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Is (7x)^1/2 an integer? [#permalink]
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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
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Last edited by Bunuel on 05 Jun 2013, 03:49, edited 3 times in total.
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Re: Is (7x)^1/2 an integer? [#permalink]
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Re: Is (7x)^1/2 an integer? [#permalink]
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13 Jan 2013, 09:01
[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
A



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Re: Is (7x)^1/2 an integer? [#permalink]
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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]
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13 Jan 2013, 09:29
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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]
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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
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]
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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 noninteger/integer. Insufficient. A.
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Re: Is (7x)^1/2 an integer? [#permalink]
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21 Dec 2013, 06:09
(1) \(\sqrt{7x}\) = 7\(\sqrt{\frac{x}{7}}\)
7 and \(\sqrt{\frac{x}{7}}\) are integers so \(\sqrt{7x}\) is an integerSufficient
(2) \(\sqrt{7x}\) = \(\sqrt{28x}/2\)
We know that \(\sqrt{28x}\) is an integer but not sure if it is divisible by 2 or notInsufficient
Hence the answer is A



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Re: Is (7x)^1/2 an integer? [#permalink]
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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]
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15 Apr 2014, 08:04
(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. tnx for this
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Re: Is (7x)^1/2 an integer? [#permalink]
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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 primefactorization 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 Final Answer: GMAT assassins aren't born, they're made, Rich
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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.



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Re: Is (7x)^1/2 an integer? [#permalink]
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15 May 2016, 09:41
sahilhanda wrote: 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. Merging topics. Please refer to the discussion above.
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Re: Is [m][square_root](7x)[/square_root][/m] an integer? [#permalink]
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sahilhanda wrote: 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. 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 A
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Re: Is (7x)^1/2 an integer? [#permalink]
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16 May 2016, 23:51
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 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



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