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

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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.

(DS04474)

Data Sufficiency
Question: 31
Page: 155
Difficulty: 650

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

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

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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.

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

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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)

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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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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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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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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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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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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GMAT 1: 370 Q29 V13 Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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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
Dinesh
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GMAT 1: 700 Q43 V42 Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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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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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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GMAT 1: 700 Q43 V42 Re: If x is a positive integer, is x^1/2 an integer  [#permalink]

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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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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

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

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