Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

(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.

Re: If x and y are positive integers is y odd? [#permalink]

Show Tags

19 Aug 2013, 08:08

Bunuel wrote:

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.

Answer: C.

Hope it's clear.

***********

Hi Bunuel,

**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, ...). **

You mentioned y=odd, 1, 3, 5, ...; and (1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient.

Are you considering y=1 as odd? can you explain how y=1 satisfies the solution.

(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.

Answer: C.

Hope it's clear.

***********

Hi Bunuel,

**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, ...). **

You mentioned y=odd, 1, 3, 5, ...; and (1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient.

Are you considering y=1 as odd? can you explain how y=1 satisfies the solution.

Well 1 is an odd number. If \(y=1=odd\) and \(x=y+1=2\), then \(\frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd\).
_________________

Re: If x and y are positive integers is y odd? [#permalink]

Show Tags

19 Aug 2013, 08:20

Bunuel wrote:

abhisheksriv85 wrote:

Bunuel wrote:

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.

Answer: C.

Hope it's clear.

***********

Hi Bunuel,

**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, ...). **

You mentioned y=odd, 1, 3, 5, ...; and (1)+(2) From (2) we cannot have case A, hence we have case B, which means \(y=odd\). Sufficient.

Are you considering y=1 as odd? can you explain how y=1 satisfies the solution.

Well 1 is an odd number. If \(y=1=odd\) and \(x=y+1=2\), then \(\frac{(y+2)!}{x!}=\frac{3!}{2!}=3=odd\).

Re: If x and y are positive integers is y odd? [#permalink]

Show Tags

10 Sep 2013, 12:38

1

This post received KUDOS

devinawilliam83 wrote:

If x and y are positive integers is y odd?

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

From (1): assume: x = y + 2, so (y + 2)!/x! = 1 = odd y can odd or even, so (1) Insufficient.

From (2): assume x = y + 1, so (y + 2)!/x! = y + 2 > 2 because y is a positive integer y can odd or even, so (2) Insufficient.

From (1) + (2): (y + 2)!/x! > 2 and (y + 2)!/x! = odd Assume: (y + 2)!/x! = 3, so (y + 2)! =x!*(2k + 1) with k >=1 If x = y + 2, so 1 = 2k + 1 (wrong because k >=1) If x = y + 1, so y + 2 = 2k + 1, therefore y = odd If x = y, so (y + 1)(y + 2) = 2k + 1 (wrong) because the product of 2 consecutive integers is even. If x = y - 1, so y(y + 1)(y + 2) = 2k + 1 (wrong) because the product is always even. Only one solution is right: If x = y + 1, so y + 2 = 2k + 1, therefore y = odd

Re: If x and y are positive integers is y odd? [#permalink]

Show Tags

16 Sep 2013, 07:17

[quote="Bunuel"]If x and y are positive integers is y odd?

I am sorry i don't want to sound dumb, i just want to know how you assumed the cases above?? also what did you mean by:"(1)+(2) From (2) we cannot have case A, hence we have case B, which means . Sufficient."

Re: If x and y are positive integers is y odd? [#permalink]

Show Tags

23 Dec 2013, 04:30

2

This post received KUDOS

skamran wrote:

also what did you mean by:"(1)+(2) From (2) we cannot have case A, hence we have case B, which means . Sufficient."

Will appreciate your help.

Let's see if I can help you out. This question involves the knowledge of few concepts.

1) The division of two factorial expressions has to be odd. When does it happen? since a factorial say y! is the product amongst y, (y-1), (y-2) and so on, this sequence will always contain a multiple of two unless y is 1 or 0 (remember 1! =1 and 0! =1). In our case x and y are positive integers thus the lowest value y can assume is 1. If y=1 then (y+2)!=3!. We will not face the zero factorial case thus we can safely get rid of it.

Back to our question: if we want to obtain an odd integer from \(\frac{(y+2)!}{x!}\) either

1. the two numbers have to be equal (entailing 1 as a quotient) EG

y=1 ---> (y+2)!=3! x!=3! - result is 1 = odd integer. In this case y is odd

y=2 (y+2)!=4! x!=4! - result is 1 = odd integer. In this case y is even.

2. or x! has to be one less than (y+2)! with y as an odd number.

EG y=3 (y+2)!=5!

x!=4! - 5! upon 4! yields 5 and y must be ODD.

Statement one is thus insufficient by itself.

2) this statement doesn't tell us anything interesting besides that x! must be different from (y+2)! y can assume both an even and an odd value. Not sufficient

1+2) Statement 2 conveys us that x! must be different from (y+2)! This said we can get rid of case one in our statement 1 analysis. Now we are sure that Y is an odd integer.

Sufficient _________________

learn the rules of the game, then play better than anyone else.

Re: If x and y are positive integers is y odd? [#permalink]

Show Tags

03 Feb 2014, 05:54

(1): \(\frac{(y+2)!}{x!}\) is odd. Clearly \(x<=(y+2)\).

Let x=y+2 then \(\frac{(y+2)!}{x!}\) =1 y can be even or odd. so (1) is insufficient

(2): \(\frac{(y+2)!}{x!} > 2\) so x has to be less than y+2

Let x=y+1 then

\(\frac{(y+2)!}{x!} > 2\)

is \(y+2>2\)

From this y can be any +ve integer. insufficient.

(1)+(2)

\(\frac{(y+2)!}{x!} > 2\) and also odd.

Continuing the analysis from above, if y+2>2 and y+2 is odd, then y has to be odd.

Next Let x=y

then \(\frac{(y+2)!}{x!} > 2\)

is (y+2)(y+1) > 2. Notice (y+2) & (y+1) are consecutive integers. their product will always be even. This contradicts (1)+(2) hence x cannot be equal to y.

similarly x cannot be equal to y-1 and so on.

x has to be equal to y+1. and y must be odd. We have a concrete answer, y is odd. (1)+(2) is sufficient. _________________

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

It's pretty easy to see the answer is (C) if you understand the concept of factorials well.

Note that a! = 1*2*3*4...(a-1)*a So a! is the product of alternate odd and even numbers. a!/b! will be an integer only if a >= b

(1) (y+2)!/x! = odd So y+2 >= x Two Cases: Case (i) If y+2 = x, (y+2)!/x! = 1 (odd). y could be odd or even. Case (ii) If y+2 > x, y+2 must be only one more than x. If y+2 is 2 more than x, you wil have two extra terms in the numerator and one of them will be even. So to ensure that (y+2)!/x! is odd, y+2 must be only 1 more than x and must be odd so all we are left in the numerator is (y+2) which must be odd. y must be odd. y could be odd or even so this statement alone is not sufficient.

(2) (y+2)!/x! is greater than 2 This means y+2 > x. But we have no idea about whether y is odd or even. Not sufficient.

Using both statements together, we know that only Case (ii) above is possible and hence y must be odd.

Case 1: (Y+2)!/X! = Odd If Y=3, then the above equation can only be odd when X= (Y+2)-1 which is 4. So you get 5!/4! =5. If Y=2, and X=4, 4!/4! = 1 which is also odd. Hence, Insufficient.

Case 2:(Y+2)!/X! is greater than 2 This holds for multiple values of Y and X. Hence clearly, Insufficient.

Together, if it is greater than 2 then it means Y has to be odd. Hence C!

_________________ If my post is helpful/correct, consider giving Kudos

Re: If x and y are positive integers is y odd? [#permalink]

Show Tags

07 Apr 2016, 11:53

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...