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 September 1st, 2010, 03:35 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, 05:32 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. $\prod_{d|N}d=\prod_{A=0}^a\prod_{B=0}^b\cdots p^Aq^B\cdots$ There are $(a + 1)(b + 1)(c + 1)\cdots$ total times that the inner product is evaluated. In $(b + 1)(c + 1)\cdots$ of them, p is raised to the 0 power; in $(b + 1)(c + 1)\cdots$ of them, p is raised to the 1 power, etc. In total p is raised to the power $(0 + 1 + \cdots + a)(b + 1)(c + 1)\cdots=a(a+1)/2\cdot(b+1)(c+1)\cdots=a(a+1)/2\cdot d(N/p^a)=a(a+1)/2\cdot d(N)/(a+1)=a/2\cdot d(N)$, where $d(N)=(a+1)(b+1)\cdots$ 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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# product of all the divisors of n

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