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Re: If x and y are positive integers is y odd? [#permalink]
04 Mar 2012, 23:37

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Expert's post

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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.

Re: If x and y are positive integers is y odd? [#permalink]
19 Aug 2013, 07:08

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.

Re: If x and y are positive integers is y odd? [#permalink]
19 Aug 2013, 07:17

Expert's post

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]
19 Aug 2013, 07:20

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]
10 Sep 2013, 11:38

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]
16 Sep 2013, 06:17

[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]
23 Dec 2013, 03:30

1

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 _________________

Either suffer the pain of discipline, or suffer the pain of regret.

If my posts are helping you show some love awarding a kudos

Re: If x and y are positive integers is y odd? [#permalink]
03 Feb 2014, 04:54

(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. _________________

Re: If x and y are positive integers is y odd? [#permalink]
13 Feb 2014, 03:40

1

This post received KUDOS

Expert's post

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

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.

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