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MGmat number properties

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MGmat number properties

by fruti_yum » Sun Sep 13, 2009 11:56 am
If x, y, and z are integers greater than 1, and (3^27)(5^10)(z) = (5^8)(9^14)(x^y), then what is the value of x?

(1) y is prime

(2) x is prime

Please let me know the method to approach this question.

I'll post the OA after some discussion!
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by prindaroy » Sun Sep 13, 2009 2:30 pm
statement B alone is sufficient; IMO

we'll come to this;

5^2 * z = 3 * x^y,

so we get that x^y = 25, 50, 75 or all multiples of 25, knowing that x is prime will show us that 5 is the only possible number that can achieve this; 15, 10 are not prime and cannot do that. So statement B alone is sufficient. Is that the OA??
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by fruti_yum » Sun Sep 13, 2009 2:52 pm
prindaroy wrote:statement B alone is sufficient; IMO

we'll come to this;

5^2 * z = 3 * x^y,

so we get that x^y = 25, 50, 75 or all multiples of 25, knowing that x is prime will show us that 5 is the only possible number that can achieve this; 15, 10 are not prime and cannot do that. So statement B alone is sufficient. Is that the OA??
Yes the OA is B.. however,what i don't understand is what does x is prime get us? I understand z has to be some multiple of 3. x^y similarly has to be some multiple of 5. but don't we already have the question stem that gives us that.. why do we need statement b?
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by PussInBoots » Wed Sep 16, 2009 2:05 pm
(3^27)(5^10)(z) = (5^8)(9^14)(x^y)
25z = 3 x^y
z = 3 * x^y / 25 -> x is multiple of 5
(2) alone says that x = 5, hence z = 3 * 5^y / 25, still not enough info cuz y = 2 and y = 4 work too. (1) narrows our choices to y = 2.

Answer is C
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by heshamelaziry » Wed Sep 16, 2009 2:54 pm
PussInBoots wrote:(3^27)(5^10)(z) = (5^8)(9^14)(x^y)
25z = 3 x^y
z = 3 * x^y / 25 -> x is multiple of 5
(2) alone says that x = 5, hence z = 3 * 5^y / 25, still not enough info cuz y = 2 and y = 4 work too. (1) narrows our choices to y = 2.

Answer is C
How did you arrive to this 25z = 3 x^y . Thanks.
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by tienvunguyen » Thu Sep 17, 2009 11:11 am
PussInBoots wrote:(3^27)(5^10)(z) = (5^8)(9^14)(x^y)
25z = 3 x^y
z = 3 * x^y / 25 -> x is multiple of 5
(2) alone says that x = 5, hence z = 3 * 5^y / 25, still not enough info cuz y = 2 and y = 4 work too. (1) narrows our choices to y = 2.

Answer is C
I do not think y matters because the question only asks for x. So I think B is the correct answer.
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by viju9162 » Fri Sep 18, 2009 2:45 am
Hi Prindaroy,

5^2 * z = 3 * x^y,

so we get that x^y = 25, 50, 75 or all multiples of 25, knowing that x is prime will show us that 5 is the only possible number that can achieve this; 15, 10 are not prime and cannot do that. So statement B alone is sufficient.

You are not considering "3" here?
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by woo » Sat Sep 19, 2009 5:40 am
It seems we all agree that modifying the question
gets us x^y=(5^2)(1/3)z

Condition 1 says y is a prime.
x must be an integer according to the question
thus, z has to be multiple of 3.
z can be 3, 6, 9, 12 and so on
If you plug in 3 in z you get x^y=(5^2). Here x=5
However, if you plug in 12 you get x^y=(5^2)(2^2)=(10^2)
Here x=10 therefore, insufficient.

Condition 2 says x is a prime.
It means that z has to eliminate either 5^2 or 1/3.
We know that z is an integer thus, z cannot
eliminate 5^2. Therefore z has to be 3.
In this case x=5. Hence sufficient.

How does it sound?
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