Even/Odd Integers

[bestofbala]

For integers x and y is x+y Even?

(1) x*x - y*y is even
(2) x-y is even

I came across this question today and I am not convinced with the answer in the explanation. I want to check with you colleagues.

[Anju@Gurome]

For integers x and y is x+y Even?

(1) x*x - y*y is even
(2) x-y is even

For any two integers x and y,

• If (x - y) is even, then (x + y) is also even.
  If (x - y) is odd, then (x + y) is also odd.

Statement 1: x² - y² = (x - y)(x + y) is even.
Hence, either (x - y) or (x + y) or both are even.
In any case, (x + y) is even.

Sufficient

Statement 2: As (x - y) is even, (x + y) must be even too.

Sufficient

The correct answer is D.

[bestofbala]

Sorry, I am not understanding the explanation for Statement 1.

If (x+y)(x-y) is Even, 3 cases are possible: Both (x+y) and (x-y) are Even, (x+y) alone is Even and (x-y) alone is Even.

Isn't this giving a chance for (x+y) to be both Even and Odd. How come this information is sufficient to answer (x+y) is Even?

[Anju@Gurome]

If (x - y) is even then (x + y) will be even and vice versa. This is true in general for any integers x and y.

Let's see why...
If (x - y) is even, then either of the following cases are possible...

• 1. x and y both even
  2. x and y are both odd

In any of the above cases, (x + y) will be even as (even + even) = even and (odd + odd) = even.

Sorry, I am not understanding the explanation for Statement 1.

If (x+y)(x-y) is Even, 3 cases are possible: Both (x+y) and (x-y) are Even, (x+y) alone is Even and (x-y) alone is Even.

Isn't this giving a chance for (x+y) to be both Even and Odd. How come this information is sufficient to answer (x+y) is Even?

You're correct that those 3 scenarios are possible (blue line).
In the first two cases, (x + y) is even so there should not be any confusion.
And in the third case, as (x - y) is even, (x + y) will be even too as I described previously. It is not possible that (x - y) is even but (x + y) is odd.

Hence, in all of the three possible cases, (x + y) will be even.

Hope that helps.

[bestofbala]

Thanks, I now see that I've not seen enough to eliminate the impossible scenario.

[Brent@GMATPrepNow]

For integers x and y is x+y Even?
(1) x*x - y*y is even
(2) x-y is even

Knowing the various rules (as Anju has demonstrated) is the best approach to these kinds of questions. However, if you don't know how to proceed in that manner, you can always test possible cases. This is more time-consuming, but still manageable within a 2-minute time frame.

For each statement, we'll examine the 4 different cases involving x and y. They are:
case a) x is even and y is even
case b) x is even and y is odd
case c) x is odd and y is even
case d) x is odd and y is odd

Target question: Is x+y even?

Statement 1: x^2 - y^2 is even
Let's examine x^2 - y^2 via the 4 cases. To do this, we'll plug in a 2 for an even number and a 1 for an odd number
case a) x is even and y is even: x^2 - y^2 = 2^2 - 2^2 = 0 (EVEN)
case b) x is even and y is odd: x^2 - y^2 = 2^2 - 1^2 = 3 (ODD)
case c) x is odd and y is even: x^2 - y^2 = 1^2 - 2^2 = -3 (ODD)
case d) x is odd and y is odd: x^2 - y^2 = 1^2 - 1^2 = 0 (EVEN)

Statement 1 tells us that x^2 - y^2 is EVEN. So, that means either case a is true or case d is true.
If case a is true, then x and y are both even, which means x+y is even.
If case d is true, then x and y are both odd, which means x+y is even.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x-y is even
Let's examine x-y via the 4 cases. To do this, we'll plug in a 2 for an even number and a 1 for an odd number
case a) x is even and y is even: x-y = 2-2 = 0 (EVEN)
case b) x is even and y is odd: x-y = 2-1 = 1 (ODD)
case c) x is odd and y is even: x-y = 1-2 = -1 (ODD)
case d) x is odd and y is odd: x-y = 1-1 = 0 (EVEN))

Statement 2 tells us that x-y is EVEN. So, that means either case a is true or case d is true.
If case a is true, then x and y are both even, which means x+y is even.
If case d is true, then x and y are both odd, which means x+y is even.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = D

If you're interested, we have a free video that further explains this technique of testing cases: https://www.gmatprepnow.com/module/gmat- ... ies?id=839

Cheers,
Brent

[GMATGuruNY]

For integers x and y is x+y Even?

(1) x² - y² is even
(2) x-y is even

Every test-take should know the following:
even+even = even.
even-even = even.
odd+odd = even.
odd-odd = even.
These are the ONLY ways to get an even result.
If one of the terms is even while the other is odd, the resulting sum or difference will be ODD.

Statement 1: x² - y² is even
Case 1: x² - y² = even-even = even.
Options for x² and y²:
...(-8)², (-6)², (-4)², (-2)², 0², 2², 4², 6², 8²...
Thus, options for x and y:
-8,- 6, -4, -2, 0, 2, 4, 6, 8...
Since all of the values here are EVEN, x+y = even+even = even.

Case 2: x² - y² = odd-odd = even.
Options for x² and y²:
...(-7)², (-5)², (-3)², (-1)², 1², 3², 5², 7²...
Thus, options for x and y:
...-7, -5, -3, -1, 1, 3, 5, 7...
Since all of the values here are ODD, x+y = odd+odd = even.
SUFFICIENT.

Statement 2: x-y = even.
Case 1: x-y = even-even = even, in which case x+y = even+even = even.
Case 2: x-y = odd-odd = even, in which case x+y = odd+odd = even.
SUFFICIENT.

The correct answer is D.