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Is x prime ?

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by rockeyb » Fri Apr 09, 2010 7:29 am
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.
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by reply2spg » Fri Apr 09, 2010 9:02 am
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.
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by kevincanspain » Fri Apr 09, 2010 9:51 am
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
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by rockeyb » Fri Apr 09, 2010 10:09 am
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?
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by chrsrook » Fri Apr 09, 2010 11:03 am
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.
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by chrsrook » Fri Apr 09, 2010 11:04 am
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.
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by gmatmachoman » Fri Apr 09, 2010 12:53 pm
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..
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by boazkhan » Fri Apr 09, 2010 1:52 pm
IMO B is the answer. What is the OA?
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by harshavardhanc » Fri Apr 09, 2010 2:09 pm
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.
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by rockeyb » Fri Apr 09, 2010 7:46 pm
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.
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by harshavardhanc » Sat Apr 10, 2010 1:40 am
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.
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Harsha
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by sanju09 » Sat Apr 10, 2010 2:55 am
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]
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by akhpad » Sun Apr 11, 2010 2:50 am
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.
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by rockeyb » Sun Apr 11, 2010 2:58 am
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 .
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by ymach3 » Sun Apr 11, 2010 3:16 am
I am stuck thinking whether (10)^1/3 or (6)^1/3 is a prime ??

could someone get me out of the ship???
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