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07 May 2021, 02:45
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A
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Difficulty:
45%
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Question Stats:
62% (01:27) correct
38%
(01:40)
wrong
based on 264
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If a and n are integers, and a^2 = 24n, then n must be divisible by which of the following?
A. 2
B. 4
C. 12
D. 18
E. 24
A. 2
B. 4
C. 12
D. 18
E. 24
08 May 2021, 07:30
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Bunuel
If a and n are integers, and a^2 = 24n, then n must be divisible by which of the following?
A. 2
B. 4
C. 12
D. 18
E. 24
\(a^2 = 24n=4*6n\);
\(a = 2*\sqrt{6n}\).
From above the least positive value of n is 6 and if it is 6, then it's divisible only by option A (2).
Answer: A.
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Hope this helps.
General Discussion
07 May 2021, 02:52
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Bunuel
a and n are integers and
\(a^2 = 24n = 2^3*3*n\)
i.e n must be a number which when multiplied with \(2^3*3\) should result in a aperfect square
CONCEPT: Perfect squares in prime factorized form have even exponents of all distinct prime factors
i.e. \(2^3*3\) is short of \(2^1*3^1\) to make \(2^3*3*n\) a perfect square
i.e. \(n_{min} = 2*3\)
i.e. among the options, only option A i.e. 2 is an essential factor of n hence
Answer: Option A
