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16 Sep 2014, 00:56
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Is \(x^2 - y^2\) an integer divisible by 8?
(1) \(x\) and \(y\) are even integers.
(2) \(x + y\) is an integer divisible by 8.
(1) \(x\) and \(y\) are even integers.
(2) \(x + y\) is an integer divisible by 8.
16 Sep 2014, 00:56
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Official Solution:
Is \(x^2 - y^2\) an integer divisible by 8?
(1) \(x\) and \(y\) are even integers.
This information is clearly insufficient. For instance, if \(x=y=0\), the answer is YES, whereas if \(x=2\) and \(y=0\), the answer is NO.
(2) \(x + y\) is an integer divisible by \(8\).
Since \(x^2 - y^2=(x+y)(x-y)\), if one of the factors is divisible by 8, then the product is as well—this is true for integers. However, we aren't given that \(x\) and \(y\) are integers. For example, if \(x=4.8\) and \(y=3.2\), while \(x+y\) is divisible by 8, \(x^2 - y^2\) is not. Not sufficient.
(1)+(2) Given that \(x\) and \(y\) are integers, and that \(x+y\) is divisible by 8, it follows that \((x+y)(x-y)\) is equal to a multiple of 8 multiplied by an integer, and is therefore divisible by 8. Sufficient.
Answer: C
Is \(x^2 - y^2\) an integer divisible by 8?
(1) \(x\) and \(y\) are even integers.
This information is clearly insufficient. For instance, if \(x=y=0\), the answer is YES, whereas if \(x=2\) and \(y=0\), the answer is NO.
(2) \(x + y\) is an integer divisible by \(8\).
Since \(x^2 - y^2=(x+y)(x-y)\), if one of the factors is divisible by 8, then the product is as well—this is true for integers. However, we aren't given that \(x\) and \(y\) are integers. For example, if \(x=4.8\) and \(y=3.2\), while \(x+y\) is divisible by 8, \(x^2 - y^2\) is not. Not sufficient.
(1)+(2) Given that \(x\) and \(y\) are integers, and that \(x+y\) is divisible by 8, it follows that \((x+y)(x-y)\) is equal to a multiple of 8 multiplied by an integer, and is therefore divisible by 8. Sufficient.
Answer: C
General Discussion
19 Jun 2015, 07:19
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Great question Bunnel..
Totally Stumped
Totally Stumped
