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February 10th, 2018, 04:36 PM  #1 
Member Joined: Oct 2017 From: Rumba Posts: 36 Thanks: 0  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? 
February 10th, 2018, 06:44 PM  #2 
Member Joined: Jan 2016 From: Athens, OH Posts: 92 Thanks: 47 
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 06:53 PM. 

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