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16 Sep 2014, 01:44
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D
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38% (02:24) correct
62%
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wrong
based on 138
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If \(x\) and \(y\) are positive integers, is \(x\) a prime number?
(1) \(|x - 2| \lt 2 - y\).
(2) \(x + y - 3 = |1-y|\).
(1) \(|x - 2| \lt 2 - y\).
(2) \(x + y - 3 = |1-y|\).
16 Sep 2014, 01:44
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Official Solution:
(1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \leq 0\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).
So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.
(2) \(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \leq 0\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.
Answer: D
(1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \leq 0\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).
So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.
(2) \(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \leq 0\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.
Answer: D
General Discussion
21 Jul 2015, 19:10
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Bunuel
Official Solution:
(1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \leq 0\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).
So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.
(2) \(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \leq 0\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.
Answer: D
(1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), or \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \leq 0\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).
So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.
(2) \(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \leq 0\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.
Answer: D
Bunuel,
Why in (1) you made X-2 = 0? I didn't understand why LHS should be the least value.
BR,
