'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

ScottTargetTestPrep · Accepted 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

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.

firas92 · 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)

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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.

Stella5200 · Answer

1. Since X^3 is div by 27 and 27 has three 3's, fair to say that x^3 has at least three 3's.

2. which means that x has at least one 3.

3. Which means that the number x can take the form of 3a (where 3 is that from pt 2 and a which is any natural number)

4. you get remainders bigger than 0 only after you cross 27, so at a = 10, 11, 12, we will notice that remianders are 3,6,9.

5. So we can see that the remainders are also a a multiple of 3.

6. So remainder has to be a multiple of 3 and less than 27.

Only Option A = 12.

Hope this clarifies above confusion. Thanks.

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

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GMAT Problem Solving (PS) Questions

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

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