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# P = p1 p2 p3 is a product of three distinct primes.

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Manager
Joined: 14 May 2005
Posts: 85
Location: San Francisco
Followers: 1

Kudos [?]: 2 [0], given: 0

P = p1 p2 p3 is a product of three distinct primes. [#permalink]  22 Oct 2006, 19:37
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50% (01:05) correct 50% (00:00) wrong based on 0 sessions
P = p1 p2 p3 is a product of three distinct primes. Determine P:

(S1) 39 is a divisor of P
(S2) 15 is a divisor of P
Senior Manager
Joined: 14 Aug 2006
Posts: 365
Followers: 1

Kudos [?]: 3 [0], given: 0

C

1) from 1 we know the number is 3* 13 but the other prime can be 2, 5, 7, 11 etc.. SO INSUFF

2) from 2 we know the number is 3 * 5 the other prime can be 2, 7, 11 etc

Combined the number has to be 3* 5* 13
Senior Manager
Joined: 08 Jun 2006
Posts: 342
Location: Washington DC
Followers: 1

Kudos [?]: 23 [0], given: 0

C

From 1 â€“ insufficient
P = P1 * P2 * P3 = 39 * k (k is an integer) = 3 * 13 * k
Two prime numbers are 3, 13. we donâ€™t know the 3rd one

From 2 â€“ insufficient
P = P1 * P2 * P3 = 15 * m (m is an integer) = 3 * 5 * k
Consider P1 = 3 and P3 = 5
Two prime numbers are 3, 5. we donâ€™t know the 3rd one

Together
3 prime numbers are 3, 13, 5
P = 3* 13*5
SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 112 [0], given: 39

P = p1 p2 p3 is a product of three distinct primes. Determine P:

(S1) 39 is a divisor of P
(S2) 15 is a divisor of P

FROM ONE

P = P1*P2*P3 = 3*13*P3........NOT SUFF

FROM TWO

P =P1*P2*P3 = 3*5*P2.....NOT SUFF

BOTH TOGETHER

P = 5*3*13....SUFF

Manager
Joined: 14 May 2005
Posts: 85
Location: San Francisco
Followers: 1

Kudos [?]: 2 [0], given: 0

Thank you fellas. The OA is C.
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