GMATPrep2: Multiple of prime factors : DS Archive
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GMATPrep2: Multiple of prime factors

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Manager
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GMATPrep2: Multiple of prime factors [#permalink]

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New post 03 Sep 2009, 02:36
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A
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Don't know how to approach the first statement..

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[Reveal] Spoiler:
B
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Re: GMATPrep2: Multiple of prime factors [#permalink]

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New post 03 Sep 2009, 06:40
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m can be written as this:
m = \(p^x\)*\(t^y\) where x and y are positives integers. (this is the decomposition of m as product of primes factors)

The question is asking if \(x>=2\)?

1) from this stament we can say that x+y > 9 : not sufficient

2) from this one: x > 3 sufficient
answer B
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Re: GMATPrep2: Multiple of prime factors [#permalink]

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New post 03 Sep 2009, 06:56
Since p and t are the only primes,
\(m=p^it^j\) where j>=1 and i>=1
If m is a multiple of \(p^2t\), i>=2 and j>=1. Since we know that j>=1, we have only to prove that i>=2.

Statement 1

\(i\times j+1>9\)(note that generally the formula for the number of primes is (j+1)(i+1) but since we can elimintae all cases with i=0 or j=0, the formula is i\times j+1...1 is added since the case j=0 and i=0 can never eliminted )

Not sufficent
consider the case i=1 and j=9...not a multiple of p^2t

Statemet 2
sufficient since i=3 and \(m=p^3t^j\) where j>=1.
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Re: GMATPrep2: Multiple of prime factors [#permalink]

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New post 03 Sep 2009, 15:02
madeinafrica wrote:
m can be written as this:
m = \(p^x\)*\(t^y\) where x and y are positives integers. (this is the decomposition of m as product of primes factors)

The question is asking if \(x>=2\)?

1) from this stament we can say that x+y > 9 : not sufficient

2) from this one: x > 3 sufficient
answer B


I liked the approach..good one..
Re: GMATPrep2: Multiple of prime factors   [#permalink] 03 Sep 2009, 15:02
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GMATPrep2: Multiple of prime factors

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