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Re: If n is an integer, is n^3 divisible by 54? [#permalink]
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Bunuel wrote:
If n is an integer, is n^3 divisible by 54?

(1) n^2 is divisible by 6.
(2) n^3 is divisible by 36.


Given: n is an integer
Required: Is n^3 divisible by 54?
54 = 3^3*2
Hence we need to show that n has at least one multiple of 3 and one multiple of 2

Statement 1: n^2 is divisible by 6.
This means n has one multiple of 3 and one multiple of 2
SUFFICIENT

Statement 2: n^3 is divisible by 36
36 = 3^2*2^2
Hence N will have at least one multiple of 3 and one multiple of 2
SUFFICIENT

Option D
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Re: If n is an integer, is n^3 divisible by 54? [#permalink]
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Here the basic logic to be applied is => K and K^n have the same prime factors
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Re: If n is an integer, is n^3 divisible by 54? [#permalink]
another way to solve the problem is that use the remainder

St 1: n must be even, and n^2 has no remainder after it divides 3, then n is divisible by 3

St 2: n must be even, and n^3 is divisible not only 9, but also 3, then n is divisible by 3

=> D
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Re: If n is an integer, is n^3 divisible by 54? [#permalink]
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