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 ...
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