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If x is a positive integer, which of the following CANNOT be

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If x is a positive integer, which of the following CANNOT be expressed as n^2, where n is an integer?

A. x^5
B. x^2 − 1
C. \(\sqrt{x^8}\)
D. x^2 + 1
E. \(\sqrt{x^5}\)
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Feb 2013, 14:23, edited 1 time in total.
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Re: If x is a positive integer, which of the following CANNOT be [#permalink]

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emmak wrote:
If x is a positive integer, which of the following CANNOT be expressed as n^2, where n is an integer?

A. x^5
B. x^2 − 1
C. \(\sqrt{x^8}\)
D. x^2 + 1
E. \(\sqrt{x^5}\)


The question basically asks: if x is a positive integer, which of the following CANNOT be a perfect square.

Now, if x=1, then options A, B, C and E ARE perfect squares, therefore by POE the correct answer must be D.

Answer: D.
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Re: If x is a positive integer, which of the following CANNOT be [#permalink]

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If we take x = 2 & calculate, we cant get the answers;
seems that x has to be taken 1 to execute all the options
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Re: If x is a positive integer, which of the following CANNOT be [#permalink]

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New post 27 Mar 2014, 16:12
emmak wrote:
If x is a positive integer, which of the following CANNOT be expressed as n^2, where n is an integer?

A. x^5
B. x^2 − 1
C. \(\sqrt{x^8}\)
D. x^2 + 1
E. \(\sqrt{x^5}\)


So I guess zero does count as a perfect square then

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PareshGmat wrote:
If we take x = 2 & calculate, we cant get the answers;
seems that x has to be taken 1 to execute all the options


Any random value of x will not help you get the answer. Even if you do not try x = 1, you can use reasoning to solve this question.

A. \(x^5\)
If x is a number with an even power, such as \(x = a^4\) (a is an integer), then \(x^5 = a^{20} = n^2\)
n will be \(a^{10}\), an integer here.

B. \(x^2 - 1\)
\(x^2 - 1 = n^2\)
You need two consecutive perfect squares. Only 0 and 1 are consecutive perfect squares. Thereafter, the distance between perfect squares keeps increasing. x needs to be a positive integers so if x = 1, n = 0 (an integer)

C. \(\sqrt{x^8}\)
\(\sqrt{x^8} = x^4 = n^2\)
n will be \(x^2\), an integer here.

D. \(x^2 + 1\)
x is a positive integer so it must be at least 1. After 1, there are no two consecutive integers. n cannot be an integer.

E. \(\sqrt{x^5}\)
If x is a number with an even power which is a multiple of 4, such as \(x = a^4\) (a is an integer), then \(\sqrt{x^5} = \sqrt{a^{20}} = a^{10} = n^2\)
n will be \(a^5\), an integer here.

Answer (D)
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New post 29 Mar 2014, 01:38
Even though Bunuel and karishma have answered this question, I am not able to digest any fundamental of this :( ....

Not sure how to crack this one... as I got this question in my Veritas prep exam today... :(
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Mountain14 wrote:
Even though Bunuel and karishma have answered this question, I am not able to digest any fundamental of this :( ....

Not sure how to crack this one... as I got this question in my Veritas prep exam today... :(


Look, the question simply asks which option CANNOT be a perfect square. In the options, x is a positive integer.

Can x^5 be a perfect square? Can x take some value such that x^5 is a perfect square? Say, x = 4. Then x^5 = 4^5 = 2^10
This is a perfect square. Hence for some value of x, x^5 could be a perfect square. Hence this is not our answer.
How do we find a value for which x^5 will be a perfect square? Perfect squares have even powers. We have x^5 which is an odd power. To get an even power, we could select x such that it already has an even power - we selected x = 2^2. Similarly, x could be 1^2 or 3^2 or 4^2 or 5^2 or 3^4 etc

Now think, can x^2 - 1 be a perfect square?
The reasoning for all the options is given in the post above.

A more intuitive approach is putting x = 1 as given by Bunuel. When x = 1, all options except option (D) results in a perfect square. So we know that all options CAN be perfect squares except (D). By elimination, answer must be (D).
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Re: If x is a positive integer, which of the following CANNOT be [#permalink]

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New post 30 Mar 2014, 20:18
VeritasPrepKarishma wrote:
Mountain14 wrote:
Even though Bunuel and karishma have answered this question, I am not able to digest any fundamental of this :( ....

Not sure how to crack this one... as I got this question in my Veritas prep exam today... :(


Look, the question simply asks which option CANNOT be a perfect square. In the options, x is a positive integer.

Can x^5 be a perfect square? Can x take some value such that x^5 is a perfect square? Say, x = 4. Then x^5 = 4^5 = 2^10
This is a perfect square. Hence for some value of x, x^5 could be a perfect square. Hence this is not our answer.
How do we find a value for which x^5 will be a perfect square? Perfect squares have even powers. We have x^5 which is an odd power. To get an even power, we could select x such that it already has an even power - we selected x = 2^2. Similarly, x could be 1^2 or 3^2 or 4^2 or 5^2 or 3^4 etc

Now think, can x^2 - 1 be a perfect square?
The reasoning for all the options is given in the post above.

A more intuitive approach is putting x = 1 as given by Bunuel. When x = 1, all options except option (D) results in a perfect square. So we know that all options CAN be perfect squares except (D). By elimination, answer must be (D).


So the one point where I get tripped up is the X^2 - 1. 0 is assumed to be a perfect square?

How can I tell from the verbage of the question that what they are asking for is determining whether or not something is a perfect square or not?

Thanks for the help.
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New post 30 Mar 2014, 21:59
dbiersdo wrote:
VeritasPrepKarishma wrote:
Mountain14 wrote:
Even though Bunuel and karishma have answered this question, I am not able to digest any fundamental of this :( ....

Not sure how to crack this one... as I got this question in my Veritas prep exam today... :(


Look, the question simply asks which option CANNOT be a perfect square. In the options, x is a positive integer.

Can x^5 be a perfect square? Can x take some value such that x^5 is a perfect square? Say, x = 4. Then x^5 = 4^5 = 2^10
This is a perfect square. Hence for some value of x, x^5 could be a perfect square. Hence this is not our answer.
How do we find a value for which x^5 will be a perfect square? Perfect squares have even powers. We have x^5 which is an odd power. To get an even power, we could select x such that it already has an even power - we selected x = 2^2. Similarly, x could be 1^2 or 3^2 or 4^2 or 5^2 or 3^4 etc

Now think, can x^2 - 1 be a perfect square?
The reasoning for all the options is given in the post above.

A more intuitive approach is putting x = 1 as given by Bunuel. When x = 1, all options except option (D) results in a perfect square. So we know that all options CAN be perfect squares except (D). By elimination, answer must be (D).


So the one point where I get tripped up is the X^2 - 1. 0 is assumed to be a perfect square?

How can I tell from the verbage of the question that what they are asking for is determining whether or not something is a perfect square or not?

Thanks for the help.


Yes, both 0 and 1 are perfect squares.

"Can you express 'this' as n^2 where n is an integer?" asks us whether we can write 'this' as square of an integer.
Square of an integer is a perfect square. So the question becomes "Can you express 'this' as a perfect square?"

Slightly convoluted verbiage is common in GMAT.
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Re: If x is a positive integer, which of the following CANNOT be [#permalink]

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New post 20 Jul 2014, 00:05
Bunuel wrote:
emmak wrote:
If x is a positive integer, which of the following CANNOT be expressed as n^2, where n is an integer?

A. x^5
B. x^2 − 1
C. \(\sqrt{x^8}\)
D. x^2 + 1
E. \(\sqrt{x^5}\)


The question basically asks: if x is a positive integer, which of the following CANNOT be a perfect square.

Now, if x=1, then options A, B, C and E ARE perfect squares, therefore by POE the correct answer must be D.

Answer: D.


Please refer to option [b] where value comes 0.
Would you please clarify whether 0 is a perfect square?
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New post 20 Jul 2014, 03:16
musunna wrote:
Bunuel wrote:
emmak wrote:
If x is a positive integer, which of the following CANNOT be expressed as n^2, where n is an integer?

A. x^5
B. x^2 − 1
C. \(\sqrt{x^8}\)
D. x^2 + 1
E. \(\sqrt{x^5}\)


The question basically asks: if x is a positive integer, which of the following CANNOT be a perfect square.

Now, if x=1, then options A, B, C and E ARE perfect squares, therefore by POE the correct answer must be D.

Answer: D.


Please refer to option [b] where value comes 0.
Would you please clarify whether 0 is a perfect square?


Yes, 0 is a perfect square 0 = 0^2.
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Re: If x is a positive integer, which of the following CANNOT be [#permalink]

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New post 16 Mar 2016, 17:12
I picked B, because I thought that 2 consecutive integers multiplied do not equal a perfect square...I did not take into consideration the possibility of 0...
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Re: If x is a positive integer, which of the following CANNOT be [#permalink]

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New post 21 May 2016, 14:21
I think the hardest part about this question is working out what is being asked.

Which of the following cannot be expressed as n^2 means which of the following cannot be perfect square when n is an integer.

Subbing in 1 for x in each example brings us to answer D.

1^2 + 1 = 2, which is not the square of any integer. D is your answer! Easy :P
Re: If x is a positive integer, which of the following CANNOT be   [#permalink] 21 May 2016, 14:21
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