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# x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of

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x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of [#permalink]

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23 Nov 2016, 05:44
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x and y are integers. $$\frac{(x^2y−1)}{2}$$ has a non-zero remainder. Which of the following must be true?

I. y is odd
II. x is even
III. xy is even

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III
[Reveal] Spoiler: OA

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x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of [#permalink]

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23 Nov 2016, 06:20
Bunuel wrote:
x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of the following must be true?

I. y is odd
II. x is even
III. xy is even

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

IMO C
[(x^2)y−1)/2] has a non-zero remainder means this expression is odd.
for this expression to be odd x^2*y has to be even
xy= even must be true
y and x can be be even or odd as long as x^2*y is even

Changing my answer as my interpretation of the equation was wrong

Last edited by Vinayak Shenoy on 23 Nov 2016, 07:50, edited 2 times in total.

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x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of [#permalink]

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23 Nov 2016, 07:14
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Bunuel wrote:
x and y are integers. $$\frac{(x^2y−1)}{2}$$ has a non-zero remainder. Which of the following must be true?

I. y is odd
II. x is even
III. xy is even

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

Given: (x²y - 1) divided by 2 has a non-zero remainder.
If (x²y - 1) DID have a zero remainder, then (x²y - 1) would be divisible by 2 (aka EVEN)
Since (x²y - 1) DOES NOT have a zero remainder, then (x²y - 1) is NOT divisible by 2 (aka ODD)
So, x²y - 1 is ODD
This tells us that x²y is EVEN
x²y will be even IF x is even, y is even, or x and y are both even
The question asks, "Which of the following MUST be true?"

I. y is odd. y COULD be odd, but it doesn't have to be. (for example, if x = 1 and y = 2, x²y is still EVEN)
II. x is even. x COULD be even, but it doesn't have to be. (for example, if x = 2 and y = 2, x²y is still EVEN)
III. xy is even. Since at least one of the variables must be even, it MUST be the case that xy is even

[Reveal] Spoiler:
C

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Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of [#permalink]

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25 Nov 2016, 08:04
GMATPrepNow wrote:
Bunuel wrote:
x and y are integers. $$\frac{(x^2y−1)}{2}$$ has a non-zero remainder. Which of the following must be true?

I. y is odd
II. x is even
III. xy is even

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

Given: (x²y - 1) divided by 2 has a non-zero remainder.
If (x²y - 1) DID have a zero remainder, then (x²y - 1) would be divisible by 2 (aka EVEN)
Since (x²y - 1) DOES NOT have a zero remainder, then (x²y - 1) is NOT divisible by 2 (aka ODD)
So, x²y - 1 is ODD
This tells us that x²y is EVEN
x²y will be even IF x is even, y is even, or x and y are both even
The question asks, "Which of the following MUST be true?"

I. y is odd. y COULD be odd, but it doesn't have to be. (for example, if x = 1 and y = 2, x²y is still EVEN)
II. x is even. x COULD be even, but it doesn't have to be. (for example, if x = 2 and y = 2, x²y is still EVEN)
III. xy is even. Since at least one of the variables must be even, it MUST be the case that xy is even

[Reveal] Spoiler:
C

Hello,

Would it not be sufficient for x to be even? If x is even, wouldn't x^2y be even regardless of y?

Thanks!

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Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of [#permalink]

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25 Nov 2016, 08:21
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FransmanPL wrote:
GMATPrepNow wrote:
Bunuel wrote:
x and y are integers. $$\frac{(x^2y−1)}{2}$$ has a non-zero remainder. Which of the following must be true?

I. y is odd
II. x is even
III. xy is even

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

Given: (x²y - 1) divided by 2 has a non-zero remainder.
If (x²y - 1) DID have a zero remainder, then (x²y - 1) would be divisible by 2 (aka EVEN)
Since (x²y - 1) DOES NOT have a zero remainder, then (x²y - 1) is NOT divisible by 2 (aka ODD)
So, x²y - 1 is ODD
This tells us that x²y is EVEN
x²y will be even IF x is even, y is even, or x and y are both even
The question asks, "Which of the following MUST be true?"

I. y is odd. y COULD be odd, but it doesn't have to be. (for example, if x = 1 and y = 2, x²y is still EVEN)
II. x is even. x COULD be even, but it doesn't have to be. (for example, if x = 2 and y = 2, x²y is still EVEN)
III. xy is even. Since at least one of the variables must be even, it MUST be the case that xy is even

[Reveal] Spoiler:
C

Hello,

Would it not be sufficient for x to be even? If x is even, wouldn't x^2y be even regardless of y?

Thanks!

Hi FransmanPL,

I believe you are misinterpreting the question.
We are asked to determine what must be true BASED ON the fact that x²y is EVEN
So, for statement 2, we should ask "Given the fact that x²y is EVEN, must it be true that x is even?" The answer is no. It need not be true that x is even.

You are treating the statements in the opposite order. So, for statement 2, you are asking, "Given the fact that x is even, must it be true that x²y is EVEN?"
This is not what the question is asking.

Does that help?

Cheers,
Brent
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Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of [#permalink]

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25 Nov 2016, 10:40
Hi Brent,

I see what you mean, I was indeed looking at it the other way around.

Thanks for the clarification!

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Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of [#permalink]

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13 Dec 2017, 13:08
Bunuel wrote:
x and y are integers. $$\frac{(x^2y−1)}{2}$$ has a non-zero remainder. Which of the following must be true?

I. y is odd
II. x is even
III. xy is even

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

Easy C.
Since ((x^2)*y -1)/2 has a non-zero remainder, the remainder has to be 1.
If (n-1)/2 has a remainder of 1 then n must be divisible by 2. Hence, (x^2)*y must be divisible by 2.
For that, either x or y has to be even or both x, y can be even. Thus,xy will always be even.
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Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of   [#permalink] 13 Dec 2017, 13:08
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