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
help
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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
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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]
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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Statement 1: (y+2)!/x! = oddsana.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
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3