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 Post subject: DS Prob
 Post Posted: Thu May 29, 2008 9:06 am 
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

Answer B

Please explain


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 Post Posted: Fri May 30, 2008 7:09 pm 
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

Answer B

Please explain

Ans: You basically have to find out if there is another "p" in m's prime box. So far, we know that m's prime box has at least one p and one t in it. They are the only distinct primes, but there could be several ps and ts....

1) tells you nothing about the number of Ps - insufficient.
2) if m is a multiple of p^3, m's prime box has at least 3 factors of p. pxpxp = p^3 = a factor of m, or m is a multiple of p^3.

So, with statement (2), we know sufficiently that m IS a multiple of (p^2)t, because we know there exists 2 Ps and one t in m's prime box. This is based on the Factor Foundation Rule, which is in the MGMAT Number Properties guide. Any possible product combination of the m's prime factors are also factors of m, or m is also a multiple of these product combinations (m would also be a multiple of p*t and p*p*p*t.

Thanks.


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 Post subject:
 Post Posted: Thu Jun 05, 2008 5:39 am 
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ManhattanGMAT Staff


Posts: 7146
Guest 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

Answer B

Please explain

Ans: You basically have to find out if there is another "p" in m's prime box. So far, we know that m's prime box has at least one p and one t in it. They are the only distinct primes, but there could be several ps and ts....

1) tells you nothing about the number of Ps - insufficient.
2) if m is a multiple of p^3, m's prime box has at least 3 factors of p. pxpxp = p^3 = a factor of m, or m is a multiple of p^3.

So, with statement (2), we know sufficiently that m IS a multiple of (p^2)t, because we know there exists 2 Ps and one t in m's prime box. This is based on the Factor Foundation Rule, which is in the MGMAT Number Properties guide. Any possible product combination of the m's prime factors are also factors of m, or m is also a multiple of these product combinations (m would also be a multiple of p*t and p*p*p*t.

Thanks.


this is an excellent explanation.
is this one of our instructors in disguise?
if not, then congratulations; you have learned the ways of the masters.


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