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Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Updated on: 03 Nov 2019, 09:23
Originally posted by MathRevolution on 26 May 2017, 04:42.
Last edited by chetan2u on 03 Nov 2019, 09:23, edited 1 time in total.
Last edited by chetan2u on 03 Nov 2019, 09:23, edited 1 time in total.
Corrected the question
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If [m]x[/m] is a positive integer, is [m]6[/m] a factor of [m]x[/m]?
(1) [m]6[/m] is a factor of [m]3x[/m]
(2) [m]12[/m] is a factor of [m]x^{2}[/m]
(1) [m]6[/m] is a factor of [m]3x[/m]
(2) [m]12[/m] is a factor of [m]x^{2}[/m]
26 May 2017, 04:42
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Official Solution:
In the original condition, there is 1 variable \((x)\). In order to match the number of variables to the number of equations, there must be 1 equation. Therefore, D is most likely to be the answer.
In the case of con 1), if \(x = 2\), 6 is not a factor of 2, so no, and if \(x = 6\), you get yes, hence it is not sufficient.
In the case of con 2), from \(x^{2} = 12t\) (\(t\)=any positive integer), you should get \(x = \sqrt{12t} = \sqrt{12(12s^{2})}\) (\(s\)=any positive integer). That is because it is said that \(x\) is a positive integer, so \(x\) should always get rid of the root. If so, from \(x=\sqrt{12(12s^{2})} = \sqrt{12^{2}s^{2}} = \sqrt{(12s)^{2}} = 12s=6(2s)\), 6 is always a factor of \(x\), hence yes, and sufficient. The answer is B.
Answer: B
In the original condition, there is 1 variable \((x)\). In order to match the number of variables to the number of equations, there must be 1 equation. Therefore, D is most likely to be the answer.
In the case of con 1), if \(x = 2\), 6 is not a factor of 2, so no, and if \(x = 6\), you get yes, hence it is not sufficient.
In the case of con 2), from \(x^{2} = 12t\) (\(t\)=any positive integer), you should get \(x = \sqrt{12t} = \sqrt{12(12s^{2})}\) (\(s\)=any positive integer). That is because it is said that \(x\) is a positive integer, so \(x\) should always get rid of the root. If so, from \(x=\sqrt{12(12s^{2})} = \sqrt{12^{2}s^{2}} = \sqrt{(12s)^{2}} = 12s=6(2s)\), 6 is always a factor of \(x\), hence yes, and sufficient. The answer is B.
Answer: B
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