If x and y are positive integers is y odd?

Question

(1)  \frac{(y+2)!}{x!}  is an odd integer.

(2)  \frac{(y+2)!}{x!}  is greater than 2.

Bunuel · Accepted Answer

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.

KarishmaB · Answer

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.

Answer (C)

harikris · Answer

very good explanation

dngboy89nh · Answer

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

Answer is C.

skamran · Answer

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."

Will appreciate your help.

gmat6nplus1 · Answer

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

gmatprav · Answer

(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.

rreiki · Answer

hello,
this is the first time I am posting. i attempted the problem twice and got it wrong both times, even after I went through the explanations, i still couldn't explain it to myself. quite literally, after many hours, this is what I think.

(1) the result is odd, so numerator and denominator have to be the same. y can be odd or even. hence, not sufficient.
(2) numerator has to be greater than denominator. y can be odd or even. hence not sufficient.

combining both conditions, result has to be odd, but if y is even, numerator is even, then result will always be even. so y is odd.

maybe I've over simplified this. after attempting many questions and getting quite a few wrong even after studying concepts, I forced myself to just spend a lot of time on one question till I got it. this is what I'll do for every wrong answer.

please do let me know if this does not make sense.

cheers

ScottTargetTestPrep · Answer

Solution:
 
 Statement One Only:

(y + 2)!/x! is an odd integer.

If y + 2 = x, then (y + 2)!/x! = 1, which is odd. Since x is 2 more than y, both x and y have the same parity, i.e., they are both odd or they are both even. Since y can be either odd or even, statement one alone is not sufficient.

Statement Two Only:

(y + 2)!/x! is greater than 2.

If y + 1 = x, then (y + 2)!/x! > 2. For example, if y = 1 and x = 2, 3!/2! = 3 > 2; if y = 2 and x = 3, 4!/3! = 4 > 2. Since x is 1 more than y, both x and y have the opposite parity, i.e., x is odd and y is even OR x is even and y is odd. Since y can be either odd or even, statement two alone is not sufficient.

Statements One and Two Together:

Using the two statements, we have: (y + 2)!/x! is an odd integer greater than 2. In that case, y + 2 > x. However, if y + 2 is at least 2 more than x, i.e., y + 2 ≥ x + 2 or y ≥ x, then (y + 2)!/x! will always be an even integer. For example, if y = 3 and x = 3, 5!/3! = 5 * 4 is even; if y = 5 and x = 4, 7!/4! = 7 * 6 * 5 is even. As you can see, when y ≥ x, there will always be an even factor left after some of the factors of (y + 2)! cancel out with all the factors of x!. Therefore, in order for (y + 2)!/x! to be an odd integer, y + 2 must be 1 more than x, i.e., y + 2 = x + 1 or y + 1 = x. Furthermore, x must be even. For example, if x = 2, y = 1 and (y + 2)!/x! = 3!/2! = 3; if x = 4, y = 3 and (y + 2)!/x! = 5!/4! = 5. As we can see, when x = y + 1 and x is even, only the factor y + 2 is left when all the other factors of (y + 2)! cancel out with all the factors of x!. Since the factor y + 2 is odd, y is also odd. The two statements together are sufficient.

Answer: C

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If x and y are positive integers is y odd?

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