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

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

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

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

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


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

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

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
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