no of factors of a number

 n=p^a * q^b * r^c.....

where p,q,r= unique prime numbers from prime factors

no of prime factors(nf) = (a+1)(b+1)(c+1)...

i.e find prime factors in power form and product of power increased by 1 

 in numbre theory nf is called tau function

proof:

let d( ) = no of factors function  

len x = prime number

then,

d(n)=2                  i.e n and 1 are factors

 d(x^n)=n+1        i.e its divisible by x^0,x^1,x^2 ... x^n so n+1 unique factors

 

so , 

 n=p^a * q^b * r^c .....

d(n)=  (a+1)(b+1)(c+1)