@ SAM Thanks On Tue, Jun 26, 2012 at 8:21 PM, SAMM <somnath.nit...@gmail.com> wrote:
> 1 /x + 1/y = 1/(n!) > * Consider N = n! , * > *The Equation becoz :-* > > 1/x + 1/y = 1/N > or (x+y)/xy = 1/N > or N( x + y ) = xy > > *Changing sides we get :-* > xy - N(x+y) = 0 > > *Adding N^2 on both sides we get :-* > xy - N( x + y) + N^2 = N^2 > or xy - Nx - Ny + N^2 = N^2 > or x(y - N) - N (y - N ) = N^2 > or (x - N) (y - N) = N^2 > > > From this equation we find that we can find the number of solution equal > to the total number of divisors of (N ^ 2) .or ( n! ^2) . > > So you need to find the divisors of the square of the n! which can be done > by finding the primes factor of the n! .... > > For example :- n! = p1^a * p2^b * ........ pn^x .... *[ p1 , p2 .. > pn are the prime factors ]* > > (n1 ^ 2) = p1^2a * p2^2b * ........ pn^2x > > So the number of divisors are *(2a + 1) * (2b +1 ) * ................ (2x > + 1) *.. You need to calculate this ... > > No need to calculate the factorial .... just need to check for the prime > factor from (2 to n ) . > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Regards Kumar Vishal _________________________________________ *http://wethecommonpeople.wordpress.com/ * *h**ttp://kumartechnicalarticles.wordpress.com/<http://kumartechnicalarticles.wordpress.com/> * _________________________________________ -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
