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24^n is divisible by 2^10
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06 Jul 2023, 04:15
06 Jul 2023, 04:15
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Question Stats:
82% (01:09) correct
17% (01:32) wrong based on 23 sessions
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\(24^n\) is divisible by \(2^{10}\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE NUMBER PROPERTIES
Quantity A |
Quantity B |
n |
3 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE NUMBER PROPERTIES
24^n is divisible by 2^10
[#permalink]
07 Aug 2023, 09:07
07 Aug 2023, 09:07
Expert Reply
OE
\(24^n\) is divisible by \(24^{10}\). Therefore we must have an integer
\(\frac{(2^3*3)^n}{2^{10}}= \) Integer
\(\frac{2^{3n}*3^n}{2^{10}}\)=Integer
This means that \(2^{3n}\) is also divisible by \(2^{10}\) and \(n\) should not be negative
Therefore \(3n \geq 10\)
\(n \geq 3.\overline{3}\)
So A > B
\(24^n\) is divisible by \(24^{10}\). Therefore we must have an integer
\(\frac{(2^3*3)^n}{2^{10}}= \) Integer
\(\frac{2^{3n}*3^n}{2^{10}}\)=Integer
This means that \(2^{3n}\) is also divisible by \(2^{10}\) and \(n\) should not be negative
Therefore \(3n \geq 10\)
\(n \geq 3.\overline{3}\)
So A > B
gmatclubot
24^n is divisible by 2^10 [#permalink]
07 Aug 2023, 09:07
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