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February 10th, 2018, 03:36 PM   #1
Joined: Oct 2017
From: Rumba

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Discrete Math question

Answer requires justification:

98000 = 2^4 * 5^3 * 7^2.

(a) Compute the number of distinct divisors of 98000
(b) Compute the number of odd divisors of 98000

Anyone got any ideas how to do this?
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February 10th, 2018, 05:44 PM   #2
Joined: Jan 2016
From: Athens, OH

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Let d(a) be the number of divisors of the positive integer a. Then d is multiplicative; i.e. if m and n are relatively prime, d(mn)=d(m)d(n). Also if p is a prime, $d(p^k)=k+1$. Now it should be obvious how to compute.

Alternatively, any divisor must be of the form $2^i5^j7^k$ where $0\leq i\leq4$, $0\leq j\leq3$ and $0\leq k\leq2$.

Last edited by johng40; February 10th, 2018 at 05:53 PM.
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