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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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10 Jan 2010, 00:55
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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.

[Reveal] Spoiler:
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

[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Dec 2012, 08:31, edited 1 time in total.
Renamed the topic, edited the question and added OA.

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

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09 Jan 2011, 09:52
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fluke wrote:
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??

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.

Hope it's clear.
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]

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16 Dec 2012, 08:37
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jmuduke08 wrote:
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 x^1/2 an integer [#permalink]

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12 Feb 2017, 14:44
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gzimmer wrote:
I am new to gmatclub so I apologize if I am using this forum incorrectly... I instinctively want to plug in numbers: The way i solved was:
Sqrt(4*1)= 2= integer but sqrt(1) is not an integer
Sqrt(4*4)= 4= integer and sqrt(4)=2=integer

so I would say statement 1 is not sufficient. Clearly my logic is flawed since statement 1 is sufficient, but can someone tell me why making up numbers for X does not work for this problem?

Highlighted part is not correct. $$\sqrt 1 = 1 = integer$$
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Re: if x is a positive integer ,is sqrt x an integer [#permalink]

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10 Jan 2010, 08:42
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.

hope that helped

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

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11 Jan 2010, 06:50
Hi Janani ,

Thanks for the explanation ,do we have a rule like we can't have a sqrt (x)=integer /2,
or its by observation ...

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

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09 Jan 2011, 09:40
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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10 Jan 2011, 01:32
good explanation Bunuel and you are right in saying that I carelessly overlooked the fact that x was a positive integer...

thanks

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

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19 Jul 2011, 06:29
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.
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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, 13:06
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, 14:31
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, 11:21
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!

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one from the official book [#permalink]

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28 May 2013, 00:26
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Hi all,
I have a question solution from the official book that appears incorrect to me.
I've added a scan as an attachment.

The solution claims that A is sufficient. But you don't always get an integer, for example:
If you take \sqrt{4x}=5 vs \sqrt{4x}=6.

what do you think?
Attachments

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Re: one from the official book [#permalink]

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28 May 2013, 00:38
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shimonam wrote:
Hi all,
I have a question solution from the official book that appears incorrect to me.
I've added a scan as an attachment.

The solution claims that A is sufficient. But you don't always get an integer, for example:
If you take \sqrt{4x}=5 vs \sqrt{4x}=6.

what do you think?

Merging similar topics. Please refer to the solutions above.

As for your doubt: if $$\sqrt{4x}=5$$, then $$x=\frac{25}{4}\neq{integer}$$, which contradicts the stem.

Similar questions:
if-x-is-a-positive-integer-is-sqrt-x-an-integer-88994.html
value-of-x-107195.html
number-prop-ds-106886.html
number-system-106606.html
odd-vs-even-trick-question-106562.html
quant-review-2nd-edition-ds-104421.html
algebra-ds-101464.html
quant-review-2nd-edition-ds-104421.html
q-31-og-12-ds-101918.html
airthmetic-ds-108287.html

Hope it helps.

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Re: one from the official book [#permalink]

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28 May 2013, 00:41
shimonam wrote:
Hi all,
I have a question solution from the official book that appears incorrect to me.
I've added a scan as an attachment.

The solution claims that A is sufficient. But you don't always get an integer, for example:
If you take \sqrt{4x}=5 vs \sqrt{4x}=6.

what do you think?

All OG13 questions with solutions are here: the-official-guide-quantitative-question-directory-143450.html

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

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29 Jun 2013, 06:03
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 $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{17}$$, ...).

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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, 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

thoughts?

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

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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

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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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

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

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10 Dec 2014, 13:21
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?

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If x is a positive integer, is x^1/2 an integer   [#permalink] 10 Dec 2014, 13:21

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