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

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.

A.

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

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

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

\(x,y\,\, \ge 1\,\,\,{\rm{ints}}\,\,\,\,\left( * \right)\)

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

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

\(\left( 2 \right)\,\,\,y + 1\,\,{\rm{prime}}\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)


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

Regards,
Fabio.

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.