Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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

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]
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? _________________

It’s been a long time, since I posted. A busy schedule at office and the GMAT preparation, fully tied up with all my free hours. Anyways, now I’m back...

Ah yes. Funemployment. The time between when you quit your job and when you start your MBA. The promised land that many MBA applicants seek. The break that every...

It is that time of year again – time for Clear Admit’s annual Best of Blogging voting. Dating way back to the 2004-2005 application season, the Best of Blogging...