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If p is positive, is p prime?

This topic has 2 expert replies and 0 member replies

If p is positive, is p prime?

Post Thu Nov 02, 2017 8:51 am
If p is positive, is p prime?

(1) p^3 has exactly 4 distinct factors
(2) p^2 - p - 6 = 0.

What's the best way to determine which statement is sufficient?

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GMAT/MBA Expert

Post Wed Nov 15, 2017 9:08 pm
Hi ardz24,

We're told that P is POSITIVE. We're asked if P is PRIME. This is a YES/NO question. To start, it's worth noting that we do NOT know whether P is an integer or not.

1) P^3 has exactly 4 distinct factors.

To start, let's focus on numbers that have just 4 distinct factors. There are two that you should be able to find relatively easily:
6 (factors are 1, 2, 3 and 6)
8 (factors are 1, 2, 4 and 8)

According to Fact 1, P^3 could be either of those 2 numbers....
IF....
P^3 = 6, then P is a NON-INTEGER and the answer to the question is NO
P^3 = 8, then P = 2 and the answer to the question is YES.
Fact 1 is INSUFFICIENT

2) P^2 - P - 6 = 0

We can factor this equation into it's pieces and solve...
P^2 - P - 6 = 0
(P-3)(P+2) = 0
P = +3 or -2
The prompt tells us that P is POSITIVE, so there's only one solution here: +3... and the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

Final Answer: B

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 Nov 09, 2017 10:12 pm
ardz24 wrote:
If p is positive, is p prime?

(1) p^3 has exactly 4 distinct factors
(2) p^2 - p - 6 = 0.

What's the best way to determine which statement is sufficient?
(1) p^3 has exactly 4 distinct factors.

Case 1: Assume that p is prime, thus, the number of factors of p^3 are 1, p, p^2, and p^3. There are four factors.
Case 2: Assume that p is non-prime; let's assume = 4, a convenient value. The number of factors of 4^3 (= 64) are 1, 2, 4, 8, 16, 32, and 64. The number of factors is not four, thus, this case is invalid.

In case, you wish to try Case 1 with some values such as 2 and 101, you may. For p = 2, the number 2^3 has 1, 2, 4, and 8 as factors (four); for p = 10, the number 101^3 has 1, 101, 101^2, and 101^3 as factors (four).

Sufficient.

(2) p^2 - p - 6 = 0

=> p^2 - 3p + 2p - 6 = 0 => p(p - 3) + 2(p - 3) = 0 => p = 3 or -2. Since p is positive, p = 3, a prime number. Sufficient.

The correct answer: D

Hope this helps!

-Jay
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