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If n is an integer, is n^3 divisible by 54?

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Bunuel
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If n is an integer, is n^3 divisible by 54? [#permalink] New post   09 Mar 2016, 12:48
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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.
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D
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Re: If n is an integer, is n^3 divisible by 54? [#permalink] New post   13 Mar 2016, 23:14
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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] New post   09 Mar 2016, 21:26
Here is my approach ]
here we need to check if N^3 is divisible by 54 so => N^3 must contain a 2 and three 3's
statement 1 => N^2 is divisible by 6 => 2,3 must be the primes of N^2
now as N^anypower and N have same prime factors => N must have 2 and three as Prime factors which in turn means N^3 will have atleast 3 two's and 3 three's Hence suffficient
Statement 2 => hence N^3 is divisible by 36 => 2,3 must be its prime factors => 2,3 must be the prime factors of N too => sufficient
Hence D is sufficient.
Here the clear trap was statement 2 as N^3 was divisible by 36 ..

I hope i am right...
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Re: If n is an integer, is n^3 divisible by 54? [#permalink] New post   09 Mar 2016, 21:49
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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.


Hi,
A good Q..

Info from Q..


1) VERY IMP- n is an integer..
2) Is n^3 div by 54

Inference--


If n is an integer and n^3 has to be div by 54, or 2*3^3..
so n has to be div by 2*3 as n^3 by 3^3 means n by 3..

Lets see the statements--


(1) n^2 is divisible by 6.
since n is an integer, n^2 by 2*3 will mean even n by 2*3..
meets our INFERENCE
Suff

(2) n^3 is divisible by 36.
n^3 by 2^2*3^2..
again as n is an integer n will be div by 2 and 3..
Again meets our INFERENCE
Suff

D
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Re: If n is an integer, is n^3 divisible by 54? [#permalink] New post   10 Mar 2016, 02:03
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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] New post   16 Apr 2017, 12:18
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]
16 Apr 2017, 12:18
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Bunuel
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