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OK I got this question wrong (I answered B but the OA is D) but I am not far of from grasping this whole integer properties bit. Can someone please clarify:

Statement 2: p^3 has exactly 4 distinct factors

What exactly does this statement mean? What should I be thinking of here? The "4 distinct factors" is not clear to me.

Please please and thank you.
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I get that B is suff. But A? even if P weren't prime say (12)^3 = P^3 = 12 is not prime. On the other hand, (2)^3 says P is prime. So a prime cubed and non prime cubed both will have 4 distinct factors (say 2 cubed has 4,8,2,1). Can someone explain why is A sufficient?
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Answer is D.

For statement 1 p^3 has exactly 4 distinct factors: 1,p,p^2,p^3. Hence prime, therefore sufficient.

For statement 2 when you factorize the equation you get x=2,3. Both are prime, therefore sufficient.
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Answer is D.

For statement 1 p^3 has exactly 4 distinct factors: 1,p,p^2,p^3. Hence prime, therefore sufficient.

For statement 2 when you factorize the equation you get x=2,3. Both are prime, therefore sufficient.

For statement 2, you get x = -2, 3.

sarahfiqbal
OK I got this question wrong (I answered B but the OA is D) but I am not far of from grasping this whole integer properties bit. Can someone please clarify:

Statement 2: p^3 has exactly 4 distinct factors

What exactly does this statement mean? What should I be thinking of here? The "4 distinct factors" is not clear to me.

Please please and thank you.

All "distinct factors" means all the positive, integer factors of a number counted only once.
Distinct factors of 125 are 1, 5, 25, and 125.

Madhavi1990
I get that B is suff. But A? even if P weren't prime say (12)^3 = P^3 = 12 is not prime. On the other hand, (2)^3 says P is prime. So a prime cubed and non prime cubed both will have 4 distinct factors (say 2 cubed has 4,8,2,1). Can someone explain why is A sufficient?
Look below.
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Bunuel
If p is positive, is p prime?

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

Let's start with 2, as it's easier.
2)
Factoring: \(p^2 – p – 6 = 0\)
Results in: \((p+2)(p-3) = 0\)
Therefore: \(p = -2\) or \(+3\)

Since P is positive, P = 3, 3 is prime. Sufficient

1)
The only time \(p ^ 3\) will have 4 distinct factors is if p is prime.

We can test this:
If \(p = 2, 2 ^ 3 = 8.\) The distinct (unique) factors of 8 are: 1, 2, 4, 8
If \(p = 3, 3 ^ 3 = 27.\) The distinct (unique) factors of 27 are: 1, 3, 9, 27
If \(p = 4, 4 ^ 3 = 64.\) The distinct (unique) factors of 64 are: 1, 2, 4, 8, 16, 32, 64
If \(p = 5, 5 ^ 3 = 125.\) The distinct (unique) factors of 125 are: 1, 5, 25, 125
If \(p = 6, 6 ^ 3 = 216.\) The distinct (unique) factors of 216 are: 1, 2, 3, 4, 6, 8, 9... and so on.

1 is also sufficient.

The answer is D.
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Answer should be B.

The catch is p is positive, but need not be an integer

Statement 1 : \(p^3\) has 4 distinct factors
if \(p\) is prime, then \(p^3\) has 4 distinct factors
BUT if \(p = x^{1/3} * y ^ {1/3}\), where x and y are prime numbers, then \(p^3 = (x^{1/3} * y ^ {1/3})^3 = xy\) still has 4 distinct factors, in this case p is not prime.

Not Suff

Statement 2: from quad eqn we can p = -2 or 3, since p is positive, p = 3, prime -> Suff

Answer (B)
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Bunuel
If p is positive, is p prime?

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

Question: Is p Prime?

Statement 1: p^3 has exactly 4 distinct factors

Number of factors = (power of 1st prime number+1)*(power of 2nd prime number+1)*(power of 3rd prime number+1)...

i.e. \(p^3\) will have 4 Factors when p is a Prime number(e.g. \(p^3=2^3\)) because (3+1) = 4

BUT if \(p = 2^{1/3}*3^{1/3}\) then \(p^3 = 2*3\) and it will still have 4 factors hence

NOT SUFFICIENT

Statement 2: p^2 – p – 6 = 0

i.e. p = 3 or -2 but given that p is positive hence
p = 3 hence
SUFFICIENT

Answer: Option B
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