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17 Nov 2021, 03:13
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If \(x\) is a non-zero integer, is \(x\) a prime number?
(1) \(|x|^{|x|} = 4\)
(2) \(|x^x| = x^2\)
(1) \(|x|^{|x|} = 4\)
(2) \(|x^x| = x^2\)
17 Nov 2021, 03:13
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Official Solution:
If \(x\) is a non-zero integer, is \(x\) a prime number?
(1) \(|x|^{|x|} = 4\)
\(x=2\) is an obvious solution but, because of the modulus, \(x=-2\) will also work. In the first case \(x\) is a prime but in the second case it's not. Not sufficient.
(2) \(|x^x| = x^2\)
Test numbers:
If \(x=1\), then \(|1^1| = 1^2\). OK.
If \(x=-1\), then \(|(-1)^{(-1)}| = (-1)^2\). OK.
If \(x=2\), then \(|2^2| = 2^2\). OK.
If \(x=-2\), then \(|(-2)^{(-2)}| =\frac{1}{4}\) but \((-2)^2=4\). NOT OK.
If \(x > 2\), then \(|x^x| > x^2\) and if \(x < -2\), then \(|x^x| < x^2\) (\(|x^x|\) will be a fraction between 0 and 1, while \(x^2\) will be positive integer).
So, \(x=2=prime\) or \(x=1\neq prime\) or \(x=-1\neq prime\).
Not sufficient.
(1)+(2) Only \(x=2\) satisfies both statements, so \(x\) is a prime number. Not sufficient.
Answer: C
If \(x\) is a non-zero integer, is \(x\) a prime number?
(1) \(|x|^{|x|} = 4\)
\(x=2\) is an obvious solution but, because of the modulus, \(x=-2\) will also work. In the first case \(x\) is a prime but in the second case it's not. Not sufficient.
(2) \(|x^x| = x^2\)
Test numbers:
If \(x=1\), then \(|1^1| = 1^2\). OK.
If \(x=-1\), then \(|(-1)^{(-1)}| = (-1)^2\). OK.
If \(x=2\), then \(|2^2| = 2^2\). OK.
If \(x=-2\), then \(|(-2)^{(-2)}| =\frac{1}{4}\) but \((-2)^2=4\). NOT OK.
If \(x > 2\), then \(|x^x| > x^2\) and if \(x < -2\), then \(|x^x| < x^2\) (\(|x^x|\) will be a fraction between 0 and 1, while \(x^2\) will be positive integer).
So, \(x=2=prime\) or \(x=1\neq prime\) or \(x=-1\neq prime\).
Not sufficient.
(1)+(2) Only \(x=2\) satisfies both statements, so \(x\) is a prime number. Not sufficient.
Answer: C
