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Updated on: 27 Oct 2023, 12:42
Originally posted by PathFinder007 on 07 May 2014, 05:44.
Last edited by Bunuel on 27 Oct 2023, 12:42, edited 2 times in total.
Last edited by Bunuel on 27 Oct 2023, 12:42, edited 2 times in total.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (01:47) correct
38%
(01:50)
wrong
based on 212
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Is x^2 > 2^x?
(1) x is a prime number.
(2) x^2 is a multiple of 9.
M28-46
(1) x is a prime number.
(2) x^2 is a multiple of 9.
M28-46
In official ans. they provide following details
x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient
I want to know how 0 can be multiple of 9.
Please clarify.
x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient
I want to know how 0 can be multiple of 9.
Please clarify.
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07 May 2014, 06:15
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PathFinder007
Is x^2>2^x?
(1) x is a prime number.
(2) x^2 is a multiple of 9.
Official answer is C. I choose B.
In official ans. they provide following details
x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient
I want to know how 0 can be multiple of 9.
Please clarify.
(1) x is a prime number.
(2) x^2 is a multiple of 9.
Official answer is C. I choose B.
In official ans. they provide following details
x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient
I want to know how 0 can be multiple of 9.
Please clarify.
Is \(x(x-2) > 0\)?
\(x(x-2) > 0\) is true when \(x < 0\) or \(x > 2\). Essentially, if \(x\) is not 0, 1, or 2, we have a YES answer to the question.
(1) \(x\) is a prime number.
If \(x=2\), then the answer is NO, but if \(x\) is some other prime, then the answer is YES. Not sufficient.
(2) \(x^2\) is a multiple of 9.
If \(x=0\), then the answer is NO, but if \(x=3\), then the answer is YES. Not sufficient.
(1)+(2) Since from (1), \(x\) is a prime, and from (2), \(x^2\) is a multiple of 9, then \(x\) can only be 3. Therefore, the answer to the question is YES. Sufficient.
Answer: C.
As for your question: zero is divisible by EVERY integer except zero itself, since 0/integer=0=integer (or, which is the same, zero is a multiple of every integer).
07 May 2014, 06:30
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Bunuel
PathFinder007
Is x^2>2^x?
(1) x is a prime number.
(2) x^2 is a multiple of 9.
Official answer is C. I choose B.
In official ans. they provide following details
x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient
I want to know how 0 can be multiple of 9.
Please clarify.
(1) x is a prime number.
(2) x^2 is a multiple of 9.
Official answer is C. I choose B.
In official ans. they provide following details
x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient
I want to know how 0 can be multiple of 9.
Please clarify.
Is x^2>2x?
Is x(x-2)>0? --> is x<0 or x>2. Basically if x is not 0, 1, or 2 we have an YES answer to the question.
(1) x is a prime number. If x=2 then the answer is NO but if x is some other prime, then the answer is YES. Not sufficient.
(2) x^2 is a multiple of 9. If x=0 then the answer is NO but if x=3, then the answer is YES. Not sufficient.
(1)+(2) Since from (1) x is a prime and from (2) x^2 is a multiple of 9, then x can only be 3. Therefore the answer to the question is YES. Sufficient.
Answer: C.
As for your question: zero is divisible by EVERY integer except zero itself, since 0/integer=0=integer (or, which is the same, zero is a multiple of every integer).
P.S. Please hide OA under the spoiler. Thank you.
Thanks. Bunnel for clarification.
