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

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Director
Joined: 03 Sep 2006
Posts: 885
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If the prime numbers p and t are the only prime factors of [#permalink]  10 Dec 2006, 00:07
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.

VP
Joined: 28 Mar 2006
Posts: 1384
Followers: 2

Kudos [?]: 19 [0], given: 0

Re: Data sufficiency [#permalink]  10 Dec 2006, 06:08
das_go wrote:
LM 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.

B !

Getting E
Am I missing something here?

OK A is INSUFF

Coming to B

(p^2)*p/m = k where k is an integer-----------(a)

but we are sked if (p^2)*t/m = k1 ( where k1 is an int)-------(b)

since p and t being co primes how can we deduce from (a) and (b) ?
Director
Joined: 01 Oct 2006
Posts: 500
Followers: 1

Kudos [?]: 18 [0], given: 0

Re: Data sufficiency [#permalink]  10 Dec 2006, 06:48
trivikram wrote:
das_go wrote:
LM 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.

B !

Getting E
Am I missing something here?

OK A is INSUFF

Coming to B

(p^2)*p/m = k where k is an integer-----------(a)

but we are sked if (p^2)*t/m = k1 ( where k1 is an int)-------(b)

since p and t being co primes how can we deduce from (a) and (b) ?

st is inusff
st2 m is multple of P^3 thus m is multiple of p^2 and t is factor of m m is a multiple of t as well
thus m is multiple of p^2*t
so b is suff
VP
Joined: 28 Mar 2006
Posts: 1384
Followers: 2

Kudos [?]: 19 [0], given: 0

Re: Data sufficiency [#permalink]  10 Dec 2006, 08:31
yogeshsheth wrote:
trivikram wrote:
das_go wrote:
LM 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.

B !

Getting E
Am I missing something here?

OK A is INSUFF

Coming to B

(p^2)*p/m = k where k is an integer-----------(a)

but we are sked if (p^2)*t/m = k1 ( where k1 is an int)-------(b)

since p and t being co primes how can we deduce from (a) and (b) ?

st is inusff
st2 m is multple of P^3 thus m is multiple of p^2 and t is factor of m m is a multiple of t as well
thus m is multiple of p^2*t
so b is suff

Thanks Yogesh. I got it

m is having co primes as factors

or m = p*t this is given

now we are aked if m is a factor of (p*t) * p which is YES and B is suff
Manager
Joined: 04 Feb 2004
Posts: 71
Location: India
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Kudos [?]: 25 [0], given: 0

1 -> INSUFF
2 -> m is a multiple of p^3. What if m = P^3 ? am I doing something wrong?
Director
Joined: 28 Dec 2005
Posts: 922
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Kudos [?]: 37 [0], given: 0

deep wrote:
1 -> INSUFF
2 -> m is a multiple of p^3. What if m = P^3 ? am I doing something wrong?

m cannot be just p^3, since it is given that m has p as well as t as it's prime factors. so, if it is p^3 * X, can may be broken down as t * Y.
Senior Manager
Joined: 20 Feb 2006
Posts: 373
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Re: Data sufficiency [#permalink]  10 Dec 2006, 11:41
LM 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.

From stem m/pt = integer

1) No useful info given ----> INSUFF

2) Tells us m/p^3 = integer so m/p*p^2 is also integer SUFF
Re: Data sufficiency   [#permalink] 10 Dec 2006, 11:41
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