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.
Is x prime ?
Source: Beat The GMAT — Data Sufficiency |
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kevincanspain
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If x is a prime number, then x^3 has 4 factors: 1, x,x^2 and x^3rockeyb 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) 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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rockeyb
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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 .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
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?
"Know thyself" and "Nothing in excess"
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.
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.
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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gmatmachoman
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Rockey bhai,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.
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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harshavardhanc
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if a number, say N, is factorized in prime factors P,Q, R......etc.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.
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.
Regards,
Harsha
Harsha
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rockeyb
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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.
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.
"Know thyself" and "Nothing in excess"
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harshavardhanc
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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: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.
Regards,
Harsha
Harsha
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sanju09
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(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. Sufficientrockeyb 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.
(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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akhpad
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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.
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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rockeyb
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Answer is B .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.
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 .
"Know thyself" and "Nothing in excess"
