Data Sufficiency

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

GMAT assassins aren't born, they're made,
Rich

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.

GMAT assassins aren't born, they're made,
Rich