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'x' is a positive integer and x^3 is divisible by 27.

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Intern
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Joined: 13 Oct 2019
Posts: 21
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'x' is a positive integer and x^3 is divisible by 27.  [#permalink]

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New post 16 Jan 2020, 07:49
3
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

61% (01:47) correct 39% (01:58) wrong based on 46 sessions

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'x' is a positive integer and \(x^3\) is divisible by 27. Which of the following could be the remainder when x is divided by 27?

A. 12
B. 16
C. 20
D. 30
E. 33
Director
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Joined: 16 Jan 2019
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WE: Sales (Other)
Re: 'x' is a positive integer and x^3 is divisible by 27.  [#permalink]

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New post 16 Jan 2020, 08:25
3
If \(x^3\) is divisible by \(27\), \(x\) must have at least one 3 in its factorization

And remainder is always less than divisor

The only answer choice with a 3 in factorization and that is less than 27 is 12

Answer is (A)

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Re: 'x' is a positive integer and x^3 is divisible by 27.  [#permalink]

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New post 18 Jan 2020, 12:37
ann1111 wrote:
'x' is a positive integer and \(x^3\) is divisible by 27. Which of the following could be the remainder when x is divided by 27?

A. 12
B. 16
C. 20
D. 30
E. 33


We can eliminate choices D and E since the remainder can’t be greater than or equal to the divisor. Let’s look at the other choices.

A. 12

Since 12 + 27 = 39 and 39^3 = (3 x 13)^3 = 3^3 x 13^3 = 27 x 13^3 is a multiple of 27, x can be 39. Since 39/27 = 1 R 12, the remainder can be 12.

Answer: A
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Re: 'x' is a positive integer and x^3 is divisible by 27.   [#permalink] 18 Jan 2020, 12:37
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