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

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If the prime numbers p and t are the only prime factors of [#permalink]  26 Oct 2009, 13:53
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If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p^2*t?

(1) m has more than 9 positive factors.
(2) m is a multiple of p^3
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Mar 2012, 03:04, edited 1 time in total.
Edited the question and added the OA
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Re: Prime factors [#permalink]  26 Oct 2009, 15:16
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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.

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Re: Prime factors [#permalink]  10 Dec 2009, 20:19
Excellent explanation but are we assuming that 'p' and 't' are different prime factors i.e. 'p' is not equal to 't'?
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Re: Prime factors [#permalink]  11 Dec 2009, 02:06
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brownybuddy wrote:
Excellent explanation but are we assuming that 'p' and 't' are different prime factors i.e. 'p' is not equal to 't'?

Yes. I think from the stem we can get this.
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Re: Prime factors [#permalink]  11 Dec 2009, 19:07
Excellent question as well as explanation.
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Re: if the prime no.s p & t [#permalink]  16 Nov 2010, 06:05
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Merging similar topics.
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Re: Prime factors [#permalink]  17 Nov 2010, 07:48
thanks bunuel good expalanation
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Re: Prime factors [#permalink]  20 Nov 2010, 05:31
Ans is B , good one Bunuel
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Re: If the prime numbers p and t are the only prime factors [#permalink]  25 Sep 2012, 10:45
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Hey
Lets look at statement 1
m has more than 9 factors
Now if p and t are the only prime factors then the other factors would be a combination of p and t either with each other or with themselves.
Now among those 9 factors, the following 2 things could happen.
1. 2 factors would be 1 and m. The other factors could be $$p, t, t^2, t^3, t^4, t^5, t^6$$. In this case the integer m is NOT a multiple of$$p^2t$$.
2. The other seven factors could have$$p^2$$. In that case m would be a multiple of $$p^2t$$
So, Insufficient.
Lets look at statement 2
If m is a multiple of$$p^3$$, then m must be a multiple of $$p^2$$. We know that m is already a multiple of t. So m must be a multiple of $$p^2t$$.
Hence Sufficient.

Hope this helps.
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Re: If the prime numbers p and t are the only prime factors [#permalink]  25 Sep 2012, 10:49
ankit0411 wrote:
If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p^2 t?

(1) m has more than 9 positive factors (2) m is a multiple of p^3

We can write $$m=p^a\cdot{t^b}$$ for some positive integers $$a$$ and $$b.$$

(1) The number of positive factors of $$m$$ is $$(a+1)(b+1)>9.$$
If $$a=1$$ and $$b>3$$ then $$m=pt^b$$ is not a multiple of $$p^2t.$$
If $$a>1$$ then the answer is yes.
Not sufficient.

(2) Obviously sufficient.

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Last edited by EvaJager on 25 Sep 2012, 10:54, edited 1 time in total.
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Re: If the prime numbers p and t are the only prime factors [#permalink]  25 Sep 2012, 10:52
souvik101990 wrote:
Hey
Lets look at statement 1
m has more than 9 factors
Now if p and t are the only prime factors then the other factors would be a combination of p and t either with each other or with themselves.
Now among those 9 factors, the following 2 things could happen.
1. 2 factors would be 1 and m. The other factors could be $$p, t, t^2, t^3, t^4, t^5, t^6$$. In this case the integer m is NOT a multiple of$$p^2t$$.
2. The other seven factors could have$$p^2$$. In that case m would be a multiple of $$p^2t$$
So, Insufficient.
Lets look at statement 2
If m is a multiple of$$p^3$$, then m must be a multiple of $$p^2$$. We know that m is already a multiple of t. So m must be a multiple of $$p^2t$$.
Hence Sufficient.

Hope this helps.

I got your second statement, but somehow I am not able to get the 1st statement.

For example you have p=2 and t=3, two prime numbers . Now the other 7 numbers can be any positive integer right ? i.e 4,6,8,9,4,6,8 isnt it ?

And the second case maybe that we have other 7 factors that include 2 and 3 as well . ex. 2,3,2,3,2,3,3,2,2 . In this case m is a multiple of p^2*t .

Is my thinking right ?
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Re: If the prime numbers p and t are the only prime factors [#permalink]  25 Sep 2012, 10:55
EvaJager wrote:
ankit0411 wrote:
If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p^2 t?

(1) m has more than 9 positive factors (2) m is a multiple of p^3

We can write $$m=p^a\cdot{t^b}$$ for some positive integers $$a$$ and $$b.$$

(1) The number of positive factors of $$m$$ is $$(a+1)(b+1)>9.$$
If $$a=1$$ and $$b>3$$ then $$m=pt^b$$ is not a multiple of $$p^2t.$$
If $$a>1$$ then the answer is yes.
Not sufficient.

(2) Obviously sufficient.

The formula you've written - (a+1)(b+1) is for the no of prime factors of a number right ? .
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Re: If the prime numbers p and t are the only prime factors [#permalink]  25 Sep 2012, 11:00
ankit0411 wrote:
EvaJager wrote:
ankit0411 wrote:
If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p^2 t?

(1) m has more than 9 positive factors (2) m is a multiple of p^3

We can write $$m=p^a\cdot{t^b}$$ for some positive integers $$a$$ and $$b.$$

(1) The number of positive factors of $$m$$ is $$(a+1)(b+1)>9.$$
If $$a=1$$ and $$b>3$$ then $$m=pt^b$$ is not a multiple of $$p^2t.$$
If $$a>1$$ then the answer is yes.
Not sufficient.

(2) Obviously sufficient.

The formula you've written - (a+1)(b+1) is for the no of prime factors of a number right ? .

NO. It is for all the positive factors of the number, including 1 and the number itself, not only prime factors.
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Re: If the prime numbers p and t are the only prime factors [#permalink]  25 Sep 2012, 11:02
Expert's post
ankit0411 wrote:
EvaJager wrote:
ankit0411 wrote:
If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p^2 t?

(1) m has more than 9 positive factors (2) m is a multiple of p^3

We can write $$m=p^a\cdot{t^b}$$ for some positive integers $$a$$ and $$b.$$

(1) The number of positive factors of $$m$$ is $$(a+1)(b+1)>9.$$
If $$a=1$$ and $$b>3$$ then $$m=pt^b$$ is not a multiple of $$p^2t.$$
If $$a>1$$ then the answer is yes.
Not sufficient.

(2) Obviously sufficient.

The formula you've written - (a+1)(b+1) is for the no of prime factors of a number right ? .

Check this: math-number-theory-88376.html It might help.
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Re: If the prime numbers p and t are the only prime factors of [#permalink]  25 Sep 2012, 11:16
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Quote:
For example you have p=2 and t=3, two prime numbers . Now the other 7 numbers can be any positive integer right ? i.e 4,6,8,9,4,6,8 isnt it ?

Note that these factors are combinations of powers of the prime factors only.
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Re: If the prime numbers p and t are the only prime factors [#permalink]  25 Sep 2012, 19:39
Quote:
Check this: math-number-theory-88376.html It might help.

Thanks Bunnuel ! I have gone through that, very valuable !
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Re: If the prime numbers p and t are the only prime factors of [#permalink]  25 Sep 2012, 19:40
souvik101990 wrote:
Quote:
For example you have p=2 and t=3, two prime numbers . Now the other 7 numbers can be any positive integer right ? i.e 4,6,8,9,4,6,8 isnt it ?

Note that these factors are combinations of powers of the prime factors only.

Thanks, got that . Took me a little while to understand the solution.
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Re: If the prime numbers p and t are the only prime factors of [#permalink]  22 Jan 2013, 02:21
phoenixgmat wrote:
If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p^2*t?

(1) m has more than 9 positive factors.
(2) m is a multiple of p^3

m = p^x * t^y where x is at least 1 and y is at least 1...
For m to be a multiple of p^2 * t then m must have at least 2 p and at least 1 t...

1. m has more than 9 factors
If m = p^1 * t^4 => number of factors = (1+1)(4+1) = 10 NOT A MULTIPLE!
If m = p^2 * t^3 => numbr of factors = (2+1)(3+1) = 12 A MULTIPLE!
INSUFFICIENT!

2. m is a multiple of p^3
Is it at least 2 factors of p? According to Statement (2) - YES!
Is it at least 1 factor of t? According to GIVEN - YES!

SUFFICIENT!

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Re: If the prime numbers p and t are the only prime factors of [#permalink]  16 Jul 2013, 07:10
IS my translation for this problem correct the given info...

we know that $$\frac{m}{p*t}$$ = Integer since p and t are different integers

The question is now framed as is$$\frac{M}{p^2 t}$$ ?( T is irrelevant for this question )
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Re: If the prime numbers p and t are the only prime factors of [#permalink]  16 Jul 2013, 07:41
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fozzzy wrote:
IS my translation for this problem correct the given info...

we know that $$\frac{m}{p*t}$$ = Integer since p and t are different integers

The question is now framed as is$$\frac{M}{p^2 t}$$ ?( T is irrelevant for this question )

Yes, the question asks whether m/(p^2t)=integer, while saying that m/(pt)=integer.
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Re: If the prime numbers p and t are the only prime factors of   [#permalink] 16 Jul 2013, 07:41

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