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If x is a positive integer, is x^1/2 an integer

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

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New post 13 Jan 2014, 01:51
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 13 Jan 2014, 01:52
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SOLUTION:

If x is a positive integer, is \(\sqrt{x}\) an integer?

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 \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...). So, as given that \(x\) is a positive integer then \(\sqrt{x}\) is either an integer itself or an irrational number.


(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.

Answer: A.

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

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New post Updated on: 31 Jul 2014, 16:18
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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.



Statement 1

\(\sqrt{4x} = k\), \(k\) is an integer
Or,\(\sqrt{x}= k/2\)

If \(k/2\) is not an integer, then \((k/2)^2\) or \(k^2/4\) is also not an integer and this implies that x is not an integer as \(x = k^2/4\).
But, it is given that \(x\) is an integer.

Therefore, \(k^2/4\) is an integer => \(k/2\) is an integer =>\(\sqrt{x}\) is an integer...........Sufficient....(B)(C)(E)


Statement 2

\(\sqrt{3x}=k\), \(k\) is not an integer
Or, \(\sqrt{x} = k/\sqrt{3}\)

Thus, \(\sqrt{x}\) = integer, when \(k\) is a multiple of \(\sqrt{3}\)
and \(\sqrt{x}\) is not an integer when \(k\) is not a multiple of \(\sqrt{3}\)

As the value of \(\sqrt{x}\) can not be uniquely determined, statement (2) is not sufficient.................(D)

Answer: (A)

Originally posted by arunspanda on 18 Jan 2014, 04:22.
Last edited by arunspanda on 31 Jul 2014, 16:18, edited 1 time in total.
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 14 Jan 2014, 07:06
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IMO A.

If x is a positive integer, is \sqrt{x} an integer?

(1) \sqrt{4x} is an integer.
(2) \sqrt{3x} is not an integer.

1) \sqrt{4x} = 2\sqrt{x} is also integer = hence sufficient.

2) if x = 27 then Yes,
if x = 4 then no 2. hence not sufficient.
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 14 Jan 2014, 11:49
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x is a positive integer, is \(\sqrt{x}\) an integer?

(1) \(\sqrt{4x}\) is an integer.
(2) \(\sqrt{3x}\) is not an integer.



Statement 1) root(4x) is an integer. root(4x) = either +2root(x) or -2root(x) . As in either case, the result is an integer, then root(x) has to be an integer as (+2 or -2) is integer. Hence Sufficient.
Statement 2) root(3x) is an integer. root(3) is not an integer.
There can be following two cases:
1) root(x) is an integer => root(3) * root(x) is not an integer as root(3) is not an integer.
2) root(x) is not an integer => root(3) * root(x) can never be an integer except in once case where x= 3.
Either case it is not sufficient.

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

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New post 12 Nov 2014, 01:36
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)

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

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New post 05 Dec 2014, 00:00
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?
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 05 Dec 2014, 04:28
Motivatedtowin wrote:
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?


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, \(\sqrt{2}*\sqrt{2}=2\).
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 07 Jan 2018, 05:45
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[quote="Bunuel"]The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x is a positive integer, is \(\sqrt{x}\) an integer?

(1) \(\sqrt{4x}\) is an integer.
(2) \(\sqrt{3x}\) is not an integer.

Given x>0 & x=Integer

Find is \(\sqrt{x}\) an integer
=> let \(\sqrt{x}=k\)---------- 'k' is integer
=> OR \(x=k^2\)
Therefore Question can be re-phrased as " IS 'x' A PERFECT SQUARE "

Statement 1 \(\sqrt{4x}\) is an integer
=> let \(\sqrt{4x}=p\) ---------------- 'p' is Integer
=> \(4x=p^2\)
=> \(x=\frac{p^2}{4}\)
=> OR \(x=(\frac{p}{2})^2\) --------- equ (1)
=> Now Since (given) x>0 & x=Integer
=> Therefore 'p' HAS TO BE 2,4,6,8....OR any even number >0
=> substituting 'p' in equ (1) makes \(x=1^2 or 2^2 or 3^2\) and so on..
=> Therefore x=Perfect Square.
=> SUFFICIENT

Statement 2 \(\sqrt{3x}\) is not an integer.
=> So \(x\neq{3, 3^3, 3^5...}\)

=> BUT x=1,2,4,5,6,9,....

=> So 'x' can be Prefect Square ( when x=4,9,16....) and 'x' NOT a Perfect Square (when x=2,5,6...)
=> Therefore INSUFFICIENT

Therefore 'A'

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

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New post 15 Apr 2018, 13:25
For the first prompt, I was able to simplify to get me to here:

√4x --> 2√x

The obvious part to me is that since the above is an integer, then x is a perfect square, totally understand that.

However, (1/2)*(1/2) = (1/4). To me, the square root of (1/4) would be (1/2), and 2*(1/2) = 1, an integer.

I know Bunuel has said:

Quote:
The square root of any positive integer is either an integer or an irrational number. So, x√=integer‾‾‾‾‾‾‾√x=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 2‾√2, 3‾√3, 17‾‾‾√17, ...).


Is there a steadfast rule that square roots of non-integers is out of scope of the GMAT? Should I just never consider square roots of fractions going forward?
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 15 Apr 2018, 22:36
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jsheppa wrote:
For the first prompt, I was able to simplify to get me to here:

√4x --> 2√x

The obvious part to me is that since the above is an integer, then x is a perfect square, totally understand that.

However, (1/2)*(1/2) = (1/4). To me, the square root of (1/4) would be (1/2), and 2*(1/2) = 1, an integer.

I know Bunuel has said:

Quote:
The square root of any positive integer is either an integer or an irrational number. So, x√=integer‾‾‾‾‾‾‾√x=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 2‾√2, 3‾√3, 17‾‾‾√17, ...).


Is there a steadfast rule that square roots of non-integers is out of scope of the GMAT? Should I just never consider square roots of fractions going forward?


Hello

NO, there is NO rule that square roots of non-integers is out of scope of GMAT. But the thing is, that in this particular question, we are specifically given that x is an integer; so for 2*√x to be an integer, x must be a perfect square integer. So we cannot plug and test the value of x as 1/4 in this particular question, because 1/4 is not an integer.
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 16 Apr 2018, 06:13
amanvermagmat wrote:
Hello

NO, there is NO rule that square roots of non-integers is out of scope of GMAT. But the thing is, that in this particular question, we are specifically given that x is an integer; so for 2*√x to be an integer, x must be a perfect square integer. So we cannot plug and test the value of x as 1/4 in this particular question, because 1/4 is not an integer.


Oh wow...I must have been getting fatigued while studying and missed that part of the prompt. Thank you!
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 20 Apr 2018, 13:36
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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If x is a positive integer, is \(\sqrt{x}\) an integer?

(1) \(\sqrt{4x}\) is an integer.
(2) \(\sqrt{3x}\) is not an integer.



Target question: Is √x an integer?

Given: x is a positive integer

Statement 1: √(4x) is an integer
IMPORTANT CONCEPT: If K is an integer, then √K will be an integer if the prime factorization of K has an even number of each prime.

Some examples:
√144 = 12 (integer), and 144 = (2)(2)(2)(2)(3)(3) [four 2's and two 3's]
√1600 = 40 (integer), and 1600 = (2)(2)(2)(2)(2)(2)(5)(5) [six 2's and two 5's]
√441 = 21 (integer), and 441 = (3)(3)(7)(7)[two 3's and two 7's]
√12 = some non-integer, and 12 = (2)(2)(3)[two 2's and one 3's]
------------------------------------------------

So, if √(4x) is an integer, then the prime factorization of 4x has an even number of each prime.
Since 4x = (2)(2)(x) we can see that the prime factorization of x must have an even number of each prime.
If the prime factorization of x has an even number of each prime, then √x must be an integer.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: √(3x) is not an integer.
There are several values of x that meet this condition. Here are two:
Case a: x = 4. This means that √(3x) = √12, which is not an integer. In this case, √x is an integer.
Case b: x = 5. This means that √(3x) = √15, which is not an integer. In this case, √x is not an integer.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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

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New post 24 Jun 2019, 17:52
Bunuel wrote:
If x is a positive integer, is \(\sqrt{x}\) an integer?

(1) \(\sqrt{4x}\) is an integer.
(2) \(\sqrt{3x}\) is not an integer.


(DS04474)


Data Sufficiency
Question: 31
Category: Algebra Radicals
Page: 155
Difficulty: 650


The Official Guide For GMAT® Quantitative Review, 2ND Edition



I though A was not sufficient because I tested values for X. If x=1 then the root of x would not be an integer but would suffice with the statement. Can anyone explain why testing the number 1 fails to evaluate the argument correctly?
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 18 Sep 2019, 03:26
Question stem tells us \(x>0\) & \(x\) = integer. Is \(\sqrt{x}\) an integer?

Constraint (1) \(\sqrt{4x}\) = integer, is also \(2\sqrt{x}\) = integer.

\(2\sqrt{x}\) = Integer?
\(2\sqrt{1}\) = YES, 2.
\(2\sqrt{2}\) = NO.
\(2\sqrt{3}\) = NO.
\(2\sqrt{4}\) = YES, 4.
\(2\sqrt{5}\) = NO.
\(2\sqrt{6}\) = NO.
\(2\sqrt{7}\) = NO.
\(2\sqrt{8}\) = NO.
\(2\sqrt{9}\) = YES, 6.

Given the constraint, filter out the "NO" cases from the table and take only the "YES" cases into account... Is \(\sqrt{x}\) an integer? Single solution: yes, \(\sqrt{x}\) is always an integer(and always a perfect square root). SUFFICIENT.

Constraint (2) \(\sqrt{3x}\)= not integer, is also \(\sqrt{3}\sqrt{x}\) = not integer.

\(\sqrt{3}\sqrt{x}\) = Integer?
\(\sqrt{3}\sqrt{1}\) = NO.
\(\sqrt{3}\sqrt{2}\) = NO.
\(\sqrt{3}\sqrt{3}\) = YES, 3.
\(\sqrt{3}\sqrt{4}\) = NO.
\(\sqrt{3}\sqrt{5}\) = NO.

Given the constraint, filter out the "YES" cases from the table and take only the "NO" cases into account... Is \(\sqrt{x}\) an integer? Two solutions: both, yes & no. \(\sqrt{1}\) & \(\sqrt{4}\) are integers, whilst \(\sqrt{2}\) & \(\sqrt{5}\) are not integers. INSUFFICIENT.
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Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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New post 06 Oct 2019, 11:30
Bunuel wrote:
SOLUTION:
as \(x\) is an integer, then \(\sqrt{x}\) can not be integer/2,


Hello,
Could you please elaborate on why the statement above holds true?

Thanks a lot!
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Re: If x is a positive integer, is x^1/2 an integer   [#permalink] 06 Oct 2019, 11:30
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