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If \(xy + x\) is a prime number, is \(xy + x\) a multiple of 3?
(1) \(xy = 3\)
(2) \(x = 2\)
(1) \(xy = 3\)
(2) \(x = 2\)
08 Nov 2021, 07:22
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Official Solution:
If \(xy + x\) is a prime number, is \(xy + x\) a multiple of 3?
There is only one prime which is a multiple of 3, 3 itself. So, the question basically asks whether \(xy + x=3\).
(1) \(xy = 3\)
The question becomes: does \(3+ x=3\)? Or: does \(x=0\)? If \(x\) were 0, then \(xy + x\) would be 0, which is NOT a prime number as given in the stem. So, \(x \neq 0\) and thus \(xy + x \neq 3\). Sufficient.
(2) \(x = 2\)
The question becomes: does \(2y+ 2=3\)? Or: does \(y=\frac{1}{2}\)? Well, we don't know that. \(y\) could be \(\frac{1}{2}\) but it might as well be some other number making \(xy + x\) a prime number, for example \(\frac{3}{2}\) or \(\frac{5}{2}\) or ... Not sufficient.
Answer: A
If \(xy + x\) is a prime number, is \(xy + x\) a multiple of 3?
There is only one prime which is a multiple of 3, 3 itself. So, the question basically asks whether \(xy + x=3\).
(1) \(xy = 3\)
The question becomes: does \(3+ x=3\)? Or: does \(x=0\)? If \(x\) were 0, then \(xy + x\) would be 0, which is NOT a prime number as given in the stem. So, \(x \neq 0\) and thus \(xy + x \neq 3\). Sufficient.
(2) \(x = 2\)
The question becomes: does \(2y+ 2=3\)? Or: does \(y=\frac{1}{2}\)? Well, we don't know that. \(y\) could be \(\frac{1}{2}\) but it might as well be some other number making \(xy + x\) a prime number, for example \(\frac{3}{2}\) or \(\frac{5}{2}\) or ... Not sufficient.
Answer: A
