If x and y are integers, is the sum of x and y even?

Hi lipsi18,

In the statement 1, either of (x+y) or (x-y) can be zero.
Hence we cannot say that x + y = 0

Is it B ?
If so, will provide my line of reasoning.

Hi subhamoycomputer,

The correct option here should be D.
However do post your reasoning as to why you thing option A is insufficient.

If x and y are integers, is the sum of x and y even?

(1) x^2 = y^2
(2) x^3 = y^3

x + y even?

Statement 1:

x^2 = y^2. This implies x = y or x =-y.

Either way x+y will be even (0 or 2x).

Statement 2:

X^3 = y^3.

Implies x = y.

So x + y is 2x, which is even.

So each statement is individually sufficient.

Answer D.

Here in both the cases we can say that both x and y must have the same odd/even nature
hence x+y=even
hence D

If x and y are integers, is the sum of x and y even?

(1) x^2 = y^2
(2) x^3 = y^3

There are 2 variables (x and y) in the original condition. In order to match the number of variables to the number of equations, we need 2 more equations. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that C is the correct answer. Using both the condition 1) and the condition 2), we get 1)=2), from which we can obtain x=y=odd or x=y=even. The answer is always yes and the conditions are sufficient. Therefore, the correct answer is D.

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

The Answer is D..Check for pairs (0,0) ...(1,1)....(-1,-1)

You are correct...
But for (1), check for |x|=|y|.. May be important in some other Q
x=1, y=-1 ; x=-1, y=1; x=1,y=1; x=-1, y=-1

Let me understand, is 0 considered even?

Because if we use x = -y
Then the answer would be 0 for option A

hi,
0 is EVEN but it is neither NEGATIVE nor POSITIVE...

Bunuel gmatbusters VeritasKarishma chetan2u

Could you share your approach?

Q is asking: Are both x and y even or odd?

For St 1, I took |x| = |y|
which is same as:
x = y and x = -y
or
x-y = 0
or
x + y = 0

Hence implying, both x and y have to be even/odd.

What is St 2 saying? How do I interpret odd powers of integer?

1) x²=y²
So, x²-y²=0
So, (x+y)(x-y)=0
Either x+y=0 where 0 is even.
Or x-y=0 where 0 is even. If the difference of two integers is even, their sum is also even.

So statement (1) is sufficient.

(2) Cube of a number preserves the sign.
So x³=y³ means x=y, say x=y=a
So the sum will always be even =2a.

So statement (2) is sufficient.

Both are statements alone are sufficient so option D is the correct answer

Posted from my mobile device

Good news, odd power are even easier to handle

Hence for Statement 2, sum of x and y = 2x or 2y = even

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