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If m^2/18 and n^3/80 are integers, where m and n are positive integers
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16 Jul 2023, 23:04
16 Jul 2023, 23:04
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If \(\frac{m^2}{18}\) and \(\frac{n^3}{80}\) are integers, where m and n are positive integers, what is the remainder when product of m and n is divided by 120?
A. 0
B. 2
C. 3
D. 5
E. None of the above
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\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
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A. 0
B. 2
C. 3
D. 5
E. None of the above
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
_________________
Re: If m^2/18 and n^3/80 are integers, where m and n are positive integers
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17 Jul 2023, 02:52
17 Jul 2023, 02:52
1
I just chose the best numbers which can satisfy the above expression to be an integer i.e. m=18 and n=80.
SO (18 x 80) is a multiple of 120 therefore dividing 18x80 by 120 will result remainder as 0.
SO (18 x 80) is a multiple of 120 therefore dividing 18x80 by 120 will result remainder as 0.
gmatclubot
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