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If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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24 Aug 2016, 11:24
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If y and z are factors of x^2, is x divisible by y? (1) y is a prime number. (2) z = y^2
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Re: If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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24 Aug 2016, 13:01
would be interesting to see the responses to this question



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Re: If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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24 Aug 2016, 21:24
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1. SUFFICIENT: Any prime number that is a factor of x^2 must also be a factor of x. If x=y*z, x^2 = (y^2)*(z^2) OR (yz)^2. 2. SUFFICIENT: z = y*y. x is divisible by z, therefor x is divisible by (y*y).
Answer D, both sufficient separately.



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Re: If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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25 Aug 2016, 15:50
Question is asking if \(\frac{x}{y}\) is an integer.
We could pick numbers
S1) \(x=2\). Hence \(x^{2}=4\). Factors of \(x^{2}\) are 1, 2, and 4. Since S1 says that \(y\) is prime, \(y\) could only be 2. Hence \(\frac{x}{y}\)=\(\frac{2}{2}=integer\) YES.
Lets pick another number to see if we can get a NO.
\(x=6.\) Hence \(x^{2}=36\). Factors of \(x^{2}\) are 1,2,3,4,6,9,12,18,and 36. Since S1 says that \(y\) is prime, \(y\) could now be 2 or 3 only. Hence in both the cases \(\frac{x}{y}=integer\). YES.
So we have tried a prime number and a composite number for \(x\) and in both the cases the answer is YES. So Statement 1 is sufficient. Answer Choices could be A or D.
S2) Since we are at A or D, we could verify if statement 2 is definitely true using information from Statement 1, and if proved answer would be D.(Because Statement 1 returned a definite YES and it cannot be contradicted)
Statement 2 says \(z=y^{2}\).
If we look into the first plugin that we did in S1, \(z\) could only be 4 meeting the condition of S1 that \(y\)=Prime=2. There was no other possible value. And it can be seen that \(z =4\) is a factor of \(x^{2}= 4\) meeting the condition in question stem also. So statement 2 is proved using statement 1.
We could try for the second plugin we did in statement 1 also. Here \(z\) could only be 4 or 9 meeting the condition of S1 that\(y=prime=2\) or \(3\). There was no other possible value. And it can be seen that \(z=4\) or \(9\) is a factor of \(x^{2}=36\), meeting the condition in question stem also. So statement 2 is proved using statement 1.
Hence answer is D.



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Re: If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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30 Aug 2016, 02:48
harsha1034 wrote: Question is asking if \(\frac{x}{y}\) is an integer.
We could pick numbers
S1) \(x=2\). Hence \(x^{2}=4\). Factors of \(x^{2}\) are 1, 2, and 4. Since S1 says that \(y\) is prime, \(y\) could only be 2. Hence \(\frac{x}{y}\)=\(\frac{2}{2}=integer\) YES.
Lets pick another number to see if we can get a NO.
\(x=6.\) Hence \(x^{2}=36\). Factors of \(x^{2}\) are 1,2,3,4,6,9,12,18,and 36. Since S1 says that \(y\) is prime, \(y\) could now be 2 or 3 only. Hence in both the cases \(\frac{x}{y}=integer\). YES.
So we have tried a prime number and a composite number for \(x\) and in both the cases the answer is YES. So Statement 1 is sufficient. Answer Choices could be A or D.
S2) Since we are at A or D, we could verify if statement 2 is definitely true using information from Statement 1, and if proved answer would be D.(Because Statement 1 returned a definite YES and it cannot be contradicted)
Statement 2 says \(z=y^{2}\).
If we look into the first plugin that we did in S1, \(z\) could only be 4 meeting the condition of S1 that \(y\)=Prime=2. There was no other possible value. And it can be seen that \(z =4\) is a factor of \(x^{2}= 4\) meeting the condition in question stem also. So statement 2 is proved using statement 1.
We could try for the second plugin we did in statement 1 also. Here \(z\) could only be 4 or 9 meeting the condition of S1 that\(y=prime=2\) or \(3\). There was no other possible value. And it can be seen that \(z=4\) or \(9\) is a factor of \(x^{2}=36\), meeting the condition in question stem also. So statement 2 is proved using statement 1.
Hence answer is D. Hi Harsha What if the value of x is sqt(2)? i.e x is not integer. Would it still be true for statement 1? x=sqt(2)> x^2=2> Factore 1,2 y could be 2 as y is prime and z could be 1 but in this case x wouldn't be divisible by y. hence, statement one is not suffice. Please correct me if I'm missing something.



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Re: If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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05 Aug 2017, 02:40
duahsolo wrote: If y and z are factors of x^2, is x divisible by y? (1) y is a prime number. (2) z = y^2 We've taken the assumption that x is an integer, and in this case both statements are individually sufficient. But the problem is that the question states that x^2 is an integer, not x. What if x is an irrational number? Let's assume \(x = \sqrt{18}.\) (1) y is a prime number. y = 3 is a factor of x^2 = 18 but not a factor of x, because it's not an integer. Insufficient. (2) z = y^2 z = 9 which is a factor of x^2 but again not a factor of x. Insufficient. Bot statements taken together are giving y=3 and z=9 are factors of x^2 but not factors of x. Again insufficient. Answer should be E.



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Re: If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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05 Aug 2017, 09:51
you cant assume x is an integer unless mentioned in the question stem. Therefore I believe the answer should be E. Experts, please comment.



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Re: If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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05 Aug 2017, 10:14
goodie wrote: duahsolo wrote: If y and z are factors of x^2, is x divisible by y? (1) y is a prime number. (2) z = y^2 We've taken the assumption that x is an integer, and in this case both statements are individually sufficient. But the problem is that the question states that x^2 is an integer, not x. What if x is an irrational number? Let's assume \(x = \sqrt{18}.\) (1) y is a prime number. y = 3 is a factor of x^2 = 18 but not a factor of x, because it's not an integer. Insufficient. (2) z = y^2 z = 9 which is a factor of x^2 but again not a factor of x. Insufficient. Bot statements taken together are giving y=3 and z=9 are factors of x^2 but not factors of x. Again insufficient. Answer should be E. Agreed totally. I also believe the same. Expert's responses will further clarify if I am really missing something. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile app



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If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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05 Aug 2017, 10:31
The question is carelessly worded, and if this were a real GMAT question, it would always tell you that x is a positive integer. As the question is written, it allows the possibility that x is some square root of an integer, say √2 or √3. But the concept of "divisibility" that is tested on the GMAT only makes sense when you're discussing integers. It's only in math far beyond GMATlevel that you'd learn what a question like "is √2 divisible by 2?" even means. Using GMAT concepts only, the question is like asking "is √2 orange?" The GMAT can't ask you questions that don't make sense, so the GMAT will always rule out the possibility that something is a noninteger in a divisibility problem. If the question states that x is a positive integer, then it's a more realistic question, and the answer is D.
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Re: If y and z are factors of x^2, is x divisible by y? (1) y is a prim [#permalink]
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05 Aug 2017, 10:49
I agree as well. It does not mention that x is an integer so x can very well be a noninteger. Sent from my iPhone using GMAT Club Forum
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