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

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'x' is a positive integer and x^3 is divisible by 27. Which of the [#permalink] New post   16 Jan 2020, 08:49
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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
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firas92
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Re: 'x' is a positive integer and x^3 is divisible by 27. Which of the [#permalink] New post   16 Jan 2020, 09:25
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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. Which of the [#permalink] New post   18 Jan 2020, 13:37
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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. Which of the [#permalink] New post   13 Feb 2021, 08:31
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firas92 wrote:
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)

Posted from my mobile device


Can someone clarify why the remainder has to be a multiple of 3?
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Re: 'x' is a positive integer and x^3 is divisible by 27. Which of the [#permalink] New post   26 Feb 2021, 22:44
wishmasterdj wrote:
firas92 wrote:
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)

Posted from my mobile device


Can someone clarify why the remainder has to be a multiple of 3?


Because x and 27 are both multiples of 3
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Re: 'x' is a positive integer and x^3 is divisible by 27. Which of the [#permalink] New post   25 Jul 2022, 04:45
Since x^3 is divisible by 27, x^3 must be in the form of 3a, where a can be any natural number.

Since the remainder is smaller than the quotient, D and E can be eliminated.

Also, since x is a multiple of 3, when it is divided by 3, the remainder will also be a multiple of 3.

Out of A, B, and C, only A is a multiple of 3,

Thus, the correct option is A.
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Re: 'x' is a positive integer and x^3 is divisible by 27. Which of the [#permalink] New post   05 Oct 2022, 13:49
I understand that x is a multiple of 3, but do not get why the remainder should be a multiple of 3 as well ((. Would be happy if someone explains.
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'x' is a positive integer and x^3 is divisible by 27. Which of the [#permalink] New post   05 Oct 2022, 15:07
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Sevil92 wrote:
I understand that x is a multiple of 3, but do not get why the remainder should be a multiple of 3 as well ((. Would be happy if someone explains.


'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

1. Since x^3 is divisible by 27, then x^3 is also divisible by 3 (x^3 is a multiple of 3);

2. Since x^3 is a multiple of 3 (and x is an integer), then x must be a multiple of 3 (how else would 3 appear in x^3? Exponentiation does not "produce" primes);

3. \(x = 27q + r\) ⇒ \(3k = 27q + r\) ⇒ \(3(k - 9q) = r\) ⇒ r is a multiple of 3.

4. The remainder (r) must be less than the divisor (27), so r cannot be 30 or 33. Thus r = 12.

Answer: A.

Hope it helps.
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Re: 'x' is a positive integer and x^3 is divisible by 27. Which of the [#permalink] New post   20 Dec 2023, 10:01
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Re: 'x' is a positive integer and x^3 is divisible by 27. Which of the [#permalink]
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