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Re: If x is a positive integer, which of the following CANNOT be [#permalink]
12 Feb 2014, 15:10

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Re: If x is a positive integer, which of the following CANNOT be [#permalink]
27 Mar 2014, 20:21

2

This post received KUDOS

Expert's post

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.

Re: If x is a positive integer, which of the following CANNOT be [#permalink]
30 Mar 2014, 18:01

3

This post received KUDOS

Expert's post

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). _________________

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

Re: If x is a positive integer, which of the following CANNOT be [#permalink]
30 Mar 2014, 21:59

Expert's post

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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