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For integers x, y, and z, x = y^2. What is the value of z? [#permalink]
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08 Apr 2014, 16:01
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For integers x, y, and z, x = y^2. What is the value of z? (1) x = z!(z−1)! (2) 12 < z < 22
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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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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)
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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 rewritten as \(x=y^2= z*(z1)!^{2}\) If x=1=y^2 then z!*(z1)! = 1 or z=1 Now x =y^2 so z*{(z1)!}^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
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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!(z1)! 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.

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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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08 Jul 2014, 06:14
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!(z1)! 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?
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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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08 Jul 2014, 06:56
Bunuel wrote: PathFinder007 wrote: Let me ask you a question. Is 1 the only possible value of x?
I got my misttake Thanks.

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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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Z = ?
Since x = y^2 we know that x is a perfect square.
Statement 1:
x = z!(z1)!
I think the hardest part of this statement is figuring out a number that satisfies the statement and is a perfect square.
Take z = 5. z!(z1)! = 5.4.3.2.1(4.3.2.1) which is not a perfect square. But, if we pick a perfect square for z we might be on to something.
z = 4.
4.3.2.1(3.2.1) is a perfect square!!! There, statement 1 is not sufficient as there are lots of different perfect squares!
Statement 2:
Not sufficient as z could be any of those numbers.
Statement 1 & 2:
Is sufficient as the only perfect square between 12 and 22 is 16. Answer C.

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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

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For integers x, y, and z, x = y^2. What is the value of z? [#permalink]
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15 Apr 2017, 14:00
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) Okay so I think I have a clearer understanding of this problem after reading some of the solutions here: It is important to pay attention to the fact that the problem reads ""integers" x,y and z etc..) knowing this it becomes clearer to see that x must be a perfect square because root x = y which must be an integer; therefore, "X" is a perfect square. Statement 1:If X is a perfect square then Z!(Z!) must be a perfect square. Though there are infinite possibilities for Z without a restriction this piece of data is insufficient. Insufficient. Statement 2: Provides us with a range for Z; however, with no established relationship between x and z we cannot, such as the equation listed in 1, we cannot solve for y. Insufficient. Statement 1 & Statement 2If we combine both elements, then we can solve the equation in statement 1 using the criteria in statement 2. Sufficient.

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