# There are 6561 balls out of them 1 is heavy.find the min. no. of times the balls have to be weighed

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.
I'd like some clarification... You mean six thousand five hundred sixty one balls? I've done problems like this before, but never on this large of a scale, so I just want to make sure there isn't a typo or something.

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For problems like this, just find the 3rd root of the number. For ex: consider 3 balls, u can find the heavier by just weighing 2 balls, so min = 1. consider 9 balls, divide them into 3 equal groups, now u can weigh 2 groups to find the group in which the heavier ball lies, and u ll be left with 3 and u know how to find out of 3. so its 3 pow 2 = 9, then 3 pow 3 = 27...... 3 pow 8 = 6561. So the ans is 8.
clear explanation :)
yuvasri Jul 20, 2010
vry clear to me
arpita Sep 10, 2010
iam not clear. 3 balls can be weighted two times. first one is to weigh with second and there after the result with third. so we r weighing two times for 3 balls. clarify my doubt plsssssssss
@sathvick13

while solving this Ques. we consider all the ball of equal weight , except the only ball which is the heavy .

ex A = 1 , B =2 , C =1
consider the above 3 balls A,B,C of weight 1,2,1 respectively . so 2 find the heavier ,

case 1 :: if u weigh 1st A,C
than when u find both the balls equal , so u'll predict that the 3rd ball B is heavy

case 2 : if u weigh 1st A,B
now when u find B heavy , then u'll predict that since only 1 ball is heavy among the 3 ball , so u dont need 2 weigh again the result with 3rd ball .
indar00 Dec 07, 2010
in case of 8 balls???
sudi Nov 19, 2013
8
Please explain how you came to that conclusion. Give a man a fish, feed him for a day, teach a man to fish, feed him for life.
The way you are asking the question the correct answer would be 3.

That is the minimum number of times the balls would have to be weighted to find the heavy ball. If lucky enough to get the ball within the first 3 weights that is...

Anything less than 3 we dont know which ball is which so 3 must be the answer.

/ McKeen
Edited Mar 30, 2010
That makes absolutely no sense. If you divide the balls up into thirds, weigh each, split the third containing the ball into thirds, and do that a third time, you'd still have 243 balls to sort through. If you're going by the logic that picking the balls one by one will get you the right ball, then the minimum would be to weigh it once.
6561 if u notice is 3^8...n can be divided by 3(prime factor) only...we keep on dividing it into sets of 3 and in 1 weighing itself you can determine which set is heavy. Divide that set into further 3 lots...keep doing that..
6561
2187 2187 2187 1st weighing
729 729 729
243 243 243
81 81 81
27 27 27
9 9 9
3 3 3
1 1 1

thnx a lot for d clarification
prem Dec 31, 2010
how in only one weighing which one is the heaviest in the group can b determined???? can someone pls xplain this.....
ankita1 Feb 21, 2011
my question is how find tat there is 3 group only
Why should we make it into 3 groups??? Cant we make it 4,5,6....etc
Only if you divide into 3 groups you can find in 1 weighing which one is heavy(If both the groups being weighed have equal weight then the third group or else the one weighing heavier). If more than 3 groups, then it takes more attempts.
divya Sep 19, 2012
if the number of balls given are not in the multiples of 3 then how to solve the problem?
smiley Mar 22, 2014
if we divide the balls into 6561 groups..then min number of attempts is 1 ie u weigh the ball in 1st attempt..my question is why should we divide into 3 groups and not 6561 groups??? o.O
If you divide 6561 groups, you have to weigh each ball at a time which is not minimum..

Let me explain:

the 1st trial is weighing balls in 3 groups each group having 2187 balls.
1st 2187 2nd 2187 3rd 2187. Lets say the 2nd group is showing heavier weight.

Then we start 2nd trial with the 2nd group of balls only because we know that this group is heavier than other two so the heavier ball must be in this group. Now we again divide this 2187 balls in 3 groups each group having 729 balls. We weigh the 3 groups and find say 3rd group as heavier.

We repeat trials 8 times till we arrive at the last stage with 1 ball in each group which means we are taking 8 attempts to find the heavier ball.