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Updated on: 12 Sep 2016, 01:32
Originally posted by RichaChampion on 12 Sep 2016, 01:15.
Last edited by Bunuel on 12 Sep 2016, 01:32, edited 1 time in total.
Last edited by Bunuel on 12 Sep 2016, 01:32, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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E
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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) \(xp = 6\)
(1) \(x^2 * y^2\) is an even number
(2) \(xp = 6\)
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12 Sep 2016, 01:16
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This is my reasoning. I believe its E. perhaps Bunuel can tell the OA.
If X and Y are positive integers and X*Y is divisible by prime number P. Is P an even number?
(1) X²∗Y² is an even number
(2) X*P = 6
Statement 1
X²∗Y² is an even number
Inferences →
One among X and Y can be even. That means either X is even or Y is even. In this case, we can't deduce whether P is even or ODD. or→
Both X and Y can be even in this case we must have P, which is a prime number, an even integer = 2.
So from this statement, we get both YES and NO. Thus, this statement is not sufficient.
Statement 1
X*P = 2 X 3 = 3 X 2 = 1 X 6 = 6 X 1
Remeber the one in red is not posisble as 1 is not a prime number. So again here we have YES and NO.
Let us see If by combination we can get anything.
By Combination we know that X and P are opposite. That means If one is odd then other is Even and Vice Versa.
A lot depend on Y now.
If X is Odd then P is Even(=2), but notice that when X is odd Y has to be even in order to maintain XY → Even.
If X is even then P will depend now on Y, but notice here that Y has no constraint now, the constraint is dismissed. Y can be Even or Odd. Thus, P can be even(=2) or odd(any prime number such as 3, 5, 7, 11, 13____).
I think the answer should be E.
If X and Y are positive integers and X*Y is divisible by prime number P. Is P an even number?
(1) X²∗Y² is an even number
(2) X*P = 6
Statement 1
X²∗Y² is an even number
Inferences →
One among X and Y can be even. That means either X is even or Y is even. In this case, we can't deduce whether P is even or ODD. or→
Both X and Y can be even in this case we must have P, which is a prime number, an even integer = 2.
So from this statement, we get both YES and NO. Thus, this statement is not sufficient.
Statement 1
X*P = 2 X 3 = 3 X 2 = 1 X 6 = 6 X 1
Remeber the one in red is not posisble as 1 is not a prime number. So again here we have YES and NO.
Let us see If by combination we can get anything.
By Combination we know that X and P are opposite. That means If one is odd then other is Even and Vice Versa.
A lot depend on Y now.
If X is Odd then P is Even(=2), but notice that when X is odd Y has to be even in order to maintain XY → Even.
If X is even then P will depend now on Y, but notice here that Y has no constraint now, the constraint is dismissed. Y can be Even or Odd. Thus, P can be even(=2) or odd(any prime number such as 3, 5, 7, 11, 13____).
I think the answer should be E.
12 Sep 2016, 01:33
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RichaChampion
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. \(x^2*y^2=\text{even}\) 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
