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Re: For integers x, y, and z, x = y^2. What is the value of z? [#permalink]
08 Apr 2014, 22:32

3

This post received KUDOS

Expert's post

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.

Re: For integers x, y, and z, x = y^2. What is the value of z? [#permalink]
08 Apr 2014, 22:55

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

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For integers x, y, and z, x=y2. What is the value of z? [#permalink]
08 Jul 2014, 05:05

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.

Re: For integers x, y, and z, x = y^2. What is the value of z? [#permalink]
08 Jul 2014, 05:14

Expert's post

PathFinder007 wrote:

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.

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? _________________