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September 1st, 2010, 04:35 AM  #1 
Newbie Joined: Aug 2010 Posts: 15 Thanks: 0  product of all the divisors of a number
Find the product of all the divisors of a number N N =a^p * b^q .... a and b are prime factors.How should I proceed to get the product of all the divisors of the number 
September 1st, 2010, 06:32 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: product of all the divisors of a number
I'm going to flip your notation: N = p^a * q^b * r^c * ..., lest I confuse myself. p and q (and r and s, to a lesser extent) are traditionally used to represent primes. There are total times that the inner product is evaluated. In of them, p is raised to the 0 power; in of them, p is raised to the 1 power, etc. In total p is raised to the power , where is the number of divisors of N. The same argument holds for every prime by changing the order of the products. Net result, I haven't seen this before. 

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