If x and y are positive integers, is x^2*y^2 even ? (1) x +

Question

If x and y are positive integers, is x^2*y^2 even ?

(1) x + 5 is a prime number

(2) y + 1 is a prime number

Bunuel · Accepted Answer

In order  x^2y^2  to be even at least one of the unknowns must be even.

(1)  x+5=prime  --> as  x  is a  positive integer  then this prime can not be the only even prime 2 (in this case  x+5=2  -->  x=-3=negative ), so  x+5=prime=odd  -->  x=odd-5=odd-odd=even . Sufficient.

(2) y + 1 is a prime number -->  y  could be 1, so odd, and we won't be sure whether  x^2y^2=even  or  y  could be even (for example 2) and then  x^2y^2=even . Not sufficient.

Answer: A.

Marshmallow · Answer

A for me.

First off, an even number squared will give an even number(2^2 = 4, 4^2 = 16 etc). An odd number squared will give an odd number (3^2 = 9, 5^2 = 25 etc). Also, for the multiplication between two numbers to come out even, at least one of the numbers must be even.

We know all primes except for 2 are odd. Since x and y are positive, the only way for x + 5 to be prime will be if x is even (i.e. if x = 1 then x + 5 = 6 which is not prime. But if x = 2 then x + 5 = 7 which is prime). Therefore, x^2 will also be even and x^2y^2 will also be even regardless of what y is. statement 1 is sufficient.

Using a similar approach for condition 2, we can set y = 1 (y + 1 = 2 which is prime) or y = 2 (y + 2 = 3 which is also prime). So we see with this condition, y can be both even or odd. So statement 2 alone is not sufficient.

scheol79 · Answer

A.

Kind of trivial but I was staring at x2y2 for 20 seconds.

edit it to x^2y^2 for others maybe?

stonecold · Answer

Excellent Question
Given info => x,y are positive integers (very important)
We need to check if x^2*y^2 is even or not.
Now x^2*y^2 will be even when either x or y or both are even
Hence we need to find => "If atleast one of x or y is even"
Statement 1
Here the least value of x+5 is 6 (as the least value of x is 1)
Here we need to remember that all the Prime numbers greater than 2 are odd.
Hence x+5 must be odd
so x must be even
BINGO
sufficient

Statement 2

Here y the least value of y+1 is 2

Let us take y = 1 (as y+1=2 which is a prime too)
if x is even => then x^2*y^2 will be even
if x is odd => then x^2*y^2 will be odd
Hence insufficient

Hence A

fskilnik · Answer

x,y\,\, \ge 1\,\,\,{m{ints}}\,\,\,\,\left( * ight)

{\left( {xy} ight)^2}\,\,\mathop = \limits^? \,\,\,{	ext{even}}\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * ight)} \,\,\,\boxed{\,\,?\,\,\,:\,\,x\,\,{	ext{even}}\,\,\,{	ext{or}}\,\,\,y\,\,{	ext{even}}\,\,\,\,}

\left( 1 ight)\,\,\left\{ \matrix{
 x + 5\,\,\,\,\mathop \ge \limits^{\left( * ight)} \,\,\,6 \hfill \cr 
 x + 5\,\,{m{prime}} \hfill \cr} ight.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x + 5\,\, = {m{odd}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x\,\,{m{even}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{m{YES}}} ightangle

\left( 2 ight)\,\,\,y + 1\,\,{m{prime}}\,\,\,\,\left\{ \matrix{
 \,{m{Take}}\,\,\left( {x,y} ight) = \left( {1,1} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{m{NO}}} ightangle \,\, \hfill \cr 
 \,{m{Take}}\,\,\left( {x,y} ight) = \left( {2,1} ight)\,\,\,\, \Rightarrow \,\,\,\left\langle {{m{YES}}} ightangle \,\, \hfill \cr} ight.

This solution follows the notations and rationale taught in the GMATH method.

Regards, 
Fabio.

ShivamGoel · Answer

1 is not prime 
So, the answer shoud be D as both x and y = 2

BrentGMATPrepNow · Answer

Target question :    Is x²y² even ?

Statement 1 : x + 5 is a prime number  
So, x+5 is a prime number greater than 5, which means x+5 must be ODD (since 2 is only even prime)
If x+5 is ODD, then  x must be EVEN 
If x is EVEN, then x²y² must be even 
The answer to the target question is  YES, x²y² IS even  
Since we can answer the  target question  with certainty, statement 1 is SUFFICIENT

Statement 2 : y + 1 is a prime number 
There are several scenarios that satisfy statement 2. Here are two: 
 Case a : x = 0 and y = 1. Notice that y+1 = 1+1 = 2, which is prime. In this case, x²y² = 0²1² = 0. So, the answer to the target question is  YES, x²y² IS even  
 Case b : x = 1 and y = 1. Notice that y+1 = 1+1 = 2, which is prime. In this case, x²y² = 1²1² = 1. So, the answer to the target question is  NO, x²y² is NOT even  
Since we cannot answer the  target question  with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If x and y are positive integers, is x^2*y^2 even ? (1) x +

Prep Toolkit

Top 5 GMAT Debrief Videos

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

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

If x and y are positive integers, is x^2*y^2 even ? (1) x +

Prep Toolkit

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

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards