9 comparisons

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8 Comparisions..Divide 6561 in 3 parts..2187 of eachKeep 2 of parts in weight machine...Take the heavy part n again divide in 3 partsIf they are eual then take the 3rd part..2187>729>243>81>27>9>3>1

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7 times.

6561 = 3 ** 8

First, divide the group of 6561 into 3 groups of 2187. Take two arbitrary groups and weigh them. Either they will be imbalanced (in which case you've identified the group with the heaviest ball), or they are equal, in which case you know the 3rd group has the heaviest ball. Split the heaviest group of 2187 into 3 groups of 729, and continue on - so first test 6561, then 2187, then 729, then 243, then 81, then 9, then 3 balls remain. You can deduce which one is heaviest at this (7th) step in which 3 balls remain.

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6541

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I have an alternate solution which makes only 1 comparison in the best case and 12 it n worst case.

split 6561 as A:B:C 3280 ,3280 and 1

==>comp 1 if A and B are equal then C is the heavy ball

else

comp 2 split 3280 as two groups of 1640. Take the heavier group for next comparison

comp 3 split 1640 as two groups of 820. Take the heavier group for next comparison

comp 4 split 820 as two groups of 410. Take the heavier group for next comparison

comp 5 split 410 as two groups of 205. Take the heavier group for next comparison

==>comp 6 split 205 as three groups of 102,102 and 1. if A and B are equal then C is the heavy ball else take the heavier group for next comparison

comp 7 split 102 as two groups of 51. Take the heavier group for next comparison

==>comp 8 split 51 as three groups of 25,25 and 1. if A and B are equal then C is the heavy ball else take the heavier group for next comparison

==>comp 9 split 25 as three groups of 12,12 and 1. if A and B are equal then C is the heavy ball else take the heavier group for next comparison

comp 10 split 12 as two groups of 6. Take the heavier group for next comparison

comp 11 split 6 as two groups of 3. Take the heavier group for next comparison

comp 12 compare 2 balls. you will find the heavy ball

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**abinash kumar**
Answered On : Sep 29th, 2011

8

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**yogita**
Answered On : Oct 2nd, 2011

Only once

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**Sumit Arora**
Answered On : Oct 6th, 2011

Its Answer is 8.

we divided the total number of balls in 3 parts.then then weighing any two parts(of balls).if the is in-balance then the heavy ball is present in heavy side or if their weight are equal then the heavy ball is present in third part.again divide the remaining balls in three parts and repeat this procedure until you find it.

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**sachin**
Answered On : Oct 10th, 2011

6561/9=729

729/3=243

243/9=27

27/9=3

3/3=1

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**Samueal**
Answered On : Jan 26th, 2012

8 times is the right answer.

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