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If \(m\) has the smallest prime number as its only prime factor, is Cube\(\sqrt{m}\) an integer?

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If \(m\) has the smallest prime number as its only prime factor, is Cube\(\sqrt{m}\) an integer?

by M7MBA » Sun Jan 17, 2021 9:10 am

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If \(m\) has the smallest prime number as its only prime factor, is Cube\(\sqrt{m}\) an integer?

(1) \(m^2\) is divisible by \(32.\)

(2) \(\sqrt{m}\) is divisible by \(4.\)

Answer: E

Source: e-GMAT
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