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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?

Re: Is (7x)^1/2 an integer? [#permalink]
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.

Re: Is (7x)^1/2 an integer? [#permalink]
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?

Re: Is (7x)^1/2 an integer? [#permalink]
13 Jan 2013, 08:29

2

This post received KUDOS

Expert's post

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 \(\sqrt{7x}=\sqrt{7*(7^{odd}*integer^2)}=\sqrt{7^{even}*integer^2}=integer\).

Re: Is (7x)^1/2 an integer? [#permalink]
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. _________________

Re: Is (7x)^1/2 an integer? [#permalink]
05 Jun 2013, 22:53

4

This post received KUDOS

Expert's post

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.

Re: Is (7x)^1/2 an integer? [#permalink]
16 Apr 2015, 09:54

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Re: Is (7x)^1/2 an integer? [#permalink]
16 Apr 2015, 10:54

Expert's post

Hi All,

This question can be solved by TESTing VALUES, but you have to really be thorough with your TESTs (and think about what X COULD be given the information in the Facts).

We're asked if Root(7X) is an integer. This is a YES/NO question.

Fact 1: Root(X/7) is an integer.

This Fact tells us that X has to be a specific type of multiple of 7....

IF.... X = 7 the Root(7/7) = 1 and is an integer Root(49) = 7 and the answer to the question is YES

IF.... X = 28 the Root(28/7) = 2 and is an integer Root(196) = 14 and the answer to the question is YES

This hints at the pattern that X will be a "perfect square times a multiple of 7', which means that the answer to the question is ALWAYS YES. You can also use prime-factorization to prove that this is the case. Fact 1 is SUFFICIENT

Fact 2: Root(28X) is an integer.

This Fact tells us that X has to be a specific type of multiple of 7....

IF.... X = 7 the Root(196) = 14 and is an integer Root(49) = 7 and the answer to the question is YES

IF.... X = 1/28 the Root(1) = 1 and is an integer Root(7/28) = 1/2 and the answer to the question is NO Fact 2 is INSUFFICIENT

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