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x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of
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Updated on: 23 Nov 2016, 08:50

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.

x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of
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23 Nov 2016, 08:14

Top Contributor

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

Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of
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25 Nov 2016, 09:04

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

Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of
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25 Nov 2016, 09:21

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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

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.

Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of
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13 Dec 2017, 14:08

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.
_________________

Cheers.

Never gonna give up!!!!

gmatclubot

Re: x and y are integers. (x^2y−1)/2 has a non-zero remainder. Which of &nbs
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13 Dec 2017, 14:08