Data Sufficiency

If x and y are integers, is the value x(y+1) even ?

1. x and y are prime numbers
2. y > 7

I went with A. Because I thought x and y could be any prime numbers from 2 to n. Apart from number 2 all prime numbers are odd. So x(y+1) will have atleast one even number making it even. Am I wrong ? The solution from GMATPrep is C.

Please let me know, I'm going to give exam on Saturday . Don't feel quite prepared.

Thanks.

crackitpal wrote:If x and y are integers, is the value x(y+1) even ?

1. x and y are prime numbers
2. y > 7

I went with A. Because I thought x and y could be any prime numbers from 2 to n. Apart from number 2 all prime numbers are odd. So x(y+1) will have atleast one even number making it even. Am I wrong ? The solution from GMATPrep is C.

Please let me know, I'm going to give exam on Saturday . Don't feel quite prepared.

Thanks.

If you pick x = 3 and y = 2, then x(y+1) = 3(3) = 9 which is odd. Only by adding in y>7 do we know that y must be an odd integer, which means that x(y+1) = x(even) = even.

Thank you very much Stuart!! I think just when your mind is supposed tobe responding, it fails to do so

maybe its too late,,maybe i just don't get it

i don't get this one..can anyone explain it in more detail pls?

Hi Stuart.

In regard to solution provided to this problem long time ago, I may be asking a silly question, but would like
to clarify your solution.

Below you say y>7 = even, but y could be 8,9,10.... and by adding 1 , it would make it even or odd,
therefore it is not only even but also odd... I think I may be missing something here,
could you please clarify at earliest.

Thanks in advance.
Parag

Stuart Kovinsky wrote:

crackitpal wrote:If x and y are integers, is the value x(y+1) even ?

1. x and y are prime numbers
2. y > 7

I went with A. Because I thought x and y could be any prime numbers from 2 to n. Apart from number 2 all prime numbers are odd. So x(y+1) will have atleast one even number making it even. Am I wrong ? The solution from GMATPrep is C.

Please let me know, I'm going to give exam on Saturday . Don't feel quite prepared.

Thanks.

If you pick x = 3 and y = 2, then x(y+1) = 3(3) = 9 which is odd. Only by adding in y>7 do we know that y must be an odd integer, which means that x(y+1) = x(even) = even.

750Goal wrote:Hi Stuart.

In regard to solution provided to this problem long time ago, I may be asking a silly question, but would like
to clarify your solution.

Below you say y>7 = even, but y could be 8,9,10.... and by adding 1 , it would make it even or odd,
therefore it is not only even but also odd... I think I may be missing something here,
could you please clarify at earliest.

Thanks in advance.
Parag

Hi,

that's in reference to why C (together) is the correct answer; only by adding y>7 to the first statement (y is prime) do we know that y must be odd.

Thanks Stuart. Another oversight.

Another way you can think of it is, the statement is asking is at least one of the following statements true: x is even, or y is odd

We know that an even number times any number is even. Since an odd plus one would be even, if either of these statements is true we know that the answer is even.

I like to break up the question before even attacking the problem. This helps me a lot.

Hi Stuart,
x can be 2 or 3 and in both case x is prime, and if y>7 it will give you two different answers. IMO the answer is e.

capthan wrote:Hi Stuart,
x can be 2 or 3 and in both case x is prime, and if y>7 it will give you two different answers. IMO the answer is e.

Combining both statements say y is a prime no greater than 7
Therefore y is odd

y+1 = even
X * ( y+1) = even irrespective of x

Stuart, I still have a doubt. If x can be odd or even, and y is odd, then there are two probable answers and thus the answer must be (E).

Got it correct. My mistake. The answer is C.