If we write down any number N in their prime factorization form as N =\( x^{a} *{y^b}*{z^c }\)and so on.
Where x,y,z are prime numbers.
Then the total no of factors of N =(a+1)(b+1)(c+1) and so on.
Let’s try it with an example, N = 20
We can prime factorize N as \( {2^2 } *{5^1}\)
We can use the above mentioned formulae to find the number of factors = \((2+1)*(1+1) =3*2= 6\)
Let’s confirm it:
The factors of 20 are1,2,4,5,10,20
So, we have a total of 6 factors
The pairs of factors that give the product N are (1,20), (2,10) ,(4,5)
So total no of pairs we got are 3.
Hence, we can generalise it as, if the no of factors is even, then the no of pairs of factors that give the product as N= (no of factors)/2
In the above example it would be 6/2 = 3.
Take another number N = 16
We can prime factorize N as \(2^4\)
No of factors = (4+1) = 5
Let’s write down the factors of 16
They are 1,2,4,8,16
So pairs of factors that give the product N are (1,16), (2,8) ,(4,4)
The total no of pairs is only 3.
We can generalise it as, if the no of factors is odd, then the no of pairs of factors that give the product as N= (no of factors +1)/2 = (5+1)/2 = 3
Hope this explanation helps.
Thanks,
Clifin J Francis