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If x is a positive integer is sqrt(x) an integer? 1)

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If x is a positive integer is sqrt(x) an integer? 1) [#permalink]

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New post 05 Nov 2005, 20:52
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A
B
C
D
E

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If x is a positive integer is sqrt(x) an integer?
1) sqrt(4x) is an integer
2) sqrt(3x) is NOT an integer

[I'm not sure if the OA is correct]

Last edited by tank on 06 Nov 2005, 00:25, edited 1 time in total.
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Re: DS-Number properties [#permalink]

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New post 05 Nov 2005, 21:06
tank wrote:
If x is a positive integer is sqrt(x) an integer?
1) sqrt(4x) is an integer
2) sqrt(3x) is an integer

[I'm not sure if the OA is correct]


from 1) sqrt(4x) = sqrt(4)*sqrt(x) so 2*sqrt(x) is an integer

sqrt can be 3/2 of 1/2 so is insuff

from 2) same stuff sqrt(3x) = sqrt(3)*sqrt(x)

so sqrt(3)*sqrt(x) is an integer

so sqrt(x) is not an integer

so i would go with B
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Re: DS-Number properties [#permalink]

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New post 05 Nov 2005, 21:09
D.

from i, if sqrt(4x) is an integer, sqrt (x) itself must be an integer because sqrt(4) results in an integer, 2. so suff....

from ii, if sqrt(3x) is an integer, sqrt (x) is not an integer because sqrt(3) doesnot result in an integer. so also suff........

probably you were wondering how it is D and not A. right.
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Re: DS-Number properties [#permalink]

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New post 05 Nov 2005, 21:10
conocieur wrote:
tank wrote:
If x is a positive integer is sqrt(x) an integer?
1) sqrt(4x) is an integer
2) sqrt(3x) is an integer
[I'm not sure if the OA is correct]

from 1) sqrt(4x) = sqrt(4)*sqrt(x) so 2*sqrt(x) is an integer
sqrt can be 3/2 of 1/2 so is insuff
from 2) same stuff sqrt(3x) = sqrt(3)*sqrt(x)
so sqrt(3)*sqrt(x) is an integer
so sqrt(x) is not an integer
so i would go with B


do not overlook that x is an integer. :wink:
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New post 05 Nov 2005, 21:54
well the OA is A. However, i don't think that is correct and here is my reasoning:

1.sqrt(4x) is an integer -> so 2 time sqrt(x) is an integer --> but sqrt(x) could be = 1.5, 2.5 3.5 or any such number so that 2sqrt(x) = 3, 5, 7 etc.
therefore just because 2sqrt(x) is an integer, x does not have to be an integer. Not Suff!

2. sqrt(3x) is an integer --> i.e. 3x is a perfect square but x could be 4/3, which would make sqrt(3x)=sqrt(4)=2 an integer. On the other hand, x could be 3 which would make sqrt(3x) = 3 an integer as well. Not Suff!

1 & 2 together: 4x and 3x is a perfect square. I can't think of any number where both these conditions are true.

Am I missing something??

PS. This question is from the OG Quant Review, DS #116
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New post 05 Nov 2005, 22:37
tank wrote:
2. sqrt(3x) is an integer --> i.e. 3x is a perfect square but x could be 4/3, which would make sqrt(3x)=sqrt(4)=2 an integer.


if x is +ve integer, how it could be 4/3????????

tank wrote:
If x is a positive integer is sqrt(x) an integer?
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New post 05 Nov 2005, 22:47
HIMALAYA wrote:
tank wrote:
2. sqrt(3x) is an integer --> i.e. 3x is a perfect square but x could be 4/3, which would make sqrt(3x)=sqrt(4)=2 an integer.


if x is +ve integer, how it could be 4/3????????

tank wrote:
If x is a positive integer is sqrt(x) an integer?


aha!!! My bad guys. Obvisouly i missed the "integer" there. Also, and more importantly i missed a "Not" in stmt 2. the Original question is:

If x is a positive integer is sqrt(x) an integer?
1) sqrt(4x) is an integer
2) sqrt(3x) is NOT an integer

This would make A, the OA, correct
..i need to go to bed..
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Re: DS-Number properties [#permalink]

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New post 05 Nov 2005, 23:11
tank wrote:
If x is a positive integer is sqrt(x) an integer?
1) sqrt(4x) is an integer
2) sqrt(3x) is an integer

[I'm not sure if the OA is correct]


X > 0, integer

(1) sqrt(4x) is an integer => sqrt 4 * sqrt x = integer = > sqrt x has to be an integer. Sufficient.

(2) sqrt(3x) is not an integer => sqrt 3 * sqrt x => sqrt 3 is obviously not an integer. If x = 1, sqrt 3 is not an integer, but sqrt 1 is. Insufficient

My answer is A
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New post 05 Nov 2005, 23:48
I got A.

1. sqrt(4x) = 2* sqrt(x).
So if sqrt(4x) is an integer sqrt(x) is also an integer.

so it will be either A or D

2. sqrt(3x) is an integer.
now sqrt(3) is not an integer.

if x = 3 , 27 , 243.... or x^a where a is an odd integer, then sqrt(3x) will be an integer.

for any other value of x, sqrt(3) will not be an integer.

so B is not sufficient.

therefore A got to be the answer.
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New post 06 Nov 2005, 00:11
D for me too

A ) sqrt(4x) = Integer take no 16,36....sqrt of these no is Integer and sqrt(x) integer ...Suff

B)
sqrt(3x) = Integer, sprt(144) = sqrt(3*48)=Integer but sqrt(48) is not integer sqrt(81) = sqrt(3*27)=Integer but sqrt(27) not Integer
sqrt(36) = sqrt(3*12)=Integer but sqrt(12) not integer... always NO so SUFF...

both are sufficient so ans D
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New post 06 Nov 2005, 00:37
cool_jonny009 wrote:
D for me too

A ) sqrt(4x) = Integer take no 16,36....sqrt of these no is Integer and sqrt(x) integer ...Suff

B)
sqrt(3x) = Integer, sprt(144) = sqrt(3*48)=Integer but sqrt(48) is not integer sqrt(81) = sqrt(3*27)=Integer but sqrt(27) not Integer
sqrt(36) = sqrt(3*12)=Integer but sqrt(12) not integer... always NO so SUFF...

both are sufficient so ans D


cool_jonny009,

with statement 2

you can use x=4
sqrt(3*4) is not an integer
but sqrt(4) is an integer

you can also use x=2
sqrt(3*2) is not an integer
and sqrt(2) is not an integer

Statement is not suff

so it A
  [#permalink] 06 Nov 2005, 00:37
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