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# M22-33

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M22-33 [#permalink]
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reddyMBA wrote:
Here, if x=y (Question didn't say that x and y are different, so we can consider this use case as well), then option (D) becomes 0/2 =0, which is not an even integer. Can you please let me know your thoughts on this?

ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. Zero is neither positive nor negative (the only one of this kind).

4. Zero is divisible by EVERY integer except 0 itself ($$\frac{x}{0} = 0$$, so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer ($$x*0 = 0$$, so 0 is a multiple of any number, x).

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 ($$x^0 = 1$$)

9. $$0^0$$ case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0), $$0^n = 0$$.

11. If the exponent n is negative (n < 0), $$0^n$$ is undefined, because $$0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}$$, which is undefined. You CANNOT take 0 to the negative power.

12. $$0! = 1! = 1$$.
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Re: M22-33 [#permalink]
After initial analysis , answer choices narrowed down to A and D. The option "0" was the caveat in the choice A which made the only possible option is D.

Thanks Bunuel for clear explanation as always
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Re: M22-33 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M22-33 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M22-33 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M22-33 [#permalink]
What if x<y? In that case, x^2-y^2 will be negative and negative numbers cant be even? Please help
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Re: M22-33 [#permalink]
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rushikam wrote:
What if x<y? In that case, x^2-y^2 will be negative and negative numbers cant be even? Please help

Negative numbers can be even or odd.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder. So, ..., -4, -2, 0, 2, 4, ... are all even integers.

An odd number is an integer that is not evenly divisible by 2. So, ..., -3, -1, 1, 3, 5, ... are all odd integers.
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Re: M22-33 [#permalink]
For Option D: (x^2−y^2)/2
My thought process was as below
(x+y)(x-y)=even*even = even and even divided by 2 (even) can be even(4/2=2) or odd (10/2=5) so I rejected D. Could you please point out the flaw in this approach ?
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Re: M22-33 [#permalink]
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SA1196 wrote:
For Option D: (x^2−y^2)/2
My thought process was as below
(x+y)(x-y)=even*even = even and even divided by 2 (even) can be even(4/2=2) or odd (10/2=5) so I rejected D. Could you please point out the flaw in this approach ?

The point is that even*even is not only even but also a multiple of 4, and a multiple of 4 divided by 2 is even.
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Re: M22-33 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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M22-33 [#permalink]
For the Answer (D), ie (x^2+y^2)/2, if x is even and greater than 0, x^2 is greater than zero. If we take y=0, we would have (x^2-0)/2, which is an even number divided by 2, which would yield an odd number,

Am I mistaken?
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Re: M22-33 [#permalink]
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w0ng3r wrote:
For the Answer (D), ie (x^2+y^2)/2, if x is even and greater than 0, x^2 is greater than zero. If we take y=0, we would have (x^2-0)/2, which is an even number divided by 2, which would yield an odd number,

Am I mistaken?

It's always helpful to test your ideas with examples. For instance, is 4/2 an odd number?
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Re: M22-33 [#permalink]
KarishmaB gmatophobia , I am missing numbers while  using trial and error method for these kinda questions. Missing number/scenarios are the reasons why I am making mistakes on these questions. What are the best ways to approach these kinda questions and how to avoid avoid missing legit  scenarios ? Please help. egmat
Bunuel wrote:
If $$x$$ and $$y$$ are even integers, which of the following must also be an even integer?

A. $$x^y$$
B. $$2\frac{x}{y}$$
C. $$\frac{x - y}{x + y}$$
D. $$\frac{x^2 - y^2}{2}$$
E. $$(x + 1)(y - 1)$$

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Re: M22-33 [#permalink]
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