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16 Sep 2014, 01:44
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If \(x\) and \(y\) are integers, is \(x = 0\)?
(1) \((-2)^x * y^3 > 0\)
(2) \((-3)^x * y^2 < 0\)
(1) \((-2)^x * y^3 > 0\)
(2) \((-3)^x * y^2 < 0\)
16 Sep 2014, 01:44
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Official Solution:
If \(x\) and \(y\) are integers, is \(x = 0\)?
(1) \((-2)^x * y^3 > 0\).
• If \(x\) is even (including 0), \((-2)^x\) is positive.
• If \(x\) is odd, \((-2)^x\) is negative.
• The sign of \(y^3\) depends on \(y\).
From this, \((-2)^x * y^3 > 0\) holds when \(x\) is even (including 0) and \(y\) is positive, as well as when \(x\) is odd and \(y\) is negative. Thus, \(x\) can be either even (including 0) or odd, depending on \(y\)'s value. Insufficient.
(2) \((-3)^x * y^2 < 0\)
The square of a number is always positive or 0. However, since the product above is not 0, \(y^2\) cannot be 0; hence, \(y^2\) is positive. This implies that \((-3)^x\) must be negative, which further implies that \(x\) must be odd. Thus, \(x\) cannot be 0. Sufficient.
Answer: B
If \(x\) and \(y\) are integers, is \(x = 0\)?
(1) \((-2)^x * y^3 > 0\).
• If \(x\) is even (including 0), \((-2)^x\) is positive.
• If \(x\) is odd, \((-2)^x\) is negative.
• The sign of \(y^3\) depends on \(y\).
From this, \((-2)^x * y^3 > 0\) holds when \(x\) is even (including 0) and \(y\) is positive, as well as when \(x\) is odd and \(y\) is negative. Thus, \(x\) can be either even (including 0) or odd, depending on \(y\)'s value. Insufficient.
(2) \((-3)^x * y^2 < 0\)
The square of a number is always positive or 0. However, since the product above is not 0, \(y^2\) cannot be 0; hence, \(y^2\) is positive. This implies that \((-3)^x\) must be negative, which further implies that \(x\) must be odd. Thus, \(x\) cannot be 0. Sufficient.
Answer: B
General Discussion
20 Aug 2023, 10:24
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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.
