Advanced Quant. In Action Problems Chapter 3 Question 9 Page

[vashist.vikas]

Dear Manhattan GMAT Team,
Could please clarify this question:

If x and y are positive integers, is y odd?
a) (y+2)!/x! is an odd number
b) (y+2)!/x! is greater than 2

I do not understand the solution of the problem, as I thought 1 was sufficient and 2 was not. When the solution said 2!/2! = 1 = odd, I got really confused! For this to happen y needs to be 0 which clearly violates the positive integer rule.

Can you tell me how y+2 might be either even or odd. I reckon it will always be odd for statement 1 to hold true. In addition x will have to be one less than y+2.

Cheers

Vikas

[thanghnvn]

is this gmatprep question? pls, check

[vashist.vikas]

My sincere apologies. I should have put this questions in the Math Strategy Guide Folder.

Cheers

Vikas

[RonPurewal]

Can you tell me how y+2 might be either even or odd. I reckon it will always be odd for statement 1 to hold true. In addition x will have to be one less than y+2.

not if the quotient is allowed to be 1.
for instance:
y = 3, x = 5 --> (3 + 2)!/5! = 5!/5! = 1
y = 4, x = 6 --> (4 + 2)!/6! = 6!/6! = 1
insufficient. in fact, x and y can be any two integers that are separated by 2 in this case.

this is admittedly somewhat of a tricky exception -- but you absolutely should be thinking about it, because of what's in the second statement.
in other words, a glance at the second statement shows that its purpose is to eliminate 1 from the possibilities. therefore, when you analyze the first statement, you should think very carefully about what happens if 1 is a possibility.

[rudransh]

Can you tell me how y+2 might be either even or odd. I reckon it will always be odd for statement 1 to hold true. In addition x will have to be one less than y+2.

not if the quotient is allowed to be 1.
for instance:
y = 3, x = 5 --> (3 + 2)!/5! = 5!/5! = 1
y = 4, x = 6 --> (4 + 2)!/6! = 6!/6! = 1
insufficient. in fact, x and y can be any two integers that are separated by 2 in this case.

this is admittedly somewhat of a tricky exception -- but you absolutely should be thinking about it, because of what's in the second statement.
in other words, a glance at the second statement shows that its purpose is to eliminate 1 from the possibilities. therefore, when you analyze the first statement, you should think very carefully about what happens if 1 is a possibility.

Friends,

so,what is the OA, IMO it is E. Below is my explanation

odd / odd = odd or even / even = odd

stmt 1 \ or \ stmt 2 \ both,
(y+2)! in numerator could be even or odd.
If (y+2)! is odd then each element in the factorial is odd thus Y is odd
if (y+2)! is even then element Y could be even or odd

so not suff E

[RonPurewal]

i'm thinking this should be (c).

these are the only ways in which the quotient of two factorials can be odd:
1/ the two factorials can be exactly the same (e.g. 6!/6!), in which case the quotient = 1 = odd.
2/ the two factorials can be 1 integer apart -- with (odd number)! on top and (even number)! on the bottom.
for instance, 7!/6! is 7. 9!/8! is 9. etc.

this can't happen in any other way -- if you have factorials that are more than 1 apart, then their quotient will be even. (try a few to see why.)

Below is my explanation

odd / odd = odd or even / even = odd

two things very, very wrong here.

first, odd/odd is impossible in the situation at hand. ALL factorials except 0! = 1! = 1 are even, so the factorial (y+2)!, where y itself is positive, must be even.

second, these statements are false: even/even doesn't have to be even. try it with simple numbers: 8/4 = 2 = even, but 12/4 = 3 = odd.
using factorials as required, e.g. 7!/6! = 7 (even/even = odd), but also 8!/7! = 8 (even/even = even).

If (y+2)! is odd then each element in the factorial is odd thus Y is odd

this is impossible. if you don't see why, write out a few factorials of odd numbers (e.g. 5! = 1*2*3*4*5, which is most definitely not odd) and look at the numbers that appear in them.

[m.francis.carver]

I think the most confusing part about the explanation is that, while (y+2)!/x! could be 1 and invalidate the assumption that x=(y+2)-1, using 2!/2! as a test case is invalid since this would require y=0 which is prohibited by the stem.

Just a poor choice of a test case in the explanation despite the fact that the premise that (y+2)=x still holds when (y+2)>2.

It took me quite a while to figure this out. MGMAT should remove the 2!/2! case (use 3!/3! and 4!/4! instead) from this explanation since it violates the stem.

[jnelson0612]

I think the most confusing part about the explanation is that, while (y+2)!/x! could be 1 and invalidate the assumption that x=(y+2)-1, using 2!/2! as a test case is invalid since this would require y=0 which is prohibited by the stem.

Just a poor choice of a test case in the explanation despite the fact that the premise that (y+2)=x still holds when (y+2)>2.

It took me quite a while to figure this out. MGMAT should remove the 2!/2! case (use 3!/3! and 4!/4! instead) from this explanation since it violates the stem.

Yes, I agree. 2!/2! does indeed violate the restrictions placed in the stem.

[ChelseaW930]

The solution in the 2nd edition says the following:

3!/3! = 1 = odd and y = 0

??? If (y+2)! = 3!, shouldn't y = 1 and not 0?

[RonPurewal]

i will pass this on to the team that edits this book. thank you.