There are 6561 balls out of them 1 is heavy.Find the min. no. of times the balls have to be weighed for finding out the haevy ball.

• Answer is 8 times. The simple logic is to divide total balls into three sets. Weigh any two sets against each other(these two must contain equal number of balls). From this we can identify which set contains the heavy ball. Divide this set into three and repeat until you find the heavier ball. Under this logic, the minimum number of weighings required turns out to be the smallest integer greater than or equal to log(n), where n is the total number of balls and the base of logarithm is 3. Or simply [log(n)/log(3)] with any base. Here, n = 6561. log 6561 / log 3 = 8
• 14 years agoHelpfull: Yes(67) No(10)
• 8
• 13 years agoHelpfull: Yes(5) No(30)
• THERE ARE 6560 BALLS OF COMMON WEIGHT, AND 1 HEAVY BALL....BUT AS WE KNOW THAT NOTHING IS HEAVY OR LIGHT, ITS ALL RELATIVE..HENCE, THE HEAVY BALL IS HEAVIER THAN 6560 BALLS.. HENCE THERE IS A NEED OF COMPARISON BETWEEN HEAVY AND LIGHT BALLS... A MINIMUM OF ONLY TWO TRIES IS REQUIRED TO VERIFY THE HEAVIER BALL...
  THERE ARE TWO SUCH CASES- ONE WE PICK THE HEAVIER BALL FIRST, AND OTHER WE PICK LIGHTER FIRST AND HEAVIER BALL SECOND...
• 14 years agoHelpfull: Yes(3) No(74)
• 8 times
  
  just divide by 3 continuously , u will get ans .
  
  like 6561/3=2187
  
  
• 8 years agoHelpfull: Yes(0) No(1)