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Math Expert V
Joined: 02 Sep 2009
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12 00:00

Difficulty:   35% (medium)

Question Stats: 59% (00:59) correct 41% (01:17) wrong based on 256 sessions

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If $$x$$ is an integer, what is the value of $$x$$?

(1) $$x^2 = x^3$$

(2) $$x$$ is both a perfect square and a perfect cube. (Note: a perfect square, is an integer that can be written as the square of an integer. For example $$16=4^2$$, is a perfect square. Similarly a perfect cube, is an integer that can be written as the cube of an integer. For example $$27=3^3$$, is a perfect cube.)

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Math Expert V
Joined: 02 Sep 2009
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Official Solution:

Statement (1) by itself is insufficient. $$x$$ can be 0 or 1.

Statement (2) by itself is insufficient. Perfect squares are integers 0, 1, 4, 9, 16, and so on. Similarly, perfect cubes are: 0, 1, 8, 27, 64... Thus $$x$$ can be 0 or 1.

Statements (1) and (2) combined are insufficient. For example: $$x$$ can still be 0 or 1.

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the hint in the question seems misleading especially when it is read very literally

hint says: Similarly a perfect cube, is an integer that can be written as the cube of some other integer-----------> this implies that 1 cannot be a perfect cube because 1 cannot be written as a cube of "some other integer" but yes it can be written as a cube of 1 itself !! , according to this hint numbers like 125 ( for 125 can be written as 5^3 ---->other integer 5 ) will fall into this definition .
however , if one does not look at the hint then it is oki because as per common knowledge 1 does qualify as perfect cube
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Joined: 18 Jul 2013
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the hint in the question seems misleading especially when it is read very literally

hint says: Similarly a perfect cube, is an integer that can be written as the cube of some other integer-----------> this implies that 1 cannot be a perfect cube because 1 cannot be written as a cube of "some other integer" but yes it can be written as a cube of 1 itself !! , according to this hint numbers like 125 ( for 125 can be written as 5^3 ---->other integer 5 ) will fall into this definition .
however , if one does not look at the hint then it is oki because as per common knowledge 1 does qualify as perfect cube

i agree with you,
this definition confused me
Intern  Joined: 09 Mar 2013
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Concentration: General Management, Entrepreneurship
GMAT 1: 550 Q43 V23 GMAT 2: 590 Q49 V22 GMAT 3: 690 Q49 V34 GMAT 4: 740 Q49 V41 Show Tags

It's a quite controversial problem.
Hope I won't face similar one on the GMAT.

http://mathforum.org/library/drmath/view/52368.html
Manager  Joined: 31 Jul 2014
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Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. $$x$$ can be 0 or 1.

Statement (2) by itself is insufficient. Perfect squares are integers 0, 1, 4, 9, 16, and so on. Similarly, perfect cubes are: 0, 1, 8, 27, 64... Thus $$x$$ can be 0 or 1.

Statements (1) and (2) combined are insufficient. For example: $$x$$ can still be 0 or 1.

I think for stmnt 2) 0,1,64,729 are there but i ignored 0 and 1 becoz u said cube and square of "other" number
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GMAT 1: 640 Q50 V26 GMAT 2: 660 Q51 V27 GMAT 3: 680 Q50 V31 GMAT 4: 710 Q50 V35 GMAT 5: 760 Q50 V42 Show Tags

I agree. The hint in (2) is really confusing. However, I don't know why I thought 1, but not 0, satisfy this condition and thus chose B as the correct answer .
Intern  Joined: 03 May 2015
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I think this is a high-quality question and I agree with explanation. Please change the definition containing "of some other integer" to " of any integer" ; then the question is right; else 0 and 1 cannot be answers to option 2, thanks.
Math Expert V
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joyandbliss wrote:
I think this is a high-quality question and I agree with explanation. Please change the definition containing "of some other integer" to " of any integer" ; then the question is right; else 0 and 1 cannot be answers to option 2, thanks.

Updated the question.
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NOTE THIS PLEASE ... the question clearly states that x is an INTEGER so x^2 = x^3 should be sufficient as 0 does not qualify as an integer and x can only take the value 1.THE question starts as if x is an INTEGER.
give me kudos if you think iam right:)

Originally posted by Vibhav10 on 13 May 2018, 11:35.
Last edited by Vibhav10 on 13 May 2018, 11:42, edited 1 time in total.
Intern  Joined: 07 Feb 2018
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Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. $$x$$ can be 0 or 1.

Statement (2) by itself is insufficient. Perfect squares are integers 0, 1, 4, 9, 16, and so on. Similarly, perfect cubes are: 0, 1, 8, 27, 64... Thus $$x$$ can be 0 or 1.

Statements (1) and (2) combined are insufficient. For example: $$x$$ can still be 0 or 1.

NOTE THIS PLEASE ... the question clearly states that x is an INTEGER so x^2 = x^3 should be sufficient as 0 does not qualify as an integer and x can only take the value 1.THE question starts as if x is an INTEGER. i may be wrong somewhere though.
give me kudos if you think iam right:)
Math Expert V
Joined: 02 Sep 2009
Posts: 55635

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Vibhav10 wrote:
Bunuel wrote:
Official Solution:

Statement (1) by itself is insufficient. $$x$$ can be 0 or 1.

Statement (2) by itself is insufficient. Perfect squares are integers 0, 1, 4, 9, 16, and so on. Similarly, perfect cubes are: 0, 1, 8, 27, 64... Thus $$x$$ can be 0 or 1.

Statements (1) and (2) combined are insufficient. For example: $$x$$ can still be 0 or 1.

NOTE THIS PLEASE ... the question clearly states that x is an INTEGER so x^2 = x^3 should be sufficient as 0 does not qualify as an integer and x can only take the value 1.THE question starts as if x is an INTEGER. i may be wrong somewhere though.
give me kudos if you think iam right:)

That's wrong.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

You should brush-up fundamentals before practising questions:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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totally unfair

i thought about 0 ...then the number has to be a perfect square , cube ...i figured 0 should not be counted ((((((((((
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Hi! just wanted to know what is 0*0?
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VP  G
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patto wrote:
Hi! just wanted to know what is 0*0?

patto

It is indeed 0.

Even $$0^3$$ = 0

It is part of the perfect squares
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