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

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?

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

firas92 · Accepted Answer

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

ScottTargetTestPrep · Answer

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

wishmasterdj · Answer

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

firas92 · Answer

Because x and 27 are both multiples of 3

GmatPoint · Answer

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.

Sevil92 · Answer

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.

Bunuel · Answer

'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.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards