bubbliiiiiiii wrote:If x and y are positive integers, is y odd?
(1) (y+2)!/x! is an odd integer.
(2) (y+2)!/x! is greater than 2.

GMATGuruNY wrote:

bubbliiiiiiii wrote:If x and y are positive integers, is y odd?
(1) (y+2)!/x! is an odd integer.
(2) (y+2)!/x! is greater than 2.

Statement 1:
Case 1: (y+2)!/x! = 1, implying that (y+2)! = x!.
It's possible that x=3 and y=1, with the result that y is ODD.
It's possible that x=4 and y=2, with the result that y is EVEN.
INSUFFICIENT.

Statement 2: (y+2)!/x! > 2
Case 2: (y+2)!/x! = 3, implying that (y+2)! = 3x!.
Here, x=2 and y=1, with the result that both sides of the equation = 3!.
In this case, y is ODD.

Case 3: (y+2)!/x! = 4, implying that (y+2)! = 4x!.
Here, x=3 and y=2, with the result that both sides of the equation = 4!.
In this case, y is EVEN.
INSUFFICIENT.

Statements combined:
Case 2 satisfies both statements.
In Case 2, y is ODD.

Case 4: (y+2)!/x! = 5, implying that (y+2)! = 5x!.
Here, x=4 and y=3, with the result that both sides of the equation = 5!.
In this case, y is ODD.

Case 5: (y+2)!/x! = 7, implying that (y+2)! = 7x!.
Here, x=6 and y=5, with the result that both sides of the equation = 7!.
In this case, y is ODD.

In every case that satisfies both statements, y is ODD.
SUFFICIENT.

The correct answer is C.

Anaira Mitch wrote:Thanks for your solution. I understand why statement 1 and statement 2 are insufficient but I am unable to understand how case 2 satisfies both statements. Please help me understand how both statements together are sufficient.

Anaira Mitch wrote:

GMATGuruNY wrote:

bubbliiiiiiii wrote:If x and y are positive integers, is y odd?
(1) (y+2)!/x! is an odd integer.
(2) (y+2)!/x! is greater than 2.

Hell Mitch,

Thanks for your solution. I understand why statement 1 and statement 2 are insufficient but I am unable to understand how case 2 satisfies both statements. Please help me understand how both statements together are sufficient.

We know that any factorial greater than 1! will be even, since it will have an even integer in it somewhere.

With that in mind, let's say that (y + 2)!/x! = n, where n is some integer. We can then say

(y + 2)! = x! * n

Since y ≥ 1, we know (y + 2) ≥ 3, which means the left side is even. That means the right side is also even, since it equals the left side.

We also can see that multiplying x! by n gives us another factorial. That means we have a few possibilities:

1:: n = 1. In that case, (y + 2)! = x!, and they could be any positive integers, odd or even.

2a:: n > 1 and (y + 2) > (x + 1). In that case, (y + 2)! = x! * (y + 2) * (y + 1) * ... * (x + 1), since n must contain all the missing numbers that get us from x! to (y + 2)!. For instance, we could have (y + 2)! = 6! and x! = 4!. Then we know 6! = 4! * 6 * 5, or (y + 2)! = x! * (y + 2) * (x + 1). Notice that this will give us (y + 2)!/x! = (y + 2) * ... * (x + 1), which MUST BE EVEN since we'll have more than one consecutive integer being multiplied on the right hand side.

2b: n > 1 and (y + 2) = (x + 1). Then we've got (y + 2)!/x! = (x + 1). (x + 1) could be even or odd.

S1 eliminates case 2a. S2 eliminates case 1. Together, we've only got case 2b. We're told that (x + 1) is odd, and since (x + 1) = (y + 2), (y + 2) is also odd, and y itself is odd.