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09 May 2009, 11:25
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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.
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05 Sep 2009, 20:00
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Is \(x^2 - y^2\) an integer divisible by 8?
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer.
(2) \(x + y\) is an integer divisible by \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2 - y^2\) is not. Not sufficient.
(1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(x-y)\) is divisible by \(8\). Sufficient.
Answer: C
(1) \(x\) and \(y\) are even integers. Clearly insufficient, consider \(x=y=0\) for an YES answer and \(x=2\) and \(y=0\) for a NO answer.
(2) \(x + y\) is an integer divisible by \(8\). Now, \(x^2 - y^2=(x+y)(x-y)\), if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that \(x\) and \(y\) are integers. If \(x=4.8\) and \(y=3.2\), \(x+y\) is divisible by \(8\), BUT \(x^2 - y^2\) is not. Not sufficient.
(1)+(2) \(x\) and \(y\) integers. \(x+y\) divisible by 8. Hence \((x+y)(x-y)\) is divisible by \(8\). Sufficient.
Answer: C
General Discussion
15 Jun 2014, 11:42
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mahendru1992
1.6 is not an integer. Divisible means that you get an integer as a result of division.
