vishal_2804 wrote:can u plz explain the logic for As x, y, and z are prime, number of factors of xyz is (1 + 1)(1 + 1)(1 + 1) = 8.
Thanks
If the prime factorization of a positive integer N is (p^a)*(q^b)*(r^c)..., the number of different positive factors of N is given by (a + 1)(b + 1)(c + 1)...
This is because any factor of N,
- may contain either p� or p¹ or p² ... or pᵃ ---> (a + 1) ways to select or not select p
may contain either q� or q¹ or q² ... or qᵇ ---> (b + 1) ways to select or not select q
may contain either r� or r¹ or r² ... or rᶜ ---> (c + 1) ways to select or not select r
Ans so on...
Note that when none of the prime factors are selected (p�, q�, r�...), we get the factor 1 and when all the prime factors are selected (pᵃ, qᵇ, rᶜ ...), we get N itself as factor.
Hence, the steps for finding out the number of different positive factors of any integer N are...
- 1. Find out the prime factorization of N
2. Increment the powers of all the prime factors by 1.
3. Multiply the incremented powers.
For example,
- Number of different positive factors of 12 = (2²)*(3¹) is (2 + 1)*(1 + 1) = 6
Number of different positive factors of 360 = (2³)*(3²)*(5¹) is (3 + 1)*(2 + 1)*(1 + 1) = 24
In this case prime factorization of xyz is (x¹)*(y¹)*(z¹)
So, number of different positive factors of xyz is (1 + 1)*(1 + 1)*(1 + 1) = 8
Hope that helps.
