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If the prime numbers p and t are the only prime factors of [#permalink]
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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
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Originally posted by phoenixgmat on 26 Oct 2009, 14:53.
Last edited by Bunuel on 21 Mar 2012, 04:04, edited 1 time in total.
Edited the question and added the OA



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Re: Prime factors [#permalink]
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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. Answer: B.
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Re: Prime factors [#permalink]
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10 Dec 2009, 21:19
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]
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11 Dec 2009, 03:06



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Re: Prime factors [#permalink]
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11 Dec 2009, 20:07
Excellent question as well as explanation.



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Re: Prime factors [#permalink]
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17 Nov 2010, 08:48
thanks bunuel good expalanation



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Re: Prime factors [#permalink]
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20 Nov 2010, 06:31
Ans is B , good one Bunuel
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Re: If the prime numbers p and t are the only prime factors [#permalink]
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25 Sep 2012, 11:45
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 2If 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]
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Updated on: 25 Sep 2012, 11:54
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. Answer B.
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Originally posted by EvaJager on 25 Sep 2012, 11:49.
Last edited by EvaJager on 25 Sep 2012, 11:54, edited 1 time in total.



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Re: If the prime numbers p and t are the only prime factors [#permalink]
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25 Sep 2012, 11:52
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]
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25 Sep 2012, 11:55
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. Answer B. 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]
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25 Sep 2012, 12:00
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. Answer B. 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]
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25 Sep 2012, 12:02



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Re: If the prime numbers p and t are the only prime factors of [#permalink]
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25 Sep 2012, 12:16



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Re: If the prime numbers p and t are the only prime factors [#permalink]
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25 Sep 2012, 20:39
Quote: 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]
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25 Sep 2012, 20:40
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]
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22 Jan 2013, 03:21
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! ANswer: B
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Re: If the prime numbers p and t are the only prime factors of [#permalink]
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16 Jul 2013, 08:10
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]
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16 Jul 2013, 08:41



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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. 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
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