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Re: If x and y are integers, and N = (x^2 – y + 3x)(2y + x), is N odd? [#permalink]

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26 Apr 2017, 07:13

1

If x and y are integers, and N = (x² – y + 3x)(2y + x), is N odd?

1) x+y is even 2) 3xy is odd

for x+y to be even, we can have x,y as either even or odd for xy to be odd, both x and y should be odd

So for 3xy is odd (x² – y + 3x)=> odd^2-odd+odd is always an odd number (2y + x) => 2*odd + odd=> even+odd= odd number product of two odd numbers is odd.

Now x+y = even consider x, y as even (x² – y + 3x)=> even^2-even+even is always an even number (2y + x) => 2*even+ even=> even+even= even number product of two even numbers is even.

consider x,y as odd, so as per above discussion for 3xy (x² – y + 3x)=> odd^2-odd+odd is always an odd number (2y + x) => 2*odd + odd=> even+odd= odd number product of two odd numbers is odd.

Hence for x+y, depending upon the the odd/even value, the result changes

So together Hence B alone is sufficient
_________________

Kudos are always welcome ... as well your suggestions

If x and y are integers, and N = (x^2 – y + 3x)(2y + x), is N odd? [#permalink]

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26 Apr 2017, 09:41

1

GMATPrepNow wrote:

If x and y are integers, and N = (x² – y + 3x)(2y + x), is N odd?

1) x+y is even 2) 3xy is odd

*kudos for all correct solutions

From 2)3xy is odd,x and y are integers ------->x and y should be odd (x^2+3x -y) & (2y+x)----> cannot be zero---->clear cut answer will be obtained hence sufficient

from 1)x+Y is even----> Both can be even or Both can be odd ----> N can be even or odd hence insufficient

Re: If x and y are integers, and N = (x^2 – y + 3x)(2y + x), is N odd? [#permalink]

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27 Apr 2017, 07:02

Top Contributor

GMATPrepNow wrote:

If x and y are integers, and N = (x² – y + 3x)(2y + x), is N odd?

1) x+y is even 2) 3xy is odd

Target question:Is N odd?

Given: N = (x² – y + 3x)(2y + x) Before we examine the statements, it might be useful SYSTEMATICALLY examine all of the possible cases we need to consider: case a: x is even, and y is even case b: x is even, and y is odd case c: x is odd, and y is even case d: x is odd, and y is odd

There are two ways to analyze each case. - We can take each case and apply the rules for evens and odd (e.g., even + odd = odd, even x even = even, etc) - We can take each case and plug in even and odd numbers for x and y. The easiest values are 1 for odd numbers and 0 for even numbers.

When we do apply either of these strategies we get: case a: x is even, and y is even. N is EVEN case b: x is even, and y is odd. N is EVEN case c: x is odd, and y is even. N is EVEN case d: x is odd, and y is odd. N is ODD

The target question ask whether N is odd. Since N is odd only when x is odd and y is odd, we can rephrase our target question... REPHRASED target question:Are x and y BOTH odd?

Okay, now onto the statements!!!

Statement 1: x+y is even If x+y is even, then there are two possible cases: - x and y are both odd, in which case, x and y ARE both odd - x and y are both even, in which case, x and y are NOT both odd Since we can answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 3xy is odd If 3xy is odd, then xy is odd, which means x and y ARE both odd Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

If x and y are integers, and N = (x^2 – y + 3x)(2y + x), is N odd? [#permalink]

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17 Nov 2017, 13:39

GMATPrepNow wrote:

GMATPrepNow wrote:

If x and y are integers, and N = (x² – y + 3x)(2y + x), is N odd?

1) x+y is even 2) 3xy is odd

Target question:Is N odd?

Given: N = (x² – y + 3x)(2y + x) Before we examine the statements, it might be useful SYSTEMATICALLY examine all of the possible cases we need to consider: case a: x is even, and y is even case b: x is even, and y is odd case c: x is odd, and y is even case d: x is odd, and y is odd

There are two ways to analyze each case. - We can take each case and apply the rules for evens and odd (e.g., even + odd = odd, even x even = even, etc) - We can take each case and plug in even and odd numbers for x and y. The easiest values are 1 for odd numbers and 0 for even numbers.

When we do apply either of these strategies we get: case a: x is even, and y is even. N is EVEN case b: x is even, and y is odd. N is EVEN case c: x is odd, and y is even. N is EVEN case d: x is odd, and y is odd. N is ODD

The target question ask whether N is odd. Since N is odd only when x is odd and y is odd, we can rephrase our target question... REPHRASED target question:Are x and y BOTH odd?

Okay, now onto the statements!!!

Statement 1: x+y is even If x+y is even, then there are two possible cases: - x and y are both odd, in which case, x and y ARE both odd - x and y are both even, in which case, x and y are NOT both odd Since we can answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 3xy is odd If 3xy is odd, then xy is odd, which means x and y ARE both odd Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

the Solution assumes N as integer, but what if N is not an integer? no info about that in the question.

Please help

Thanks

If x and y are integers, and N = (x² – y + 3x)(2y + x), then N must be an integer. Here's why:

If x is an integer, then x², and 3x must also be integers. If y is an integer, then 2y must also be an integer. Also, the sums and differences of integers are also integers.

So, N = (x² – y + 3x)(2y + x) = (some integer - some integer + some integer)(some integer + some integer) = (an integer)(an integer) = integer