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If \(m\) has the smallest prime number as its only prime factor, is \((\sqrt{m})^3\) an integer?

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by VJesus12 » Thu Oct 07, 2021 1:58 am

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If \(m\) has the smallest prime number as its only prime factor, is \((\sqrt{m})^3\) 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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