'x' is a Positive Integer and x^3 is Divisible by 27 GMAT Problem Solving

byRituparna Nath Content Writer at Study Abroad Exams

Question: '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?

'x' is a positive integer and x^3 is divisible by 27?’ - is a topic of the GMAT Quantitative reasoning section. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have good knowledge of qualitative skills. The GMAT Quant section consists of 31 questions in total. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:
Given in the question that x is a positive integer and x^3 is divisible by 27. It is asked which of the given options can be the remainder when x is divided by 27.
We know that \(x^3\)is divisible by 27
\(x^3\)= k. 27
Where k is some integer.
Taking cube root on both sides
x = \(\sqrt[3]{k}\)* 3
Therefore it can be clearly seen that x is a multiple of 3 or we can say x has one of the factors = 3
We also know that the remainder is always less than the divisor.
Therefore 30 and 33 cannot be the remainder when x is divided by 27
Now the only number among 12,16 and 20 which is divisible by 3 and less than 27 is 12
Therefore 12 could be the remainder when x is divided by 27
The correct option is A.

Correct Answer: A

Approach Solution 2:

Explanation:
It is important to know that the remainder is always less than the divisor.
Therefore option D and E can be eliminated as they are greater than 27
As \(x^3\)is divisible by 27
X must have a factor of 3
Since 12 + 27 = 39 and 39^3 = (13 * 3)^3 = 3^3 * 13^3 = 27 * (13^3)
Is a multiple of 27
Since 39 / 7 gives remainder 12
Correct answer is 12, option A.

Correct Answer: A

Approach Solution 3:

It is given in the question that x^3 is divisible by 27.
So, x^3 is also divisible by 3 (x^3 is a multiple of 3);

If x^3 is a multiple of 3 (and x is an integer), then x must be a multiple of 3

Solving the equation,

x=27q+rx=27q+r
That implies 3k=27q+r3k=27q+r
That implies 3(k−9q)=r3(k−9q)=r
That implies r is a multiple of 3.

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

Correct Answer: A

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