8 ball puzzle..to select lighter ball from them

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Question: Just Example : "Given 8 cue balls , one is weighing lesser than other 7. Find that(light weight) ball using just 2 chances on balance weight." How to find a general solution to these kind of question? How to divide this set of balls? Is there any finer method or general formula? Solution: Divide in group such that each group contain elements equal to certain power of 3 which is nearest to the total number of elements .. for example if total 25 elements are given divide in groups of 9 element each so there will be 2 groups of 9 elements and 1 group of 3 elements .. each group will be further divide into 3 groups of 3 element each .... coming to given problem of 8 cue balls .. divide into 3 groups .. 2 groups of 3 elements each .. say A and B 1 group of 2 element .. put group A and B in balance ... there are 3 cases .. case 1) if side with group A is lighter then remove the groups from balance and choose any 2 balls from group and put on balance .. if any one of them is lighter then u will directly know but if the balance is balanced then its the third ball (the one which u didnt put on balance) .. case 2) if side with B on it is lighter .. apply the same logic as above .. case 3) if balance is balanced then its the group 3 .. take the given 2 balls and put it on balance ... u r done :) POST YOUR OPINION IF YOU HAVE BETTER SOLUTION

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Question:
Just Example : "Given 8 cue balls , one is weighing lesser than other 7. Find that(light weight) ball using just 2 chances on balance weight."

How to find a general solution to these kind of question? How to divide this set of balls? Is there any finer method or general formula?

Solution:
Divide in group such that each group contain elements equal to certain power of 3 which is nearest to the total number of elements ..

for example if total 25 elements are given divide in groups of 9 element each so there will be 2 groups of 9 elements and 1 group of 3 elements .. each group will be further divide into 3 groups of 3 element each ....

coming to given problem of 8 cue balls .. divide into 3 groups ..
2 groups of 3 elements each .. say A and B
1 group of 2 element ..

put group A and B in balance ... there are 3 cases ..
case 1) if side with group A is lighter then remove the groups from balance and choose any 2 balls from group and put on balance .. if any one of them is lighter then u will directly know but if the balance is balanced then its the third ball (the one which u didnt put on balance) ..

case 2) if side with B on it is lighter .. apply the same logic as above ..

case 3) if balance is balanced then its the group 3 .. take the given 2 balls and put it on balance ... u r done :)

POST YOUR OPINION IF YOU HAVE BETTER SOLUTION