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If m has the smallest prime number as its only prime factor, is an in

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If m has the smallest prime number as its only prime factor, is an in  [#permalink]

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New post 24 Mar 2019, 09:24
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

38% (03:05) correct 62% (02:20) wrong based on 13 sessions

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

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

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

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Re: If m has the smallest prime number as its only prime factor, is an in  [#permalink]

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New post 24 Mar 2019, 10:49
PriyankaPalit7 wrote:
If \(m\) has the smallest prime number as its only prime factor, is \(\sqrt[3]{m}\) an integer?

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

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


Given: m has the smallest prime number as its ONLY prime factor
To find: Is \(\sqrt[3]{m}\) an integer? (Yes/No Question)

Statement (1): \(m^2\) is divisible by 32 which means m can be 16, 32, 64, 128 etc.
If m=16=\(2^4\) => smallest prime factor is 2 and only prime factor but \(\sqrt[3]{16}\) is NOT an integer..............NO
If m=64= \(2^6\) => smallest prime factor is 2 and only prime factor and \(\sqrt[3]{64}\) is 4 and it is an integer............YES

Inconsistent answers.......NOT SUFFICIENT!


Statement (2): \(\sqrt{m}\) is divisible by 4 which means m=16, 64....etc
Use example used in statement (1)

Inconsistent answers.......NOT SUFFICIENT!


Statement (1) + (2) gives m=32, 64 etc and by solving these values of n, you get different answers. NOT SUFFICIENT!

Thus, answer is option E
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Re: If m has the smallest prime number as its only prime factor, is an in   [#permalink] 24 Mar 2019, 10:49
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