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If the prime numbers p and t are the only prime factors of

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Re: If the prime numbers p and t are the only prime factors of [#permalink] New post 17 Aug 2014, 07:59
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Re: If the prime numbers p and t are the only prime factors of [#permalink] New post 17 Aug 2014, 14:39
Bunuel wrote:
phoenixgmat wrote:
I would appreciate some help with:

If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p²t?
1) m has more than 9 positive factors.
2) m is a multiple of p³

some explanations to both statements would be great!
thx a lot


We are told that p and t are the ONLY prime factors of m. It could be expressed as m=p^x*t^y, where x and y are integers \geq{1}.

Question: is m a multiple of p^2*t. We already know that p and t are the factors of m, so basically question asks whether the power of p, in our prime factorization denoted as x, more than or equal to 2: so is x\geq{2}.

(1) m has more than 9 positive factors:

Formula for counting the number of distinct factors of integer x expressed by prime factorization as: n=a^x*b^y*c^z, is (x+1)(y+1)(z+1). This also includes the factors 1 and n itself.

We are told that (x+1)(y+1)>9 (as we know that m is expressed as m=p^x*t^y)
But it's not sufficient to determine whether x\geq{2}. (x can be 1 and y\geq{4} and we would have their product >9, e.g. (1+1)(4+1)=10.) Not sufficient.

(2) m is a multiple of p^3
This statement clearly gives us the value of power of p, which is 3, x=3>2. So m is a multiple of p^2t. Sufficient.

Answer: B.


I could answer the question in 1.5 min. But I have never known the formula for counting the number of distinct factors of a integer. Thanks a ton, Bunuel :-D :-D :-D
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Re: If the prime numbers p and t are the only prime factors of   [#permalink] 17 Aug 2014, 14:39
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