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# For each of the following, could the answer be an integer if x is an

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Manager
Joined: 05 Nov 2012
Posts: 146
For each of the following, could the answer be an integer if x is an  [#permalink]

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06 Nov 2012, 09:43
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Question Stats:

100% (01:03) correct 0% (00:00) wrong based on 44 sessions

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For each of the following, could the answer be an integer if x is an integer greater than 1?

(1) $$x^{10} + x^{(–10)}=$$

(2) $$x^{\frac{1}{6}} + x^{\frac{1}{2}} =$$

This is a quant question from one of Manhattan's Flash cards

For each of the following, could the answer be an integer if x is an integer greater than 1?

b) x^1/6 + x^1/2 =

His explanation is
Yes. This is equivalent to 6th \sqrt{x} x + \sqrt{x} , so if x has an integer
sixth root this will be an integer. For example, if x equals 64, the sixth root of x is 2, and the square root is 8.

question was if x is integer greater than 1. What if 6th root is not an integer.... sum won't be an integer right? and in explanation he uses "will be an integer" (in bold) how is it possible?
Manager
Status: *Lost and found*
Joined: 25 Feb 2013
Posts: 121
Location: India
Concentration: General Management, Technology
GMAT 1: 640 Q42 V37
GPA: 3.5
WE: Web Development (Computer Software)
Re: For each of the following, could the answer be an integer if x is an  [#permalink]

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04 May 2013, 04:42
Amateur wrote:
This is a quant question from one of Manhattan's Flash cards

For each of the following, could the answer be an integer if x is an integer greater than 1?

b) x^1/6 + x^1/2 =

His explanation is
Yes. This is equivalent to 6th \sqrt{x} x + \sqrt{x} , so if x has an integer
sixth root this will be an integer. For example, if x equals 64, the sixth root of x is 2, and the square root is 8.

question was if x is integer greater than 1. What if 6th root is not an integer.... sum won't be an integer right? and in explanation he uses "will be an integer" (in bold) how is it possible?

Well the question states that can the sum be an integer not must. In other words, I think the question is asking, is it possible for the given expression to have integer values for integer values of x. So we have to establish cases where the value can be an integer. The basic explanation would be,

if x = m^6 where m is an integer, then x^(1/6) = m
also, x^(1/2) = m^3. Hence there will always be an m for which the above expression will be an integer. Hence m belongs to the soln set {2^6,3^6,.....} for which the expression can be an integer.

So the explanation provided, that the sixth root needs to be an integer is accurate, as if the value x^1/6 is not an integer, the expression wont be.

Hope the solution is satisfactory Please correct me if I am wrong!

Regards,
Arpan
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Feed me some KUDOS! *always hungry*

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Joined: 29 Apr 2017
Posts: 4
Re: For each of the following, could the answer be an integer if x is an  [#permalink]

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01 Sep 2017, 01:43
the answer will always be greater than as 1 as
square root of any integer greater than 1 will more than 1.
Director
Joined: 27 May 2012
Posts: 665
Re: For each of the following, could the answer be an integer if x is an  [#permalink]

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13 Jan 2019, 00:03
1
Amateur wrote:
For each of the following, could the answer be an integer if x is an integer greater than 1?

a) $$x^{10} + x^{(–10)}=$$

b) $$x^{\frac{1}{6}} + x^{\frac{1}{2}} =$$

This is a quant question from one of Manhattan's Flash cards

For each of the following, could the answer be an integer if x is an integer greater than 1?

b) x^1/6 + x^1/2 =

His explanation is
Yes. This is equivalent to 6th \sqrt{x} x + \sqrt{x} , so if x has an integer
sixth root this will be an integer. For example, if x equals 64, the sixth root of x is 2, and the square root is 8.

question was if x is integer greater than 1. What if 6th root is not an integer.... sum won't be an integer right? and in explanation he uses "will be an integer" (in bold) how is it possible?

Dear Moderator,
Found this DS question in the PS section, hope you will do the needful. Thank you.
_________________

- Stne

Math Expert
Joined: 02 Sep 2009
Posts: 52161
Re: For each of the following, could the answer be an integer if x is an  [#permalink]

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13 Jan 2019, 00:07
stne wrote:
Amateur wrote:
For each of the following, could the answer be an integer if x is an integer greater than 1?

a) $$x^{10} + x^{(–10)}=$$

b) $$x^{\frac{1}{6}} + x^{\frac{1}{2}} =$$

This is a quant question from one of Manhattan's Flash cards

For each of the following, could the answer be an integer if x is an integer greater than 1?

b) x^1/6 + x^1/2 =

His explanation is
Yes. This is equivalent to 6th \sqrt{x} x + \sqrt{x} , so if x has an integer
sixth root this will be an integer. For example, if x equals 64, the sixth root of x is 2, and the square root is 8.

question was if x is integer greater than 1. What if 6th root is not an integer.... sum won't be an integer right? and in explanation he uses "will be an integer" (in bold) how is it possible?

Dear Moderator,
Found this DS question in the PS section, hope you will do the needful. Thank you.

__________________________
Done. Thank you.
_________________
Re: For each of the following, could the answer be an integer if x is an &nbs [#permalink] 13 Jan 2019, 00:07
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