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655-705 (Hard)|   Algebra|      
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PathFinder007
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Is x^2 > 2^x? (1) x is a prime number. (2) x^2 is a multiple of 9. 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.
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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)!

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55% (hard)

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62% (01:46) correct 38% (01:45) wrong based on 269 sessions

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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

Show Spoiler
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.

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C
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Bunuel
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Is x^2 > 2^x? (1) x is a prime number. (2) x^2 is a multiple of 9. 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.
Show more

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).
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Is x^2 > 2^x? (1) x is a prime number. (2) x^2 is a multiple of 9. 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.

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.
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Thanks. Bunnel for clarification.
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Is x^2 > 2^x? (1) x is a prime number. (2) x^2 is a multiple of 9. 07 May 2014, 06:36
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Actually for statement 2, you can do it even without using 0.

For ex: If x=3, then x^2 = 9 is a multiple of 9 and satisfies the equation.
But, if x=6, then x^2 = 36 is also a multiple of 9, but does not satisfy the equation ( 6^2 is not greater than 2^6).

Hence, not sufficient.
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Is x^2 > 2^x? (1) x is a prime number. (2) x^2 is a multiple of 9. 07 May 2014, 06:43
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Bunuel
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).
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So does this mean in case of any DS question if it involves multiple of integer type information we should consider 0 as a case to test?
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Is x^2 > 2^x? (1) x is a prime number. (2) x^2 is a multiple of 9. 07 May 2014, 08:00
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Bunuel
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).
So does this mean in case of any DS question if it involves multiple of integer type information we should consider 0 as a case to test?
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If a variable can be 0, then yes.
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Is x^2 > 2^x? (1) x is a prime number. (2) x^2 is a multiple of 9. 07 May 2014, 08:24
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Thanks Bunuel. This is going into my notes as a caveat...:)
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Is x^2 > 2^x? (1) x is a prime number. (2) x^2 is a multiple of 9. 21 Mar 2025, 08:55
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