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 February 10th, 2018, 03:36 PM #1 Member   Joined: Oct 2017 From: Rumba Posts: 39 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, 05:44 PM #2 Member   Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 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. Tags discrete, math, question Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post nwicole Number Theory 5 October 23rd, 2014 02:49 PM mgk501 Real Analysis 2 March 21st, 2013 12:23 PM amyporter17 Applied Math 1 November 17th, 2010 07:23 AM peiyilee Applied Math 0 March 24th, 2008 08:46 AM

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