Question

If n is an integer, is n^{3} divisible by 54?

1. n^{2} is divisible by 6.

2. n^{3} is divisible by 36.

Correct Answer : Choice D. Each statement is independently sufficient to answer the question.

Explanatory Answer

For “is” questions in DS, we need to answer with a clear Yes or a clear No. If the data in the statements does not lead to arriving at a definite Yes or No, the data is insufficient.

We know from the question stem that n is an integer.

I. Let us look at statement 1 alone : n^{2} is divisible by 6.

If n^{2} is divisible by 6, then n^{2} is divisible by both 2 and 3 – the prime factors of 6.

But, we know that n is an integer.

Therefore, n^{2} will be of the form p_{1}^{a} * p_{2}^{b}, where p_{1} and p_{2} are prime factors of n and a and b are even.

Hence, we can deduce that when n^{2} is expressed in terms of its prime factors, the power of 2 and 3 in it will be even.

So, n^{2} will be divisible by both 2^{2} and 3^{2}. Hence, n is divisible by 2 and 3.

If n is divisible by 2 and 3, then n^{3} will be divisible by 2^{3} and 3^{3} or by 216.

If n^{3} is divisible by 216, it will certainly be divisible by any factor of 216 – and therefore by 54.

Statement 1 alone is sufficient.

Answer is either choice A or choice D.

II. Let us look at statement 2 alone : n^{3} is divisible by 36.

For integer n, when n^{3} is expressed in terms of its prime factors, the powers of the prime factors will be multiples of 3.

So, if n^{3} is divisible by 36 or 2^{2} * 3^{2}, we can deduce that n^{3} is actually divisible by 2^{3} and 3^{3} as the power of 2 and 3 should be a multiple of 3.

If n^{3} is divisible by 2^{3} and 3^{3}, it is divisible by 2^{3} and 3^{3} or by 216.

Statement 2 alone is sufficient.

Because each of these statements is independently sufficient, Choice D is the answer.

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