optimizing the solution a little bit more:

For N = 9 (3*3*3) = 27
For N= 12 (3^4) = 81
For N= 15 (3^5) = 243

and generalizing the solution it looks that if we are going to divide the 
given N the solution is: 3^(N/3)*3^(N%3). I think we can also prove it.

I hope it helps u. Could u plz explain...which kind of question are u 
looking for mathematical, algorithmic or riddle. As internet is full of all 
these problems....being specific will help us to suggest better

-rspr
http://techie-rs.blogspot.in/



On Friday, 10 August 2012 16:03:52 UTC+5:30, Sajal Jain wrote:
>
> For the first case,
> you have 9 circles (set S). Break them into two sets s1 and s2 such that 
> s1 U s2 = S with |s1| = 4 circles and |s2| = 5 circles.
> For the first set s1, you will come back to the original configuration at 
> times 4,8,12,16,*20*,24... multiple of 4. Similarly for set s2 you will 
> come back to the original configuration at times 5,10,15*,20*,25.. multiples 
> of 
> 5. You can infer from this that the answer should be lcm(|s1|,|s2|).
>
> So for a general case, the solution should be the max{lcm of 
> (|s1|,|s2|,...)} with s1 U s2 U s3.. =S and intersection is null. 
>
> I hope you get the explanation.
> Read Kristofer's solution again once or twice and you will get it surely. 
> He also suggests on what shall be done to calculate the answer for a 
> general case.
>
> --
> Cheers,
> Sajal Jain
>
>
>
> On Thu, Aug 9, 2012 at 11:40 PM, Bonethug <[email protected]<javascript:>
> > wrote:
>
>> Ppl pls look at this question and help me out here, pls!!!! I need a 
>> better explanation, didn't get a hang of the answer kristofer wrote! 
>>
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