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16 Sep 2014, 00:49
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Is \((x - y)*(x + y)\) an even integer?
(1) \(x\) and \(y\) are integers
(2) \(x + y\) is even
(1) \(x\) and \(y\) are integers
(2) \(x + y\) is even
16 Sep 2014, 00:49
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Official Solution:
Is \((x - y)*(x + y)\) an even integer?
(1) \(x\) and \(y\) are integers.
If both of them are even or both of them are odd, then \((x - y)*(x + y)=even*even=even\) but if one of them is even and another is odd, then \((x - y)*(x + y)=odd*odd=odd\). Not sufficient.
(2) \(x+y\) is even.
If \(x=y=1\) then the answer is YES but if \(x=1.8\) and \(y=0.2\), then the answer is NO, since in this case \((x-y)*(x+y)\) is not an integer at all. Not sufficient.
(1)+(2) Since from (1) \(x\) and \(y\) are integers and from (2) \(x+y\) is even then either \(x\) and \(y\) are both even or both odd. In either case \((x - y)*(x + y)=even*even=even\) is even. Sufficient.
Answer: C
Is \((x - y)*(x + y)\) an even integer?
(1) \(x\) and \(y\) are integers.
If both of them are even or both of them are odd, then \((x - y)*(x + y)=even*even=even\) but if one of them is even and another is odd, then \((x - y)*(x + y)=odd*odd=odd\). Not sufficient.
(2) \(x+y\) is even.
If \(x=y=1\) then the answer is YES but if \(x=1.8\) and \(y=0.2\), then the answer is NO, since in this case \((x-y)*(x+y)\) is not an integer at all. Not sufficient.
(1)+(2) Since from (1) \(x\) and \(y\) are integers and from (2) \(x+y\) is even then either \(x\) and \(y\) are both even or both odd. In either case \((x - y)*(x + y)=even*even=even\) is even. Sufficient.
Answer: C
13 Dec 2014, 02:11
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Bunuel
Can you please explain how x=y=1 gives the equation an even integer... Thanks
