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If the prime numbers p and t are the only prime factors of
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Updated on: 21 Mar 2012, 04:04
Updated on: 21 Mar 2012, 04:04
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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
(1) m has more than 9 positive factors.
(2) m is a multiple of p^3
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.
Last edited by Bunuel on 21 Mar 2012, 04:04, edited 1 time in total.
Edited the question and added the OA
If the prime numbers p and t are the only prime factors of
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26 Oct 2009, 16:16
26 Oct 2009, 16:16
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Expert Reply
If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p²t?
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 \(x\)) is 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 \(n\) 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 \((x+1)(y+1)>9\) (since \(m=p^x*t^y\))
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.
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 \(x\)) is 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 \(n\) 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 \((x+1)(y+1)>9\) (since \(m=p^x*t^y\))
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.
General Discussion
Re: Prime factors
[#permalink]
10 Dec 2009, 21:19
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'?
