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the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be square of an integer and therefore sqrt (x) is an integer . i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer

friends please help me in pointing out whee i am going wrong.

if root(4x) is an integer fact. Then 2 * root(x) is an integer. So what!!!!!!!!!! This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer.

So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer. if x=4(an integer) root(x)=2 and 2*2=4 is also an integer. So, statement one would be true for two values of x (2.25 and 4). root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer.

What's wrong with my explanation??

You forgot that x is a positive integer, so \(\sqrt{x}\) cannot equal to \(\frac{integer}{2}\). Generally \(\sqrt{integer}\) is either an integer or an irrational number.

Complete solution:

If x is a positive integer, is sqrt(x) an integer

If \(x=integer\), is \(\sqrt{x}=integer\)?

(1) \(\sqrt{4x}\) is an integer --> \(2\sqrt{x}=integer\) --> \(2\sqrt{x}\) to be an integer \(\sqrt{x}\) must be an integer or integer/2, but as \(x\) is an integer, then \(\sqrt{x}\) can not be integer/2, hence \(\sqrt{x}\) is an integer. Sufficient.

(2)\(\sqrt{3x}\) is not an integer --> if \(x=9\), condition \(\sqrt{3x}=\sqrt{27}\) is not an integer satisfied and \(\sqrt{x}=3\) IS an integer, BUT if \(x=2\), condition \(\sqrt{3x}=\sqrt{6}\) is not an integer satisfied and \(\sqrt{x}=\sqrt{2}\) IS NOT an integer. Two different answers. Not sufficient.

Bunuel - can you further explain your explanation for Statement 1? I am confused because if x is a positive integer, sqrt(x) can equal an integer/2 if for example x was equal to 4. 4 is a positive integer and the square root of 4 is equal to 4/2. I think I am missing the overall takeaway, can you help clarify? Thanks!

4/2=2 and is not a fraction (integer/2 means reduced fraction).
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]

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10 Jan 2010, 07:42

if x is a positive integer ,is sqrt (x) an integer

(1) sqrt(4x) is an integer .

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

the explanation says since sqrt (4x) is an integer ,it follows that 4x must be square of an integer and so x must be square of an integer and therefore sqrt (x) is an integer . i was trying to solve it this way sqrt(4x) =integer ,=> 2. sqrt (x) =integer => sqrt (x) =integer/2 =integer or non integer for example if 2 .sqrt (x) = 4 ,=> sqrt (x) =2 and so sqrt (x) is integer but if 2. sqrt (x) =3 ,=> sqrt (x) =3/2 and so sqrt (x) is non integer yes. sqrt(x) is an integer.

how i worked it out : sqrt(4x) is an integer ==> 2 * sqrt(x) is an integer

in order that the product of 2 and sqrt(x) be an integer, sqrt(x) must either be 1) an integer 2) exactly half of an integer. i.e a number like 0.5, 1.5, 2.5 etc etc you worked that out as well. sqrt(x) = integer/2

we also know x is an integer. is there any integer whose square root is a half of an integer ? no!

therefore the only other alternative is sqrt(x) is a whole integer.

Re: if x is a positive integer ,is sqrt x an integer [#permalink]

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09 Jan 2011, 08:40

I didn't get the explanation:

if root(4x) is an integer fact. Then 2 * root(x) is an integer. So what!!!!!!!!!! This doesn't mean that root(x) should be an integer. Because root(x) can be 1.5 and yet won't distort the fact that 2 * 1.5 = 3 is an integer.

So, if x = 2.25(a non integer) root(x)=1.5 and 2*1.5=3 is an integer. if x=4(an integer) root(x)=2 and 2*2=4 is also an integer. So, statement one would be true for two values of x (2.25 and 4). root(2.25) is 1.5, not an integer. root(4) is 2, an integer. This statement is insufficient to conclude whether root(x) is an integer.

What's wrong with my explanation??
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]

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19 Jul 2011, 05:29

Quote:

then sqrt(x) can not be integer/2

I think sqrt(x) can be integer/2 as long as (integer/2) itself is an integer i.e. that integer is multiple of 2. I think that is what bunuel meant. And the answer remains same.
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Re: if x is a positive integer ,is sqrt (x) an integer (1) [#permalink]

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03 Jan 2012, 12:06

Oh nice problem. Took quite some time to answer it but got A. Explanation by Bunuel is more than sufficient to understand the solution.
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Re: if x is a positive integer ,is sqrt (x) an integer (1) [#permalink]

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03 Jan 2012, 13:31

Both equations and number plugging helps here.

1. sqrt(4x) = integer, that is 2 x sqrt(x) = integer. For this equation to be true, sqrt(x) has to be an integer. SUFFICIENT. If number plugging, use 4x4 and 4x9 combinations 2. sqrt(3x) = frac. Here, sqrt(3) x sqrt(x) = frac, that is frac x sqrt(x) = frac. Difficult to determine if this relationship can infer sqrt(x) as integer. So, let's go with number plugging. sqrt(3x4) = 12 satisfies, and sqrt(4) = integer. sqrt(3x5) = 15 satisfies, but sqrt(5) = frac. So, insufficient.

+1 for A.
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]

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15 Dec 2012, 10:21

Bunuel - can you further explain your explanation for Statement 1? I am confused because if x is a positive integer, sqrt(x) can equal an integer/2 if for example x was equal to 4. 4 is a positive integer and the square root of 4 is equal to 4/2. I think I am missing the overall takeaway, can you help clarify? Thanks!

The square root of any positive integer is either an integer or an irrational number. So, \(\sqrt{x}=\sqrt{integer}\) cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...).

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

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05 Jul 2014, 02:40

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Re: If x is a positive integer, is x^1/2 an integer [#permalink]

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12 Nov 2014, 00:36

Hello Easiest way i can think of: S1-\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer We know that \(\sqrt{4}\) is 2 - int. so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here) Sufficient (12TEN) or (ABCDE)

S2- \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer We know that \(\sqrt{3}\) is not an integer But Non-Integer * Integer OR Non-integer = Non- integer So Insufficient (1E) or (AD)

Re: If x is a positive integer, is x^1/2 an integer [#permalink]

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04 Dec 2014, 23:00

deeuk wrote:

Hello Easiest way i can think of: S1-\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer We know that \(\sqrt{4}\) is 2 - int. so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here) Sufficient (12TEN) or (ABCDE)

S2- \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer We know that \(\sqrt{3}\) is not an integer But Non-Integer * Integer OR Non-integer = Non- integer So Insufficient (1E) or (AD)

thoughts?

Bunuel could you please tell me if this approach is correct/incorrect?

Hello Easiest way i can think of: S1-\(\sqrt{4x}\) is an integer =\(\sqrt{4}*\sqrt{x}\) has to be integer We know that \(\sqrt{4}\) is 2 - int. so\(\sqrt{x}\) has to be int ( because Non Integer*integers is ALWAYS a Non integer, which is not the case here) Sufficient (12TEN) or (ABCDE)

S2- \(\sqrt{3x}\) is NOT an integer = \(\sqrt{3}*\sqrt{x}\) is not an integer We know that \(\sqrt{3}\) is not an integer But Non-Integer * Integer OR Non-integer = Non- integer So Insufficient (1E) or (AD)

thoughts?

Bunuel could you please tell me if this approach is correct/incorrect?

No. That's not correct.

(Non-integer)*(integer) could yield an integer. For example, 1/2*2 = 1.

(Non-integer)*(non-integer) could also yield an integer. For example, \(\sqrt{2}*\sqrt{2}=2\).
_________________

If x is a positive integer, is x^1/2 an integer [#permalink]

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10 Dec 2014, 12:21

Thank you for the correction But now i am wondering since \(\sqrt{4}\) is 2, doesn't \(\sqrt{x}\) have to be an integer as well. I mean doesnt x have to be a perfect square?

gmatclubot

If x is a positive integer, is x^1/2 an integer
[#permalink]
10 Dec 2014, 12:21

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