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Joined: 02 Sep 2009
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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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Question Stats: 58% (01:28) correct 42% (01:35) wrong based on 295 sessions

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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

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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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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

Originally posted by Vinayak Shenoy on 23 Nov 2016, 07:20.
Last edited by Vinayak Shenoy on 23 Nov 2016, 08:50, edited 2 times in total.
GMAT Club Legend  V
Joined: 12 Sep 2015
Posts: 4017
x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of  [#permalink]

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1
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

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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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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

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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Joined: 12 Sep 2015
Posts: 4017
Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of  [#permalink]

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Top Contributor
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

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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Hi Brent,

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

Thanks for the clarification!
Intern  B
Joined: 11 Nov 2017
Posts: 11
Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of  [#permalink]

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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

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

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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

$$\frac{(x^2y−1)}{2}$$

So, $$x^2y - 1$$ must be odd as it's not divisible by 2.

even - 1 = odd.

$$x^2y$$ = even.

xy must be even.

C is the correct answer. x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of   [#permalink] 21 Feb 2019, 09:44
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