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# Disagree with OA if x^2 is a positive integer, is x a

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29 Jul 2005, 17:00
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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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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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29 Jul 2005, 23:45
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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30 Jul 2005, 08:20
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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30 Jul 2005, 08:59
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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30 Jul 2005, 11:15
it is a question of denotaion of square root

it seems to me: sqrt(4) = +/- 2
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30 Jul 2005, 14:04
I'm also getting (E) on this one. What's the OA?
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30 Jul 2005, 21:39
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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31 Jul 2005, 06:12
OA is C. But I think it's E
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31 Jul 2005, 07:18
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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31 Jul 2005, 08:03
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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31 Jul 2005, 09:35
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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31 Jul 2005, 11:35
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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31 Jul 2005, 16:40
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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31 Jul 2005, 18:18
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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01 Aug 2005, 02:57
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
01 Aug 2005, 02:57
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