Number of Positive Divisors of an Integer: Proof

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SUMMARY

The discussion centers on proving the formula for the number of positive divisors of an integer based on its prime factorization. Specifically, if an integer n is expressed as n = p1^v1 * p2^v2 * ... * pk^vk, the number of positive divisors d(n) is calculated using the formula d(n) = (v1 + 1)(v2 + 1)...(vk + 1). The example provided involves the integer 2^r * 3^s, where the number of positive integral divisors is (r + 1)(s + 1). Participants emphasize the importance of understanding the structure of divisors in relation to their prime factors.

PREREQUISITES
  • Understanding of prime factorization
  • Familiarity with the concept of divisors
  • Basic knowledge of mathematical notation
  • Experience with integer properties
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Mathematicians, students studying number theory, educators teaching divisor functions, and anyone interested in the properties of integers and their factors.

Gear300
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Edit: I'm shifting to a more general question:
If the prime factorization of an integer n is given by

n = p1v1p2v2⋅⋅⋅pkvk

then what would be a proof for the number of positive divisors of n being

d(n) = (v1 + 1)(v2 + 1)⋅⋅⋅(vk + 1)
 
Last edited:
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Gear300 said:
I am supposed to find the number of positive integral divisors of 2r3s. The number is (r+1)(s+1). I tried a number of ways, but I'm not getting the intended answer. Any help?

Hi Gear300! :wink:

Just write out the typical divisor …

what does it look like? :smile:
 
In each divisor pik appear with 0 <= k <= vi
 

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