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Re: GMATPrep2: Multiple of prime factors [#permalink]
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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