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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?

(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.

(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?

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

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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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.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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

Is [m][square_root](7x)[/square_root][/m] an integer? [#permalink]

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15 May 2016, 09:36

Is \(\sqrt{(7x)}\) an integer?

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

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

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. EITHER statement BY ITSELF is sufficient to answer the question. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. EITHER statement BY ITSELF is sufficient to answer the question. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

Merging topics. Please refer to the discussion above.
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Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. EITHER statement BY ITSELF is sufficient to answer the question. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

Hi lets see the what info Q has.. Is \(\sqrt{(7x)}\) an integer?.. possible - a) if x ia an integer and of type y^2*7.. b) if x is a fraction of type y^2/7..

lets see the statements (1) \(\sqrt{(x/7)}\) is an integer. this tells us that x = y^2*7, where y is an integer, SAME as case (a) above Suff

(2) \(\sqrt{(28x)}\)is an integer. this tells us that \(\sqrt{(4*7*x)}\)is an integer. so x = y^2/28.. if y is multiple of 2, ans is YES if y is not a multiple of 2, ans is No Insuff

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

A: let's say \(\sqrt{\frac{x}{7}}\) = m

\(\sqrt{\frac{x*7}{7*7}}\) = m

\(\sqrt{7x}\)/7 = m

=> \(\sqrt{7x}\) = 7*m => \(\sqrt{7x}\) is an integer. (m is given to be an integer)

A is sufficient.

B: let's say \(\sqrt{28x}\) = k

2*\(\sqrt{7x}\) = k

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

Here the answer depends on the parity (evenness or oddness) of k and parity of k is unknown. Hence B is not sufficient.

Final answer: A

gmatclubot

Re: Is (7x)^1/2 an integer?
[#permalink]
16 May 2016, 23:51

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