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September 1st, 2010, 04:35 AM   #1
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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
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September 1st, 2010, 06:32 AM   #2
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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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