value of z

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value of z

by jain2016 » Tue Mar 15, 2016 7:10 am
For all integers x,y, and z, x=y^2. What is the value of z?

1) x = z!(z-1)!

2) 12<z<22

OAC

Hi Experts ,

Please explain.

Many thanks in advance.

SJ

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by [email protected] » Tue Mar 15, 2016 9:13 am
Hi jain2016,

This question can be solved with a mix of TESTing VALUES and 'brute force.'

We're told that X, Y and Z are all INTEGERS and that X = Y^2. That last piece of information is interesting - it proves that X can ONLY be a PERFECT SQUARE. We're asked for the value of Z.

1) X = Z!(Z-1)!

Since X can only be a perfect square, there are only certain values of Z that will fit this equation.

IF....
Z = 1
1!(0)! = 1
X = 1, which is a perfect square
The answer to the question is 1

IF....
Z = 4
4!(3)! = 144
X = 144, which is a perfect square
The answer to the question is 4
FACT 1 is INSUFFICIENT

Notice how in these two TESTs, Z had to be a PERFECT SQUARE!!! That pattern will come in handy later on.

2) 12 < Z < 22

With this Fact, Z could be any integer from 13 to 21, inclusive.
FACT 2 is INSUFFICIENT

Combined, we know
Z is a perfect square
12 < Z < 22

The ONLY value of Z that 'fits' both Facts is 16
Combined, SUFFICIENT

Final Answer: C

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by jain2016 » Tue Mar 15, 2016 9:41 am
Combined, we know
Z is a perfect square
12 < Z < 22

The ONLY value of Z that 'fits' both Facts is 16
Combined, SUFFICIENT
Hi Rich ,

Many thanks for your reply.

One thing is that if we solve the above part , then this will take a long time because we have to check from 13 to 21 inclusive.

Can you please explain how we can manage to do.

Please explain.

Many thanks in advance.

SJ

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by John fran kennedi » Tue Mar 15, 2016 11:14 am
Hi guys,
how did you get 144? did you get it by multiplying 4 by 3, then square its result?

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by [email protected] » Tue Mar 15, 2016 3:33 pm
Hi jain2016,

To save time on this question, you'll have to recognize the patterns that I pointed out in my original post (re: X is a PERFECT SQUARE; with Fact 1, Z MUST be a PERFECT SQUARE). If you don't recognize those patterns, then this question would take a long time to solve.

The reason why Z must be a perfect square takes a bit of an explanation:

Since X = Y^2 and we were told that X, Y and Z are INTEGERS, whatever value Y equals leads to an X that is a perfect square. For example, if Y=2, then X=4.

In Fact 1, we're told that X = Z!(Z-1)!....

Perfect Square = Z!(Z-1)!

By definition, a perfect square is a number that has factors that are 'paired.' For example: 25 = (5)(5) and 225 = (3)(3)(5)(5).

Z! will have all the same 'pieces' as (Z-1)!.... EXCEPT for the Z itself.

For example... IF Z = 4.... 4!(3)! = (4)(3)(2)(1)(3)(2)(1). Notice how there are two 3s, two 2s and two 1s? There's just one 4, but that 4 can be rewritten as (2)(2). So each factor comes paired. Taking this logic further, the ONLY way for Z!(Z-1)! to be a perfect square is if Z is a perfect square (so we could break the Z down into a factor that shows up paired).

This logic works with much larger numbers too, BUT they have to be perfect squares (re: Z = 9, 16, 25, etc.). NO other integers will work here. Once you have that deduction, combining Facts won't require much additional work at all - the ONLY perfect square in that range is 16.

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by Matt@VeritasPrep » Thu Mar 17, 2016 10:47 pm
It's probably easiest to factor it algebraically:

x = z! * (z - 1)!

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

x = z * (z-1)!²

Since we also know that x = y², we now have

y² = z * (z - 1)!²

Since the left side is a perfect square, the right side is also a perfect square. But this is only possible if z and (z-1)!² are BOTH perfect squares. So S1 tells us z is a perfect square, and we're pretty close to the answer with just that statement. Adding the second, we realize that z = 16 is the only square in that range, and S1+S2 together solve the problem.