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16 Sep 2014, 01:22
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Is \(\frac{xyz}{12}\) an integer?
(1) \(x\), \(y\), \(z\) are consecutive integers
(2) \(x\) and \(z\) are prime numbers
(1) \(x\), \(y\), \(z\) are consecutive integers
(2) \(x\) and \(z\) are prime numbers
16 Sep 2014, 01:22
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Official Solution:
Is \(\frac{xyz}{12}\) an integer?
(1) \(x\), \(y\), \(z\) are consecutive integers
It's important to note that \(x\), \(y\), and \(z\) being consecutive integers doesn't necessarily mean \(x < y < z\). They could be in any order. Now, if \(x = 0\), \(y = 1\), and \(z = 2\), then the answer is YES. However, if \(x = 1\), \(y = 2\), and \(z = 3\), the answer is NO. Not sufficient
(2) \(x\) and \(z\) are prime numbers
This condition alone is clearly insufficient, as we have no information about \(y\).
(1)+(2) Again, note that \(x\), \(y\), \(z\) can be in any order. If \(x = 3\), \(y = 4\), and \(z = 5\), the answer would be YES. However, if \(x = 5\), \(y = 6\), and \(z = 7\), the answer would be NO. Not sufficient.
Answer: E
Is \(\frac{xyz}{12}\) an integer?
(1) \(x\), \(y\), \(z\) are consecutive integers
It's important to note that \(x\), \(y\), and \(z\) being consecutive integers doesn't necessarily mean \(x < y < z\). They could be in any order. Now, if \(x = 0\), \(y = 1\), and \(z = 2\), then the answer is YES. However, if \(x = 1\), \(y = 2\), and \(z = 3\), the answer is NO. Not sufficient
(2) \(x\) and \(z\) are prime numbers
This condition alone is clearly insufficient, as we have no information about \(y\).
(1)+(2) Again, note that \(x\), \(y\), \(z\) can be in any order. If \(x = 3\), \(y = 4\), and \(z = 5\), the answer would be YES. However, if \(x = 5\), \(y = 6\), and \(z = 7\), the answer would be NO. Not sufficient.
Answer: E
08 Oct 2023, 02:56
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
