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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, 22:24

1

This post received KUDOS

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).

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, 16:50

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 plug-in 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 plug-in 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.

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, 03:48

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 plug-in 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 plug-in 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.

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 GMAT-level 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 non-integer 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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