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For integers x and y, when x is divided by y, the remainder

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For integers x and y, when x is divided by y, the remainder is odd. Which of the following must be true?

A. x is odd
B. xy is odd
C. x and y share no common factors other than 1
D. The sum x + y is odd
E. At least one of x and y is odd
[Reveal] Spoiler: OA

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Last edited by Bunuel on 28 Feb 2013, 05:52, edited 1 time in total.
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Re: For integers x and y, when x is divided by y, the remainder [#permalink]

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For integers x and y, when x is divided by y, the remainder is odd. Which of the following must be true?

A. x is odd. Not necessarily true. Consider x=4 and y=3.

B. xy is odd. Not necessarily true. Consider x=4 and y=3.

C. x and y share no common factors other than 1. Not necessarily true. Consider x=3 and y=6.

D. The sum x + y is odd. Not necessarily true. Consider x=1 and y=3.

E. At least one of x and y is odd. If both x and y were even, then the remainder when x is divided by y would be even, not odd. Hence at least one of x and y is odd.

Answer: E.

Hope it's clear.
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Re: For integers x and y, when x is divided by y, the remainder [#permalink]

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New post 28 Feb 2013, 07:02
For integers x and y, when x is divided by y, the remainder is odd. Which of the following must be true?

A. x is odd. May be. not must.
B. xy is odd. x= 6, y =5 R= 1. xy is not odd
C. x and y share no common factors other than 1. x=12 , y=9. r= 3. common factor = 3, 1
D. The sum x + y is odd. x= 3, y=5. r= 3. x+y = 8 even
E. At least one of x and y is odd. correct
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Re: For integers x and y, when x is divided by y, the remainder [#permalink]

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Re: For integers x and y, when x is divided by y, the remainder [#permalink]

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New post 17 Mar 2016, 05:46
Here we the easiest way out is use the examples
Now by using 3/11 ; 3/9 and 3/12 => all options can be eliminated except the last one => E
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For integers x and y, when x is divided by y, the remainder [#permalink]

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New post 21 Aug 2017, 05:22
stonecold wrote:
Here we the easiest way out is use the examples
Now by using 3/11 ; 3/9 and 3/12 => all options can be eliminated except the last one => E


I have got one more;

let X=nY+R (n= some quotient, R=remainder)

or X=nY+2k+1 (2k+1=odd remainder)

Now if Y is even;

X=nY+2k+1=even+even+1=odd

and if Y is odd;

X=(odd or even)+(even)+1
=(odd or even)

So,
at least 1 of X and Y will always be Odd.


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For integers x and y, when x is divided by y, the remainder   [#permalink] 21 Aug 2017, 05:22
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