Question

Let \[n \in Z ^{+}\] and let

n = \[p _{1}^{k1} + p _{2}^{k2} + p _{3}^{k3} ..... + p _{n}^{kn}\]

be its prime decomposition. How many divisors n has.

Answer 1

How many divisors does n have?*

Answer 2

So this is NOT a product?

Answer 3

I'm assuming not. Whatever n is equal to, the number of divisors of n have to be determined.

Answer 4

\[p _{1}^{k1} + p _{2}^{k2} + p _{3}^{k3} ..... + p _{n}^{kn}\] is not a prime decomposition i don't think

Answer 5

Oops, its actualy multiplication, not addition.

Answer 6

In that case, the number of divisors \(d\) with this factorization is
$$
d(n) = (k_1 + 1) (k_2 + 1) \cdots (k_n + 1),
$$