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If x and y are positive integers and xy is divisible by prime number p . Is p an even number?

1 x^2 * y^2 is an even number

2 xp = 6

Here is a solution for that problem.

If x and y are positive integers and xy is divisible by prime number p. Is p an even number?

Notice that as given that p is a prime number and the only even prime is 2, then the question basically asks whether p=2.

(1) x^2 * y^2 is an even number --> xy=even (this means that at least one of the unknowns is even). We have that some even number is divisible by prime number p, not sufficient to say whether p=2, for example if xy=6 then p can be ether 2 or 3.

(2) xp = 6 --> since x is a positive integer and p is a prime number then either x=2 and p=3 (answer NO) or x=3 and p=2 (answer YES). Not sufficient.

(1)+(2) If y=6 then xy=even, so the first statement is satisfied irrespective of the value of x and thus we have no constraints on its value. So from (2) x can take any of the two values 2 or 3, which means that p can also take any of the two values 2 or 3, respectively. Not sufficient.

make sure that the posted question is correct in wording. i think that there has to some information about the x,y &z. if we say that x,y & z are integers, then, statement 1: Product of x^2 * y^2 is even doesn't make out any solution. because it doesnot tell anything about z. whether it ll be divisible by Q or not.

Statement 2: z * Q = 6

this statement doesn't say anything about the x & y.

so, using both statement alone or together we cant answer the question.

correct my explanation if wrong. _________________

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Attitude determine everything. all the best and God bless you.

S1: for x^2 *y^2 to be even any of x or y should be even or both can be even. This does not help nail the value of Q as z could be odd or even and based on that Q could be even or odd S2 : this shows that either of z and Q can be 2 or 3

Together too they don't answer the question as z could be odd or even and based on that Q could be even or odd.

I don't understand why z or Q has to be equal to 2; can't Q be 11 and z equal 6/11?

even if we assume that there is no mention of the fact that x, y and z are integers and your solution is correct the answer still remains E. However the explanation provided by GMAT club can be questioned in that case. In all probability it should have been mentioned that all variables in this question should be assumed to be integers.

I got it down to an unsure C or E and guessed E. :/ _________________

I'm trying to not just answer the problem but to explain how I came up with my answer. If I am incorrect or you have a better method please PM me your thoughts. Thanks!

That 40*x2*y2 is even is evident even without looking at stmt 1. Stmt 1 is insuff as it doesn't say anything abt z or q z*q = 6 means either z or q is 2 or 3. If z is 2 then q is 3, an odd prime. But if z is 3 then q is 2, an even prime. But stmt 2 doesn't say anything conclusively. So stmt 2 insuff

Together, they don't indicate the value of z or q either

In order to know that, we must figure out what z is (or q, but they dont usually give that directly. S1: Says nothing about the denominator. S2: gives us Z and Q together, but we don't know which is 3 and which is 2.

I kinda forgot that E/E and E/O can yield an integer if the numerator is a multiple of the denominator. The GMAT Club questions are great memory joggers!

Basically asking whether Q=2 Statement 1 not very useful since it is mum on what Q could be. Not sufficient. Statement 2 implies that Q is 2 or 3. Not sufficient.

Combining leaves us still unsure since x^2y^=even could mean that x^2y^2 could be a multiple of 2 or 3, thus masking the identity of Q, which could still be a 2 0r 3. _________________

x^2.y^2=even mean the multiple of these two is an integer (can be a factor of 3) so 40*even integer = even INTEGER

Q*z=6 if Q is an prime number it is also an integer thus z must be an integer so when we take these facts together Q can be (1,2,3, NOT 6 THAT IS WHY Z IS AN INTEGER)

1,2,3 Q can be so both statements together insufficient