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

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

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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 20 Oct 2010, 13:47, edited 1 time in total.
Edited the stem.
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Re: Number properties [#permalink]

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New post 19 Oct 2010, 10:36
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.
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Re: Number properties [#permalink]

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shrive555 wrote:
If x and y are positive integers, is x2y2 even ?

(1) x + 5 is a prime number

(2) y + 1 is a prime number


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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Re: Number properties [#permalink]

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New post 20 Oct 2010, 13:45
A.

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

edit it to x^2y^2 for others maybe?
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Re: If x and y are positive integers, is x^2*y^2 even ? (1) x + [#permalink]

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Re: If x and y are positive integers, is x^2*y^2 even ? (1) x + [#permalink]

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Re: If x and y are positive integers, is x^2*y^2 even ? (1) x + [#permalink]

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New post 01 Jan 2017, 13:55
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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Re: If x and y are positive integers, is x^2*y^2 even ? (1) x +   [#permalink] 01 Jan 2017, 13:55
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