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Disagree with OA if x^2 is a positive integer, is x a [#permalink ]
29 Jul 2005, 17:00

Disagree with OA
if x^2 is a positive integer, is x a positive integer?
1. sqrt x^2 is an integer
2. sqrt x^2 =x

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Re: positive integer [#permalink ]
29 Jul 2005, 19:20

mirhaque wrote:

Disagree with OA if x^2 is a positive integer, is x a positive integer? 1. sqrt x^2 is an integer 2. sqrt x^2 =x

Let us take x^2 = 16 so x = +/- 4

1) +/- 4 both are integers so A does not help

2) agaiin +/- 4 does not help

both together does not help either. So E? THis is too easy or i am missing some thing..

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Agree with E
From stem X can be +/-
From A) same case
From B) identical to A)
When combined it doesn't help much cause in both X is squared.

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I'l go with C...
X^2=integer.... X=integer?
1) Sqrt(X^2)=integer....i.e Sqrt (4)= 4, x could be +- 4...Insuff
2) Sqrt(X^2)=X...
OK so we know that sqrt(x^2) is positive...so from (2) we know X is positive...but we dont know if SQRT(X^2) is an integer or not...Insuff....
combining them together...Sufficient...

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fresinha12 wrote:

I'l go with C... X^2=integer.... X=integer? 1) Sqrt(X^2)=integer....i.e Sqrt (4)= 4, x could be +- 4...Insuff 2) Sqrt(X^2)=X... OK so we know that sqrt(x^2) is positive...so from (2) we know X is positive...but we dont know if SQRT(X^2) is an integer or not...Insuff.... combining them together...Sufficient...

if 2) Sqrt(X^2)=X..., How does it mean X is positive? can you eloborate on that..

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it is a question of denotaion of square root
it seems to me: sqrt(4) = +/- 2

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I'm also getting (E) on this one. What's the OA?

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Quote:

sqrt(x^2)=integer....i.e sqrt (16)= 4, x could be +- 4...Insuff

the above expression is not correct. Sqrt(X^2) is always +ve.

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OA is C. But I think it's E

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read my explanation...it should be C...

sqrt(x^2) is always positive....

mirhaque wrote:

OA is C. But I think it's E

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fresinha12 wrote:

read my explanation...it should be C...

sqrt(x^2) is always positive....

mirhaque wrote:

OA is C. But I think it's E

Hi Freisinha12,

Can you eloborate why sqrt(x^2) is always positive, i am not getting it...

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fresinha12 wrote:

OK so we know that sqrt(x^2) is positive...so from (2) we know X is positive...but we dont know if SQRT(X^2) is an integer or not...Insuff.... combining them together...Sufficient...

Agree with fresinha12.

sqrt(x^2) = |x|.

and |x | = x only when x is positive or 0.But from the stem x^2 is positive so x can't be 0. So x is definitely positive.

and from (1) x is an integer.

So using (1) and (2) x is a positive integer.

HMTG.

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i thought A) as well b/c in gmatland sqrt n is always nonnegative, is it ?

HMTG, how do you know that sqrt x^2=|x| ? x can denote a negative number as well.

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christoph wrote:

i thought A) as well b/c in gmatland sqrt n is always nonnegative, is it ?

mirhaque wrote:

OA is C. But I think it's E

C cannot be OA because the question asking here is whether x is a positive integer or not?

statement i says sqrt x^2 is an integer, then x must be positive integer since it cannot be a negative integer. so statement i is suff.

from ii, sqrt x^2 =x. here x could be a positive integer or a positive nuber (fraction). so not suff.

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HIMALAYA wrote:

statement i says sqrt x^2 is an integer, then x must be positive integer since it cannot be a negative integer. so statement i is suff.

sqrt() always refers to the nonnegative root.

Let x = -3. x^2 = 9, sqrt(x^2) = 3, which is an integer. But x is not a positive integer.

This is why we need S2 as well.

C is correct.

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I would pick C.
B tells us that X is positive:
B tells us "sqrt x^2 =x" therefore "x= sqrt x^2 "
we know that sqrt can be only positive
therefore x is positive
A tells us that X is integer
both concludes that X is positive and integer