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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]

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17 Aug 2014, 08:59
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Re: If the prime numbers p and t are the only prime factors of [#permalink]

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17 Aug 2014, 15:39
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.

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
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Re: If the prime numbers p and t are the only prime factors of [#permalink]

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20 Sep 2015, 00:43
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Re: If the prime numbers p and t are the only prime factors of   [#permalink] 20 Sep 2015, 00:43

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