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

Question

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

BrentGMATPrepNow · Accepted Answer

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

BrentGMATPrepNow · Answer

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

Answer:  C

Vinayak Shenoy · Answer

IMO C
  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

FransmanPL · Answer

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!

FransmanPL · Answer

Hi Brent,

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

Thanks for the clarification!

Madnov2017 · Answer

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.

KSBGC · Answer

\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.

achloes · Answer

Bunuel   BrushMyQuant   gmatophobia

I don't understand how (x^2)(y) is even.

This is how I've solved it - please let me know where I've gone wrong?

Many thanks in advance!

BrushMyQuant · Answer

Given that x and y are integers and  \frac{(x^2y−1)}{2}  has a non-zero remainder. And we need to find which of the following must be true?

Any number when divided by 2 can give only two remainders 
 
 → Remainder of 0 when the number is Even and
→ Remainder of 1 when the number is Odd

Since  x^2*y - 1  when divided by 2 gives non-zero remainder
=>  x^2*y - 1  when divided by 2 gives 1 remainder
=>  x^2*y - 1  is an odd number
=>  x^2*y  is even

Now, this will be true when at least one of x or y is even.

I. y is odd 
Now this doesn't have to be true as both x and y can be even or y can even and x can be odd
=>   NOT NECCASARILY TRUE

II. x is even 
Now this doesn't have to be true as both x and y can be even or x can odd and y can be even
=>   NOT NECCASARILY TRUE

III. xy is even
This will be true in any case as either x or y is even or both are even
Making xy as even for sure.
=>   TRUE

So,  Answer will be C 
Hope it helps!

Watch the following video to MASTER Even Odd Numbers

https://www.youtube.com/watch?v=6WX-7UwJUnw

BrushMyQuant · Answer

achloes  Correcting the steps in your method:

x^2y - 1  when divided by 2 gives non-zero remainder (which is 1 as explained in my solution above)
=>  x^2y - 1  = 2*k + 1  
=>  x^2y - 1  = odd
=>  x^2y  = even

Hope it helps!

Following video will help you understand how to frame the equations for a remainder problem

https://www.youtube.com/watch?v=P0S6j2uTaj4

bumpbot · Answer

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

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