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Integers ..

by bubbliiiiiiii » Wed Aug 26, 2015 12:56 am
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.

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by GMATGuruNY » Wed Aug 26, 2015 3:39 am
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.
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by [email protected] » Wed Aug 26, 2015 9:08 am
Hi bubbliiiiiiii,

This prompt can be solved by TESTing VALUES (as Mitch has showcased). There are some Number Properties built into this prompt though; recognizing those patterns can help make the work go a little faster...

1) Since we're told that Y is a POSITIVE INTEGER, (Y+2)! will ALWAYS be EVEN....

For example:
Y = 1, (Y+2)! = 3! = 6
Y = 2, (Y+2)! = 4! = 24
Y = 3, (Y+2)! = 5! = 120
Etc.

Since the numerator of those fractions will always be even, you can focus on what the denominator would have to be to satisfy each of the two given Facts.

In Fact 1, to end up with an ODD INTEGER, the only possible option is to make the fraction equal 1 - so the denominator MUST be equal the numerator.

In Fact 2, to end up with a number GREATER than 2, the numerator has to be MORE than double the denominator (which won't be difficult to accomplish, since we're dealing with factorials - there are limitless possibilities).

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by Anaira Mitch » Wed Feb 22, 2017 7:22 pm
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.
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.
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by GMATGuruNY » Thu Feb 23, 2017 3:49 am
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.
When the statements are combined, (y+2)!/x! = an odd integer greater than 2.
Cases 2, 4 and 5 satisfy this constraint.
In each of these cases, solving for x and y yields an odd integer value for y.
Implication:
When the statements are combined, y must be odd.
Thus, the answer to the question stem is YES, and the two statements combined are SUFFICIENT.
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by Matt@VeritasPrep » Thu Mar 02, 2017 5:26 pm
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.
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by Matt@VeritasPrep » Thu Mar 02, 2017 5:27 pm
My approach is probably more conceptual than you'd want to get under test conditions, though. Try a few numbers and you can reach the cases easily (and more quickly) through trial and error.
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