavy, H1 is the odd coin and is heavier.
  
  ? If H2 is heavy, H2 is the odd coin and is heavier.
  
  o If (H3, H4, X) is heavier side on the scale, either H3 or
  H4 is heavier or L1 is lighter. Weight H3 against H4
  
  ? If both are equal, L1 is the odd coin and is lighter.
  
  ? If H3 is heavy, H3 is the odd coin and is heavier.
  
  ? If H4 is heavy, H4 is the odd coin and is heavier.
  
  3. The remaining coin X is the odd one. Weigh X against the
  anyone coin used in initial weighing.
  
  o If both are equal, there is some error.
  
  o If X is heavy, X is the odd coin and is heavier.
  
  o If X is light, X is the odd coin and is lighter.
• 11 years agoHelpfull: Yes(3) No(5)
• Two iterations : log9/log3 = 2
• 14 years agoHelpfull: Yes(2) No(9)
• Two weighings are enough bcoz, we split the coins into 3 equal sets.if we weigh any two sets against each other,there are three chances 1.both are equal 2.one set weigh more or less(3) from that we can find the odd set and it consists of three coins,now weigh one coin against the any one of other coin.the same three possibilities exist
• 8 years agoHelpfull: Yes(0) No(0)