How many positive integers are factors of a given number?

Question

I've been trying to find / generate a formula for the following problem:

• Given a number, how many positive integers are factors of this number.

In practice, you could simply build a table as such (lets assume the number is 36):

1 | 36
2 | 18
3 | 12
4 | 9
6 | 6

Thus there are 9 positive integers that are factors to 36.

This method seems like it would be taxing if the number was large. So I was thinking that if you knew the prime factorization for a number such as $\,2 \cdot 2 \cdot 3 \cdot 3 = 36, \,$ there should be a formula for the number of unique combinations you can multiply these prime factors together. That's where I am stuck. Thanks in advance!

Answer 1

Hint: if the prime factorization of $\,n\in\Bbb N\,$ is

$$n=\prod_{k=1}^rp_k^{a_k}$$

then the number of divisors of $\,n\,$ is

$$\prod_{k=1}^r(a_k+1)$$

Answer 2

Let's say you have a number n. Let $n=\ p_1^r.p_2^m.p_3^n$ then the number of positive factors is (r+1)(m+1)(n+1). (I do not know how to make $r_1,r_2$ in exponent, so I used r,m,n. In your case, 36=$2^2.3^3$ so (2+1)(2+1)=9.

Answer 3

Consider this: Every positive integer has a unique prime factorisation, so choosing integers is the same as choosing prime factorisations. A prime factorisation divides another prime factorisation if and only if the exponents of each prime in the former are less than or equal to their counterparts in the latter.

So the number of divisors is the number of ways of choosing exponents satisfying that condition. For each prime $p^a$, I can choose any exponent between $0$ and $a$ inclusive – that's $a + 1$ choices. So, in conclusion: if \[n = \prod_{k = 1}^{N} p_k^{a_k}\] then \[\mathrm{\#factors}(n) = \prod_{k = 1}^N (a_k + 1)\]