Question

How to find the no of factors of a number using prime factorization?

Answer 1

Prime factorization of 840 = 23×3 × 5 ×7. So, the number of factors of 840 = (3+1)(1+1)(1+1)(1+1) = 32. The number of pairs that will yield unique positive integral solutions for this equation = no. of factors /2 = 32/2 = 16. Because, for every pair, say, 4 x 210, we obtain unique solutions for x and y.

Answer 2

Your Question :---- How to find the no of factors of a number using prime factorization ?

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Basic formula related to factors of a number :------

Let a number N, such that,

N= p^aq^br^c

Where, p, q and r are prime factors of the number n.

Where, p, q and r are prime factors of the number n.a, b and c are non-negative powers/ exponents

Than:----

Number of factors of N = (a+1)(b+1)(c+1). (Ans).

$\huge\blue{ADDITIONAL\:INFORMATION}$ :---

==>> Product of factors of N = N^(No. of factors/2)

==>> Sum of factors = ( p^0+p^1+...+p^a) ( q^0+ q^1+....+q^b) (r^0+r^1+...+r^c)/ (p^a-1)(q^b-1)(r^c-1)