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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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24 Mar 2019, 09:24
00:00

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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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

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

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

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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