Official Solution:


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. \(x^2*y^2=\text{even}\) means that \(xy=\text{even}\) (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 either 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 \(xy=\text{even}\), 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.


Answer: E
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I think this is a high-quality question and I agree with explanation.

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

Statement 1 says that (xy)^2 is even, which means that xy is even. Now we know for sure that xy is divisible by 2 and may or may not be divisble by any other prime number.

So can we not say that the only prime number that surely divides xy is 2, and so p is surely an even number. Other prime numbers apart from 2 (e.g. 3,5 etc) may or may not divide xy (depending on what xy is => 6 or 10 etc.), but 2 surely does. So I think the answer should be A. Experts could you please clarify my thought here?

You say yourself that p may be any other prime but 2, so why is it necessarily 2?
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Hi Bunuel,

So, here's my confusion -

Per statement 1, xy could have the values 2,4,6,8 etc.
The only prime number that constantly divides all the values of xy is 2. P could be other prime numbers in case xy > 2, so I had concluded that P "should" be 2 to satisfy all cases. I understand your explanation in the original post that since there could be other values of prime P in different cases, we cannot conclude that p = 2. Thanks !

Hi Bunuel

I have a question here:

For statement 1: if xy is even ; then one of the prime factor of xy must be 2. So p can be 2 or 2 and some other prime numbers. But one value of p must be 2. So why is this statement not sufficient? The question does not ask whether the only value that p can take is an even prime number (2) ?

Can you please explain I am a bit confused there.

p is some specific number. We know from the stem that it's prime. So, p could be 2, 3, 5, 7, ... The question asks whether p =2.

Now, from (1) p COULD be 2 but it also COULD be 3. We don't know for sure. That's why (1) is not sufficient.
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Bunuel : Thanks for you reply: I get it since p can have other values as well it is not sufficient.

I think this is true. It should be A.

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The correct answer is E not A. From (1) p COULD be true but it COULD be any other prime, so we cannot say for sure that p = 2.
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I think this is a high-quality question and I agree with explanation.