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

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Folks. Not sure if this was ever posted before if u saw this [#permalink] New post 30 Jan 2004, 21:09
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
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 [#permalink] New post 30 Jan 2004, 21:26
(n1+1)(n2+1)...(nk+1)

|{1, 2, 2^2} X { 1, 3, 3^2}| = 9
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 [#permalink] New post 31 Jan 2004, 15:00
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.
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 [#permalink] New post 31 Jan 2004, 16:47
Thanks rakesh1239. grrrrrrr
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 [#permalink] New post 31 Jan 2004, 19:58
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"

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 [#permalink] New post 31 Jan 2004, 20:00
Hey Vivek r u not watching cricket? whats happening with our players :arh
  [#permalink] 31 Jan 2004, 20:00
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