@Sammm: Suppose that N has prime factorization N = p1^e1 * p2^e2 * ... * pn^en
where ^ indicates exponentiation. Then for a and b to be coprime, a must contain all or none of the factors of each prime, and similarly for b. Thus, a is the product of some subset of the pi^ei and b is the product of the complementary subset. If in your example you regard (33,8) as different from (8,33), then there are 2^n coprime factors (a,b); otherwise there are 2^(n-1) coprime factors. In your example above, N = 2^3 * 3 * 11, so n = 3. The factors are (1,264), (3,88), (8,33), (11,24), and their reverses, if you count the reverses as distinct. Here I have used that 1 is considered coprime to every integer. Dave On Oct 28, 9:51 am, SAMMM <[email protected]> wrote: > If a natural number N is given such that N = a × b where a and b are > the factors of N. How many such sets of (a, b) can be formed in which > the selection of the two numbers a and b is distinctly different if N > = 8 × 33 and the distinct factors should be Prime to each other ? -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
