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# If X and Y are positive integers , is y odd ? (1)

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Joined: 25 Dec 2010
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If X and Y are positive integers , is y odd ? (1) [#permalink]

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02 Oct 2011, 15:02
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95% (hard)

Question Stats:

30% (03:06) correct 70% (02:50) wrong based on 40 sessions

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If X and Y are positive integers , is y odd ?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-integers-is-y-odd-128602.html
[Reveal] Spoiler: OA

Last edited by shashankp27 on 04 Oct 2011, 11:53, edited 1 time in total.
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Posts: 2011
Re: If X and Y are positive integers , is y odd ? (1) [#permalink]

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03 Oct 2011, 03:24
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shashankp27 wrote:
If X and Y are positive integers , is y odd ?

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

1)
x=4; y=2
x=3; y=1
Not Sufficient.

2)
x=1; y=1
x=1; y=2
Not Sufficient.

Combining both;
y = odd integer greater than 0;
x = y+1
Sufficient.

Ans: "C"
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Re: If X and Y are positive integers , is y odd ? (1) [#permalink]

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20 Nov 2016, 10:32
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If x and y are positive integers is y odd?

(1) (y+2)!/x! = odd. Notice that $$\frac{(y+2)!}{x!}=odd$$ can happen only in two cases:

A. $$(y+2)!=x!$$ in this case $$\frac{(y+2)!}{x!}=1=odd$$. For this case $$y$$ can be even: $$y=2=even$$ and $$x=4$$: $$\frac{(y+2)!}{x!}=\frac{24}{24}=1=odd$$;

B. $$y=odd$$ and $$x=y+1$$, in this case $$\frac{(y+2)!}{(y+1)!}=y+2=odd$$. For example, $$y=1=odd$$ and $$x=y+1=2$$: $$\frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd$$ (basically all the cases like: 3!/2!, 5!/4!, 7!/6!, ... --> y+2=odd, 3, 5, 7, ... --> y=odd, 1, 3, 5, ...).

Not sufficient.

(2) (y+2)!/x! is greater than 2. Clearly insufficient: consider $$y=1=odd$$ and $$x=2$$ OR $$y=2=even$$ and $$x=1$$.

(1)+(2) From (2) we cannot have case A, hence we have case B, which means $$y=odd$$. Sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-integers-is-y-odd-128602.html
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If X and Y are positive integers , is y odd ? (1) [#permalink]

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20 Nov 2016, 10:27
This is a great Question.
Testing our knowledge on factorial theory.

So we are given that both x and y are integers
And we need to get whether or not y is odd

Lets see
Statement 1
(y+2)!/x! => odd
99!/99!=> 1 => odd => y is odd
98!/98!=1=> odd=> y is even
Hence using test cases we can say that this statement is insufficient.
Lets look at statement 2
(y+2)!/x! is greater than 2
hmm
101!/2! is greater than 2=> y is 99 which is odd
100!/2! is also greater than 2=> y is 98 which is even
Hence insufficient
Combing the two statements
(y+2)!/x! to be a odd number >2
101!/100!=> y=99=> odd
100!/99!=> notes this is not odd
Hmm
Intersting
Here we know that every other number in the factorial series is even number.
so if y will be even then y+2 will be even too and no matter what value of x we take => (y+2)!/x! will always be even (constraints two the combining the two statements)
Hence y must be odd
Hence C

This Solution is pretty messed up for my liking.
I think Using test cases really helps in this one. Doesn't it
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Re: If X and Y are positive integers , is y odd ? (1) [#permalink]

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20 Nov 2016, 10:49
shashankp27 wrote:
If X and Y are positive integers , is y odd ?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-x-and-y-are-positive-integers-is-y-odd-128602.html

took me some time to test some additional values for x and y
1.
y=1, x=2 -> 1*2*3/1*2 = 3 - odd
y=2, x=4 -> 1*2*3*4/1*2*3*4 = 1 - odd
2 different options - not sufficient.

2.
y=1, x=2 -> 1*2*3/1*2 = 3 > 2
y=2, x=1 -> 1*2*3*4/1 = 24 > 2
2 outcomes - not sufficient.

1+2
must be odd, and greater than 2.
y=1, x=2
y=3, x=4
y=5, x=6

in all cases, y is odd.

Re: If X and Y are positive integers , is y odd ? (1)   [#permalink] 20 Nov 2016, 10:49
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