IMO D

rockeyb wrote:If x is + ve , is x prime ?

(1) x^3 has exactly 4 distinct positive integer factors .

(2)x^2 - x - 6 = 0 .

source : Kaplan.

rockeyb wrote:If x is + ve , is x prime ?

(1) x^3 has exactly 4 distinct positive integer factors .

(2)x^2 - x - 6 = 0 .

source : Kaplan.

If x is a prime number, then x^3 has 4 factors: 1, x,x^2 and x^3

(1) If x is not a prime number, then it is either 1 or has at least 3 factors.
If x =1, x^3 has 1 factor. If x has at least 3 factors (1, y, x), then x^3 has more than 4 factors (1, y, y^2, y^3, x^3, ...)
Thus x must be a prime number
SUFF

(2) This quadratic equation has only 1 positive root: 3. Thus x=3
SUFF

kevincanspain wrote: If x is a prime number, then x^3 has 4 factors: 1, x,x^2 and x^3

(1) If x is not a prime number, then it is either 1 or has at least 3 factors.
If x =1, x^3 has 1 factor. If x has at least 3 factors (1, y, x), then x^3 has more than 4 factors (1, y, y^2, y^3, x^3, ...)
Thus x must be a prime number
SUFF

(2) This quadratic equation has only 1 positive root: 3. Thus x=3
SUFF

This is exactly the approach that I used and got the same answer . But I would have to say D is not the correct answer . This is the reason for posting this question here .

As per the Official explanation cube root of x dose not have to be an integer and still can have at least 4 distinct factors.

This is what I find hard to digest. Can you explain?

I think the answer is B.

Statement 1:

if X were 3^1/3, then x^3=6 and 6 has 4 factors but not prime.
if X were 3, then 3^3=27, which has 4 factors and 3 is prime.

Insufficient.

Statement 2:

Solving the quadratic, X= 3 or -2. Since it is given that X is positive, X=3 and is prime.

I meant 6 in statement 1.

I think the answer is B.

Statement 1:

if X were 6^1/3, then x^3=6 and 6 has 4 factors but not prime.
if X were 3, then 3^3=27, which has 4 factors and 3 is prime.

Insufficient.

Statement 2:

Solving the quadratic, X= 3 or -2. Since it is given that X is positive, X=3 and is prime.

rockeyb wrote:If x is + ve , is x prime ?

(1) x^3 has exactly 4 distinct positive integer factors .

(2)x^2 - x - 6 = 0 .

source : Kaplan.

Rockey bhai,

I understand, this one is damn tricky..Not that easy to pick it !!

Coming to your query let X^3 = 10.. So X will not be a integer. but 10 has only 4 factors (1,2,5,10)

So A wont be sufficient.

As posted by others B is very much sufficcient X=3..

IMO B is the answer. What is the OA?

rockeyb wrote:If x is + ve , is x prime ?

(1) x^3 has exactly 4 distinct positive integer factors .

(2)x^2 - x - 6 = 0 .

source : Kaplan.

if a number, say N, is factorized in prime factors P,Q, R......etc.

or N = (P)^p * (Q)^q * (R)^r

then the number of factors of N = (p+1)(q+1)(r+1)

Statement 1 :

we are given that X^3 has 4 factors ( 2*2 , 1*4)

so, X^3 can be represented as :

either P * Q

or 1 * P^3

{P and Q being prime numbers)

so, we can only deduce that X will be prime when it is the second case. If it is the first, then we can't.

Therefore, statement 1 is insufficient.

Statement 2 has been shown to be sufficient by itself.

Therefore, answer is B.

Thanks for your response guys .

OA = B .

@Harsha excellent explanation as usual thanks man .

But I would say picking numbers here like gmatmachoman and chrsrook will be a much faster and quicker technique , I agree with
gmatmachoman its not that easy to pick this trick up and more often than not even the best in the business will fall in the trap .

So kudos to Kaplan for this excellent question.

rockeyb wrote:Thanks for your response guys .

OA = B .

@Harsha excellent explanation as usual thanks man .

But I would say picking numbers here like gmatmachoman and chrsrook will be a much faster and quicker technique , I agree with
gmatmachoman its not that easy to pick this trick up and more often than not even the best in the business will fall in the trap .

So kudos to Kaplan for this excellent question.

it's only the explanation which looks long. If you know this concept, the answer can be found out orally for statement 1.

rockeyb wrote:If x is + ve , is x prime ?

(1) x^3 has exactly 4 distinct positive integer factors .

(2)x^2 - x - 6 = 0 .

source : Kaplan.

(1) A number having positive integer factors is a positive integer by itself, so x is a positive integer. Now, just remember the fact that, a prime has exactly 2 distinct positive integer factors, square of a prime has exactly 3 distinct positive integer factors, cube of a prime has exactly 4 distinct positive integer factors, and so on. Hence, if the cube of a positive integer, x, has exactly 4 distinct positive integer factors, then the positive integer, x, is prime. Sufficient

(2) The negative root in this case shall not be considered as the stem confirms x > 0. Finally x is 3, hence prime. Sufficient


[spoiler]gooDquestion[/spoiler]

but that's the trap here, (1) is not sufficient in fact


[spoiler]unBreakaBle[/spoiler]

Answer must be D

Statement 1:
No of factors, including 1 and themselves, is one more than the power of prime no. This is the way we calculate no of factors.

2^3 has 4 factors 1,2,4,8
3^3 has 4 factors
10^n = 2^n * 5^n has (n+1)(n+1) factors
25^n = 5^(3n) has (3n+1) factors

So,

X^3 is has exactly 4 factors only when X is prime. Sufficient.

Statement 2: Sufficient.

akhp77 wrote:Answer must be D

Statement 1:
No of factors, including 1 and themselves, is one more than the power of prime no. This is the way we calculate no of factors.

2^3 has 4 factors 1,2,4,8
3^3 has 4 factors
10^n = 2^n * 5^n has (n+1)(n+1) factors
25^n = 5^(3n) has (3n+1) factors

So,

X^3 is has exactly 4 factors only when X is prime. Sufficient.

Statement 2: Sufficient.

Answer is B .

Try x^3 = 10 or x^3 = 6 each have at least 4 factors and still they are not prime .

I agree that cube root of 6 or 10 will not be an integer but it will not be prime either.

Also look at Harsa's explanation above will help make things clear .

I am stuck thinking whether (10)^1/3 or (6)^1/3 is a prime ??

could someone get me out of the ship???