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Re: M05-16 [#permalink]
Bunuel wrote:
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



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.
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Re: M05-16 [#permalink]
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ZXSohamGMAT2020 wrote:
Bunuel wrote:
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



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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Re: M05-16 [#permalink]
Bunuel wrote:
ZXSohamGMAT2020 wrote:
Bunuel wrote:
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



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.


Bunuel : Thanks for you reply: I get it since p can have other values as well it is not sufficient.
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Re: M05-16 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M05-16 [#permalink]
pretty high quality question. need to be very careful and easy to make silly mistakes. definitely this shud go in repeat list!
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Re: M05-16 [#permalink]
Expert Reply
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M05-16 [#permalink]
Hey Bunuel, I did not mean question was not clear. It was perfect the way it was and I just appreciated the way it tested the candidate.

Bunuel wrote:
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M05-16 [#permalink]
I think this is a high-quality question and I agree with explanation.
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