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# Folks. Not sure if this was ever posted before if u saw this

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Intern
Joined: 16 Jan 2004
Posts: 14
Location: Indonesia
Folks. Not sure if this was ever posted before if u saw this [#permalink]

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30 Jan 2004, 22:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Folks. Not sure if this was ever posted before if u saw this one.

How many positive divisors does a number M have, if M can be expressed as:

M = (p1^n1)(p2^n2)(p3^n3).....(pn^nk)

px = Prime divisor
nx = Power of prime divisor

eg M = 12 = 2^2 x 3^2
Manager
Joined: 25 Jan 2004
Posts: 92
Location: China

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30 Jan 2004, 22:26
(n1+1)(n2+1)...(nk+1)

|{1, 2, 2^2} X { 1, 3, 3^2}| = 9
Manager
Joined: 26 Dec 2003
Posts: 227
Location: India

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31 Jan 2004, 16:00
Bhai, lets say the factors of x=a^p * b^q *c^r then the number of divisors = (p+1) * (q+1) * (r+1). For example 500= 5^3 * 2^2 then the no of divisors will be (3+1) (2+1) = 12 , They are 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250 and 500. Hope it helps.
SVP
Joined: 16 Oct 2003
Posts: 1801

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31 Jan 2004, 17:47
Thanks rakesh1239. grrrrrrr
Senior Manager
Joined: 23 Aug 2003
Posts: 461
Location: In the middle of nowhere

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31 Jan 2004, 20:58
Rakesh..gr8888 explanation.

Vivek.
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"

Manager
Joined: 26 Dec 2003
Posts: 227
Location: India

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31 Jan 2004, 21:00
Hey Vivek r u not watching cricket? whats happening with our players
31 Jan 2004, 21:00
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