Advanced Quant Strategy Guide Ques 9 Chapter 3

[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

[jnelson0612]

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

Sure, just plug in numbers! Note that the (y+2)! is always going to be even; we need to play with the x! to that (y+2)!/x! is always odd.

Again, the statement says that (y+2)!/x! is an odd number. We must find numbers that make this statement true (and that are positive integers as stated in the problem stem.

If y=1, then (1+2)! is 3!, or 6. If x=2, then 2!=2. Thus 6/2 is 3, fitting the statement. In this case y is odd.

If y=2, then (2+2)! is (4)!, or 24. Let's say that x=4, thus 24/4! is 24/24 or 1, fitting the statement. In this case y is even