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# If y and z are factors of x^2, is x divisible by y? (1) y is a prim

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Manager
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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, 12:24
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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, 14:01
would be interesting to see the responses to this question
Intern
Joined: 24 Aug 2016
Posts: 28
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
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, 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.

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, 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.

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, 03:40
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, 10:51
you cant assume x is an integer unless mentioned in the question stem. Therefore I believe the 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, 11:14
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.

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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, 11:31
2
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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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, 11:49
I agree as well. It does not mention that x is an integer so x can very well be a non-integer.

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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] 05 Aug 2017, 11:49
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# If y and z are factors of x^2, is x divisible by y? (1) y is a prim

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