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by Sunny22uk » Thu Aug 28, 2008 6:20 pm
we can split 6^m to 2^m x 3^m
that means we need to >= number of 2's and equal number of 3's in the numerator than the denominator in order for the faction to be an integer
In this Q, you can do a quick divisibilty test and find out how many 3's are there, since 3's are rare in the sequence, we don't have to worry about number of 2's.

we have 5 3's so answer should be 5
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by kshankker » Thu Aug 28, 2008 6:52 pm
Iam not getting it...is the question is (92*103)/6^m.......then wat is the use of N and K........
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by sumithshah » Sat Sep 06, 2008 10:31 pm
I also thought about it for a while until I was like dhu!

The N and K is used for the numerator - n&k = n * (n=1).... *(k)

Makes sense now? The numerator is in the form of n&k NOT n*k ( where n and k are integers
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ans was clearly 5

by karthikgmat » Sat Sep 06, 2008 11:04 pm
mathematical solving of this ques is easy...
its 5
logically we have to get numbers which have factors as 2& 3
and 6 there will be 5 so M=5
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