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

Question

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

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

umadurga · Answer

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

VyshakhR1995 · Answer

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

--->B

BrentGMATPrepNow · Answer

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

Answer: B

https://www.youtube.com/watch?v=MyG1K3ee69w

hellosanthosh2k2 · Answer

Hi  GMATPrepNow

I have a question

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

Please help

Thanks

BrentGMATPrepNow · Answer

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

Does that help?

Cheers,
Brent

hellosanthosh2k2 · Answer

Hi  GMATPrepNow ,

Thanks for reply. Actually I misread the question as N = (x² – y + 3x)/(2y + x) instead of N = (x² – y + 3x)(2y + x).

Sorry for confusion. I must have been asleep while solving

Posted from my mobile device

BrentGMATPrepNow · Answer

I'll take some responsibility on that too. I could have formatted the question better.

Cheers,
Brent

Fdambro294 · Answer

Given that x and y are integers, we can apply the following Properties regarding the Even/Odd nature of the terms

N  =  ( x^2   - y + 3x) (2y + x)

Property 1:  raising a variable such as X to a Power will not change the Even/Odd nature of the variable
Thus, for purposes of the Even/Odd analysis, we can say:
X will have the same even/odd nature as --->    x^2

Property 2:   Multiplying a variable such as X by any ODD Integer will not change the even/odd nature of the variable
Thus, for purposes of the Even/Odd analysis, we can say:
3X will have the same even/odd nature as ---->    X

Finally, regardless of the even/odd nature of Y, the term (2Y) will always be EVEN

so, the expression that is equal for N will have the same even/odd nature as the follownig:

N  =  (x  -  y  +  x) (EVEN  +  x)
or
N  =  (2x  -  y) (EVEN + x)
or
N  =   (EVEN  -  y) (EVEN  +  x)

now, we can analyze the statements much easier

statement 1:   x + y  =  Even

case 1:   x = even  ;   y = even
N = (Even - even) (Even + even)  =  (even)(even) = EVEN

case 2:   x = odd  ;  y  =  odd  
N  =  (Even -- odd) (Even  + odd)  =  (odd) (odd)  =  ODD

thus, N can be either EVEN or ODD
statement 1 is NOT SUFFICIENT

statement 2:   3xy = Odd

x = Odd  ;   y = Odd     is the only possibiliity

therefore,   N  =  (Even - y) (Even + x)   will have only ONE Even/Odd outcome

statement 2 is sufficient ALONE

*B*

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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745 with Only Self-Prep and GMAT Club

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My Rewards