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kiran120680
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If m has the smallest prime number as its only prime factor, is Updated on: 20 Apr 2019, 10:17
Originally posted by kiran120680 on 19 Apr 2019, 08:48.
Last edited by kiran120680 on 20 Apr 2019, 10:17, edited 1 time in total.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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85% (hard)

Question Stats:

46% (02:33) correct 54% (02:18) wrong based on 197 sessions

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If m has the smallest prime number as its only prime factor, is Cube√m an integer?

(1) m^2 is divisible by 32

(2) √m is divisible by 4
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E
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If m has the smallest prime number as its only prime factor, is 20 Apr 2019, 01:47
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kiran120680
If m has the smallest prime number as its only prime factor, is Cube√m an integer?

(1) m2 is divisible by 32

(2) √m is divisible by 4
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\(\sqrt[3]{m}\), where \(m=2^x\)..
So, \(\sqrt[3]{2^x}\) will be an integer when x is multiple of 3..

(1) \(m^2\) is divisible by 32 or \(2^5\).
So if x is 10, answer is no, but if x is 15 answer is yes.

(2) √m is divisible by 4
So m is multiple of 16 or 2^4..
So if x is 20, answer is no, but if x is 24 answer is yes

Combined..
So if x is 20, answer is no, but if x is 60, answer is yes
Insufficient

E
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If m has the smallest prime number as its only prime factor, is 20 Apr 2019, 10:15
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kiran120680
If m has the smallest prime number as its only prime factor, is Cube√m an integer?

(1) m2 is divisible by 32

(2) √m is divisible by 4
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Hope Statment 1 worngly mentioned.. please do correct.
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If m has the smallest prime number as its only prime factor, is 20 Apr 2019, 10:17
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mangamma
kiran120680
If m has the smallest prime number as its only prime factor, is Cube√m an integer?

(1) m2 is divisible by 32

(2) √m is divisible by 4

Hope Statment 1 worngly mentioned.. please do correct.
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Edited.. Thank you :thumbup:
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If m has the smallest prime number as its only prime factor, is 20 Apr 2019, 18:40
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I didn't understand the question stem.Can someone explain this?
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If m has the smallest prime number as its only prime factor, is 28 Feb 2023, 04:33
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kiran120680
If m has the smallest prime number as its only prime factor, is Cube√m an integer?

(1) m^2 is divisible by 32

(2) √m is divisible by 4
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Solution:
Pre Analysis:
  • m has the smallest prime number as its only prime factor
  • This means that upon prime factorizing, we can write \(m=2^n\)
  • We are asked if \(\sqrt[3]{x}=\sqrt[3]{2^n}=2^{\frac{n}{3}}\) is integer or not
  • For \(2^{\frac{n}{3}}\) to be integer, n has to be multiple of 3

Statement 1: m^2 is divisible by 32
  • According to this statement, \(m^2=32k\)
    \(⇒(2^n)^2=2^5\times k\)
    \(⇒2^{2n}=2^5 k\)
    \(⇒2^{2n-5}=k\)
  • The value of n can be 6 (multiple of 3) or 8 (not multiple of 3) as well
  • Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: √m is divisible by 4
  • According to this statement, \(\sqrt{m}=4q\)
    \(⇒\sqrt{2^n}=2^2 q\)
    \(⇒2^{\frac{n}{2}}=2^2 q\)
    \(⇒2^{\frac{n}{2}-2}=q\)
  • The value of n can be 6 (multiple of 3) or 8 (not multiple of 3) as well
  • Thus, statement 2 alone is also not sufficient and we can eliminate option B

Combining:
  • Even after combining, we can still have the same 2 possible values n = 6 or 8

Hence the right answer is Option E
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