If x and y are positive integers is y odd?

very good explanation

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.

[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."

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

(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.

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

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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