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Is (7x)^1/2 an integer?

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Is (7x)^1/2 an integer? [#permalink] New post 13 Jan 2013, 07:22
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Is \sqrt{7x} an integer?

(1) \sqrt{\frac{x}{7}} is an integer

(2) \sqrt{28x} is an integer

[Reveal] Spoiler:
Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

Thanks!
Alex
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Jun 2013, 02:49, edited 3 times in total.
Renamed the topic and edited the question.
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Re: Is (7x)^1/2 an integer? [#permalink] New post 13 Jan 2013, 07:58
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Is \sqrt{7x} an integer?

Notice that we are not told that x is an integer.

(1) \sqrt{\frac{x}{7}} is an integer. Given that \sqrt{\frac{x}{7}}=integer --> square it: \frac{x}{7}=integer^2 --> x=7*integer^2. So, \sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer. Sufficient.

(2) \sqrt{28x} is an integer. If x=\frac{1}{28}, then \sqrt{7x}=\frac{1}{2}\neq{integer} BUT if x=0, then \sqrt{7x}=0=integer. Not sufficient.

Answer: A.
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Re: Is (7x)^1/2 an integer? [#permalink] New post 13 Jan 2013, 08:01
[quote="alexpavlos"]Is \sqrt{7x} an integer?

(1) \sqrt{\frac{x}{7}} is an integer

(2) \sqrt{28x} is an integer



for sqt 7x to be an integer , x has to have the form 7^a*z^b, where a is an odd integer and z is an integer and b a an even intiger.

from 1

x is in the form 7^a*z^b

from 2
2sqt 7x is an integer....this means that sqt 7x is in the form k/2 however , it says nothing about the form of x itself... thus insuff

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Re: Is (7x)^1/2 an integer? [#permalink] New post 13 Jan 2013, 08:16
Bunuel wrote:
Is \sqrt{7x} an integer?

Notice that we are not told that x is an integer.

(1) \sqrt{\frac{x}{7}} is an integer. Given that \sqrt{\frac{x}{7}}=integer --> square it: \frac{x}{7}=integer^2 --> x=7*integer^2. So, \sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer. Sufficient.

(2) \sqrt{28x} is an integer. If x=\frac{1}{28}, then \sqrt{7x}=\frac{1}{2}\neq{integer} BUT if x=0, then \sqrt{7x}=0=integer. Not sufficient.

Answer: A.



Thanks sorry for asking again... But could you please explain if my method/ logic was completely wrong? ie if we break out 28 int primes we see that it has 2,2,7 so x must have at least one 7? therefore we can assume that \sqrt{7x} should be integer?

Thanks again!
Alex
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Re: Is (7x)^1/2 an integer? [#permalink] New post 13 Jan 2013, 08:29
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alexpavlos wrote:
Bunuel wrote:
Is \sqrt{7x} an integer?

Notice that we are not told that x is an integer.

(1) \sqrt{\frac{x}{7}} is an integer. Given that \sqrt{\frac{x}{7}}=integer --> square it: \frac{x}{7}=integer^2 --> x=7*integer^2. So, \sqrt{7x}=\sqrt{7*(7*integer^2)}=7*integer=integer. Sufficient.

(2) \sqrt{28x} is an integer. If x=\frac{1}{28}, then \sqrt{7x}=\frac{1}{2}\neq{integer} BUT if x=0, then \sqrt{7x}=0=integer. Not sufficient.

Answer: A.



Thanks sorry for asking again... But could you please explain if my method/ logic was completely wrong? ie if we break out 28 int primes we see that it has 2,2,7 so x must have at least one 7? therefore we can assume that \sqrt{7x} should be integer?

Thanks again!
Alex


If x is NOT an integer you cannot make its prime factorization.

If we were told that x IS an integer, then \sqrt{28x}=2\sqrt{7x}=integer. Thus x must be of the form 7^{odd}*integer^2, so in this case \sqrt{7x}=\sqrt{7*(7^{odd}*integer^2)}=\sqrt{7^{even}*integer^2}=integer.

Hope it's clear.
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Re: Is (7x)^1/2 an integer? [#permalink] New post 26 May 2013, 20:17
alex1233 wrote:
Is \sqrt{7x} an integer?

(1) \sqrt{\frac{x}{7}} is an integer

(2) \sqrt{28x} is an integer



Can someone please explain why point 2 is not correct? What I did is that 28 is factored into => 2*2*7 so therefore I concluded x must have at least one 7. so if x has one seven then \sqrt{7x} must be an integer. Am I doing something wrong here?

Thanks!
Alex



Quote:
\sqrt{7x} = integer ? or 7x= int^2 ?



(1) \sqrt{\frac{x}{7}} is an integer ---- lets say b
x/7 = b^2 --> where b is an integer
x= 7 *b^2
7x = (7*b) ^2
hence 7x=int^2--- sufficient


(2) \sqrt{28x} is an integer ---- lets say c

28x= c^2
7x=c^2/4 = (c/2)^2
If C is odd , then ans is No and if C is even then yes . ( If you knew that x was an integer , then this would have sufficed)
Insufficient.
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Re: Is (7x)^1/2 an integer? [#permalink] New post 05 Jun 2013, 02:55
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Re: Is (7x)^1/2 an integer? [#permalink] New post 05 Jun 2013, 22:53
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alex1233 wrote:
Is \sqrt{7x} an integer?

(1) \sqrt{\frac{x}{7}} is an integer

(2) \sqrt{28x} is an integer


From F.S 1 , we know that \sqrt{\frac{x}{7}} = Integer(I)--> \sqrt{\frac{x*7}{7*7}} = \sqrt{7x}*\frac{1}{7} = I--> \sqrt{7x} = 7*I. Thus, an integer. Sufficient.

From F.S 2, we know that \sqrt{28x} = Integer(I) --> \sqrt{7*4x} = 2*\sqrt{7x} = I.
Thus, \sqrt{7x} =\frac{I}{2}. Depending on I being odd/even, the given expression will be a non-integer/integer. Insufficient.

A.
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Re: Is (7x)^1/2 an integer? [#permalink] New post 21 Dec 2013, 05:09
(1) \sqrt{7x} = 7\sqrt{\frac{x}{7}}

7 and \sqrt{\frac{x}{7}} are integers so \sqrt{7x} is an integer---Sufficient

(2) \sqrt{7x} = \sqrt{28x}/2

We know that \sqrt{28x} is an integer but not sure if it is divisible by 2 or not---Insufficient

Hence the answer is A
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Re: Is (7x)^1/2 an integer? [#permalink] New post 03 Feb 2014, 05:29
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(1) \sqrt{\frac{x}{7}} is integer.

Let this integer be p. then

\sqrt{\frac{x}{7}} = p

\frac{x}{7} = p^2

x=7*p^2

7x=7^2*p^2

\sqrt{7x}=7*p an integer. (1) is sufficient

(2) \sqrt{28x} is integer.

Let this integer be q.

\sqrt{28x} = q

\sqrt{4*7x} = q

2*\sqrt{7x} = q

\sqrt{7x} = \frac{q}{2}

It is given that q is integer, but \frac{q}{2} may or may not be an integer. not sufficient.

A is the soln.
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Re: Is (7x)^1/2 an integer? [#permalink] New post 15 Apr 2014, 07:04
(1) \sqrt{\frac{x}{7}} is integer.

Let this integer be p. then

\sqrt{\frac{x}{7}} = p

\frac{x}{7} = p^2

x=7*p^2

7x=7^2*p^2

\sqrt{7x}=7*p an integer. (1) is sufficient

(2) \sqrt{28x} is integer.

Let this integer be q.

\sqrt{28x} = q

\sqrt{4*7x} = q

2*\sqrt{7x} = q

\sqrt{7x} = \frac{q}{2}

It is given that q is integer, but \frac{q}{2} may or may not be an integer. not sufficient.

A is the soln.

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Re: Is (7x)^1/2 an integer? [#permalink] New post 15 Apr 2014, 07:09
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gmatonline wrote:
(1) \sqrt{\frac{x}{7}} is integer.

Let this integer be p. then

\sqrt{\frac{x}{7}} = p

\frac{x}{7} = p^2

x=7*p^2

7x=7^2*p^2

\sqrt{7x}=7*p an integer. (1) is sufficient

(2) \sqrt{28x} is integer.

Let this integer be q.

\sqrt{28x} = q

\sqrt{4*7x} = q

2*\sqrt{7x} = q

\sqrt{7x} = \frac{q}{2}

It is given that q is integer, but \frac{q}{2} may or may not be an integer. not sufficient.

A is the soln.

tnx for this


I think this is the post you are looking for: rules-for-posting-please-read-this-before-posting-133935.html#p1096628

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Re: Is (7x)^1/2 an integer?   [#permalink] 15 Apr 2014, 07:09
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