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For integers x, y, and z, x=y2. What is the value of z?

1) x=z!(z−1)! (2) 12<z<22

Given \(x = y^2\) What is the relevance of y here? The statements only give x and z and we need the value of z. y is introduced here to tell you that x is a perfect square. so all its prime factors have even powers.

1) x=z!(z−1)!

When will x be a perfect square? Whenever z is a perfect square. We know that z! = z * (z - 1)!

\(x = z * (z - 1)! * (z - 1)! = z * (z - 1)!^2\) (z -1)! is already squared. So for x to be perfect square, z should also be a perfect square. z could be 1 or 4 or 9 or 16 or 25 and so on...

(2) 12<z<22 z could be 13 or 14 or 15 and so on...

Using both together, between 12 and 22, there is only one perfect square i.e. 16. So z must be 16.

For integers x, y, and z, x = y^2. What is the value of z? [#permalink]

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08 Apr 2014, 23:55

Mountain14 wrote:

For integers x, y, and z, x=y2. What is the value of z?

1) x=z!(z−1)! (2) 12<z<22

Sol:St 1 can be re-written as \(x=y^2= z*(z-1)!^{2}\)

If x=1=y^2 then z!*(z-1)! = 1 or z=1 Now x =y^2 so z*{(z-1)!}^2 will also need to be a perfect square So if z=25 then we have x=y^2 = 25*(24!)^2 or 25*(24*23*22*21.....1)^2 or 5^2(24*23*22*21....1)^2

So Z can have any value 1,4,9,16,25 and so on and therefore St 1 is not sufficient

from St 2 we know that z is in between 12 and 22 but there is no relation given between x and z or y and z. Therefore not sufficient

Combining both statement we get that z=16.

Ans is C

Good Question
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

For integers x, y, and z, x=y2. What is the value of z? [#permalink]

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08 Jul 2014, 06:05

VeritasPrepKarishma wrote:

Mountain14 wrote:

For integers x, y, and z, x=y2. What is the value of z?

1) x=z!(z−1)! (2) 12<z<22

Given \(x = y^2\) What is the relevance of y here? The statements only give x and z and we need the value of z. y is introduced here to tell you that x is a perfect square. so all its prime factors have even powers.

1) x=z!(z−1)!

When will x be a perfect square? Whenever z is a perfect square. We know that z! = z * (z - 1)!

\(x = z * (z - 1)! * (z - 1)! = z * (z - 1)!^2\) (z -1)! is already squared. So for x to be perfect square, z should also be a perfect square. z could be 1 or 4 or 9 or 16 or 25 and so on...

(2) 12<z<22 z could be 13 or 14 or 15 and so on...

Using both together, between 12 and 22, there is only one perfect square i.e. 16. So z must be 16.

Answer (C)

Hi Karishma,

I have a query here.

x = y^2. i.e. x is a perfect square. ( but in question where this is telling this is a perfect square. if I will take y =1 then x=y^2 = 1. Now this can not be perfect square bcz 1 is not having any prime factor)perfect square for Here x=z!(z-1)! so if I will take 1 the 1!*0! = 1*1 = 1 we can say 1 is square no. so x=y^2. x=1^2.

For integers x, y, and z, x=y2. What is the value of z?

1) x=z!(z−1)! (2) 12<z<22

Given \(x = y^2\) What is the relevance of y here? The statements only give x and z and we need the value of z. y is introduced here to tell you that x is a perfect square. so all its prime factors have even powers.

1) x=z!(z−1)!

When will x be a perfect square? Whenever z is a perfect square. We know that z! = z * (z - 1)!

\(x = z * (z - 1)! * (z - 1)! = z * (z - 1)!^2\) (z -1)! is already squared. So for x to be perfect square, z should also be a perfect square. z could be 1 or 4 or 9 or 16 or 25 and so on...

(2) 12<z<22 z could be 13 or 14 or 15 and so on...

Using both together, between 12 and 22, there is only one perfect square i.e. 16. So z must be 16.

Answer (C)

Hi Karishma,

I have a query here.

x = y^2. i.e. x is a perfect square. ( but in question where this is telling this is a perfect square. if I will take y =1 then x=y^2 = 1. Now this can not be perfect square bcz 1 is not having any prime factor)perfect square for Here x=z!(z-1)! so if I will take 1 the 1!*0! = 1*1 = 1 we can say 1 is square no. so x=y^2. x=1^2.

I just want to know why ans. can not be A.

Thanks.

Let me ask you a question. Is 1 the only possible value of x?
_________________

Re: For integers x, y, and z, x = y^2. What is the value of z? [#permalink]

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Re: For integers x, y, and z, x = y^2. What is the value of z? [#permalink]

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20 Oct 2016, 06:00

For integers x,y, and z, x=y^2. What is the value of z? (1) x=z!(z−1)! (2) 12<z<22

We know X is a perfect square. 1) x=z!(z−1)! = perfect square

take cases when z=0, the factorial of negative is undefined so Z cannot be zero when z=1 then x=1!0! = 1- which is a perfect square when z=2 then x=2!1! = 2 - which is not a perfect square when z=3 then x=3!2! = 3*2^2 - not a perfect square again when z=4 then x=4!3! = 4*3^2*2^2=2^2*3^2*2^2 - ok! this is a perfect square. You should have noticed by now that you will never get a perfect square unless z itself is a perfect square. so again when Z=9 then x=9!8! =9*8^2*7^2*......*2^2=3^2*8^2*7^2*......*2^2 - here also we have a perfect square so, Z=1,4,9,16,25,36............ NOT SUFFICIENT.

(2) 12<z<22 so z= 13,14,15,16,17,18,19,20 and 21 Clearly, NOT SUFFICIENT

(1) and (2) Together Compare the values , You will get a unique value for Z i.e. Z=16

BOTH SUFFICIENT Ans C

gmatclubot

Re: For integers x, y, and z, x = y^2. What is the value of z?
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20 Oct 2016, 06:00

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