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OG What is the value of the integer P?

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AbeNeedsAnswers Master | Next Rank: 500 Posts Default Avatar
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OG What is the value of the integer P?

Post Wed Aug 23, 2017 9:19 pm
What is the value of the integer p ?

(1) Each of the integers 2, 3, and 5 is a factor of p.
(2) Each of the integers 2, 5, and 7 is a factor of p.

E

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GMAT/MBA Expert

Post Fri Feb 16, 2018 4:56 pm
Hi All,

We're told that P is an integer. We're asked for the value of P. This question can be solved with a little logic (and TESTing VALUES as needed);

1) Each of the integers 2, 3, and 5 is a factor of P.

The information in Fact 1 essentially tells us that P is a multiple of 30 (since 2x3x5 = 30 and since all of those numbers are factors of P, then P has to be a multiple of all of those numbers). Thus, P could be 0, 30, 60, 90, etc.
Fact 1 is INSUFFICIENT

2) Each of the integers 2, 5, and 7 is a factor of p.

The information in Fact 2 essentially tells us that P is a multiple of 70 (since 2x5x7 = 70 and since all of those numbers are factors of P, then P has to be a multiple of all of those numbers). Thus, P could be 0, 70, 140, 210, etc.
Fact 2 is INSUFFICIENT

Combined, we know:
P is a multiple of 30
P is a multiple of 70
There are lots of numbers that are a multiple of 30 and 70 (0, 210, 420, 630, etc.)
Combined, INSUFFICIENT

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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Post Thu Aug 24, 2017 10:28 am
AbeNeedsAnswers wrote:
What is the value of the integer p ?

(1) Each of the integers 2, 3, and 5 is a factor of p.
(2) Each of the integers 2, 5, and 7 is a factor of p.

E
ASIDE:
For questions involving factors (aka "divisors"), we can say:
If k is a divisor of N, then k is "hiding" within the prime factorization of N
Consider these examples:
3 is a divisor of 24 because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a divisor of 70 because 70 = (2)(5)(7)
And 8 is a divisor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a divisor of 630 because 630 = (2)(3)(3)(5)(7)

Conversely, we can say that, if k is "hiding" within the prime factorization of N, then N is a multiple of k
Examples:
24 = (2)(2)(2)(3) <--> 24 is a multiple of 3
(2)(5)(7) <--> 70 is a multiple of 5
330 = (2)(3)(5)(11) <--> 330 is a multiple of 6

----NOW ONTO THE QUESTION-------------
Target question: What is the value of the integer p ?

Statement 1: Each of the integers 2, 3, and 5 is a factor of p.
So, p = (2)(3)(5)(possibly other primes)
This tells us that p is a multiple of 30.
There are infinitely many values of p that satisfy statement 1.
For example, p could equal 30 or p could equal 60 (etc)
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: Each of the integers 2, 5, and 7 is a factor of p.
So, p = (2)(5)(7)(possibly other primes)
This tells us that p is a multiple of 70.
There are infinitely many values of p that satisfy statement 2.
For example, p could equal 70 or p could equal 140 (etc)
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that p = (2)(3)(5)(possibly other primes)
Statement 2 tells us that p = (2)(5)(7)(possibly other primes)
When we COMBINE the statements, we can conclude that p = (2)(3)(5)(7)(possibly other primes)
In other words, p is a multiple of 210.
There are infinitely many values of p that satisfy this condition.
For example, p could equal 210 or p could equal 420 (etc)
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent

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