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by sana.noor » Sat Nov 09, 2013 10:00 pm
If x and y are positive integers is y odd?

(1) (y+2)!/x! = odd
(2) (y+2)!/x! is greater than 2

OA is C
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by [email protected] » Sun Nov 10, 2013 1:34 am
Hi sana.noor,

While this is a crazy "looking" DS question, you can beat it by TESTing simple Values and/or using Number Properties:

We're told that X and Y are positive integers. We're asked if Y is odd? This is a YES/NO question.

Since this question deals with factorials, I'm going to write down the first few for easy reference:
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120

Fact 1: (Y+2)!/X! = ODD

Since the simplest odd number is 1, let's TEST Values:
X = 3
Y = 1
The answer to the question is YES

X = 4
Y = 2
The answer to the question is NO
Fact 1 is INSUFFICIENT

Fact 2: (Y+2)!/X! > 2

Since > 2 does NOT necessarily mean an integer, we have lots of options to TEST Values:
X = 1
Y = 1
The answer to the question is YES

X=1
Y=2
The answer to the question is NO
Fact 2 is INSUFFICIENT

Combined, we know that (Y+2)! is EVEN. For (Y+2)!/X! = ODD, we'd have...
Even/? = ODD

This means that the denominator must be EVEN and for (Y+2)!/X! = ODD INTEGER > 2
Since we're dealing with factorials, this will only happen under certain conditions (not all) when the numerator is 1 more than the denominator.
X = 2, Y = 1 and the answer is YES
X = 4, Y = 5 and the answer is YES
X = 6, Y = 7 and the answer is YES

Combined, SUFFICIENT

Final Answer: C

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by theCodeToGMAT » Sun Nov 10, 2013 1:35 am
To find: y is odd

Statement 1:
(y+2)!/x! = odd
=> (y+2)!/x!
==> (x+1)(x+2).....(y)(y+1)(y+2)
Since, the resultant is ODD... that means (x+1)(x+2).....(y)(y+1)(y+2) is ODD
We also know that
ODD x EVEN = EVEN
So, the resultant must contain only either (y+2) or "1"..
We cannot conclude whether "y" is ODD
INSUFFICIENT


Statement 2:
(y+2)!/x!
=> (x+1)(x+2).....(y)(y+1)(y+2)
the result can be ODD or EVEN
We cannot comment of any value
INSUFFICIENT

Combining...
Resultant is greater than "1"
So, "y" is ODD
SUFFICIENT

Answer [spoiler]{C}[/spoiler]
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by GMATGuruNY » Sun Nov 10, 2013 5:01 am
sana.noor wrote:If x and y are positive integers is y odd?

(1) (y+2)!/x! = odd
(2) (y+2)!/x! is greater than 2

OA is C
Statement 1: (y+2)!/x! = odd
Let (y+2)!/x! = 1, implying that y+2 = x.

If x=3, then y=1.
Result: y is ODD.

If x=4, then y=2.
Result: y is EVEN.
INSUFFICIENT.

Statement 2: (y+2)!/x! is greater than 2
If (y+2)!/x! = 3, then 3 is included in (y+2)! but not in x!, implying the following:
(y+2)! = 3*2*1.
x! = 2*1.
In this case, y+2 = 3, so y=1.
Result: y is ODD.

If (y+2)!/x! = 4, then 4 is included in (y+2)! but not in x!, implying the following:
(y+2)! = 4*3*2*1
x! = 3*2*1.
In this case, y+2 = 4, so y=2.
Result: y is EVEN.
INSUFFICIENT.

Statements combined:
If (y+2)!/x! = 3, then y is ODD, as shown above.

If (y+2)!/x! = 5, then 5 is included in (y+2)! but not in x!, implying the following:
(y+2)! = 5*4*3*2*1
x! = 4*3*2*1.
In this case, y+2 = 5, so y=3.
Result: y is ODD.

If (y+2)!/x! = 7, then 7 is included in (y+2)! but not in x!, implying the following:
(y+2)! = 7*6*5*4*3*2*1
x! = 6*5*4*3*2*1.
In this case, y+2 = 7, so y=5.
Result: y is ODD.

The cases above illustrate that y must be ODD.
SUFFICIENT.

The correct answer is C.
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