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

Question

(1) x = z!(z−1)!
(2) 12 < z < 22

KarishmaB · Accepted Answer

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)

WoundedTiger · Answer

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

PathFinder007 · Answer

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.

Bunuel · Answer

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

PathFinder007 · Answer

I got my misttake :oops:

Thanks.

Dirky · Answer

Z = ?

Since x = y^2 we know that x is a perfect square.

Statement 1:

x = z!(z-1)!

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!(z-1)! = 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.

deepak268 · Answer

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

Nunuboy1994 · Answer

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 2

If we combine both elements, then we can solve the equation in statement 1 using the criteria in statement 2. Sufficient.

NigerianBankScammer · Answer

I don't get it. If we work backwards, we can make any number work with this. Pick 18 for Z.
18!*17! =355,687,428,096,000 = x
Let that number be x

Square that and you get y.

Why isn't the answer E?

amanvermagmat · Answer

Hi

Square of x is NOT y 
Its given that square of y is x... so x must be a perfect square

CAMANISHPARMAR · Answer

The key point for this question is to realize that z needs to be a perfect square and when you realize that this question becomes very easy!

Option C is the correct answer.

TheUltimateWinner · Answer

Hi  Bunuel 
What if the word "integers" is removed from question stem?
Thanks__

Posted from my mobile device

Basshead · Answer

(1) We can rewrite statement 1 as:

x = z * (z-1)! * (z-1)! 
 x = z * (z-1)!^2

Since  x = y^2 , z must be a perfect square.

However, we can't determine the value of z; INSUFFICIENT.

(2) Clearly insufficient.

(1&2) Combined, there is only one perfect square between 12 and 22. z must be 16. SUFFICIENT.

Answer is C.

dave13 · Answer

VeritasKarishma for ST 1 when you write x = z * (z - 1)! * (z - 1)! = z * (z - 1)!^2 (z - 1)!^2 is perfect square but Z itself is z

following this pattern x = z * (z - 1)! * (z - 1)! = z * (z - 1)!^2 then i can assume that Z! can be equal 17! since 17!*16! so here 16 is perfect square but it is not Z, cause Z is 17! what am i missing

KarishmaB · Answer

When z = 17,

x = 17! * 16!
For x to be a perfect square, all its prime factors should have an even power.

We know that 17! = 17 * 16 * 15 * ....1 so we can write 17! as 17 * 16! too.

x = 17 * 16! * 16!

Now we have two 16! so we know that they are ok to make a perfect square. But we have another 17 which is not a perfect square. So x is not a perfect square.

Instead, if z were 16,

x = 16! * 15! = 16 * 15! * 15!
So we have two 15! which is fine. Also 16 = 4 * 4

x = 4 * 4 * 15! * 15! = (4 * 15!)^2 
Here x is a perfect square.

ocelot22 · Answer

There ought to be a subtype called "abstraction in question stem " subtype. whenever you see some serious math in the question stem, try some simplified versions of what is going on to try to uncover a pattern. Thats what the test taker is really wanting you to do

notice x = y^2 is there to tell us x is a perfect square.

(1)
Ok lets try some small values of z

if z = 3, Z!(z-1)! = 3!2! = 3x2x2 = 3*2^2
if z = 4, Z!(z-1)! = 4*3*2*3*2=4(2*3)^2 since 4 is perfect square, we have perfect square
if z=5 ................5*4*3*2*1*4*3*2*1 = 5(4*3*2*1)^2
if z = 6.............................................6(5*4*3*2*1)^2
if z = 7...........................................7(6*5*4*3*2)^2
if z = 8 ..........................................8(7*6*5*4*3*2)^2
if z = 9..........................................9(8*7*6*5*4*3*2)^2 a perfect square so we can notice two things fronm this little exercise

1) z!(z-1)! = z((z-1)!)^2)
2) n!(n-1)! will only be a perfect square only if z is a perfect square (1) is clearly NS

(2)
12<z<22 clearly NS

(1) and (2) Ok, between 12 and 22 there is only one perfect square. that number if 16. So, we are sufficient

OA is C

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

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