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If j is a positive integer, is (j^3-27)^2(j^3+1)^3 odd?

Expert replies
by GMATinsight » Mon Oct 12, 2015 11:04 pm
If j is a positive integer, is (j^3-27)^2(j^3+1)^3 odd?

(1) j + 2 is even

(2) 2j is even

Answer: Option A
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Source: — Data Sufficiency |

by GMATinsight » Mon Oct 12, 2015 11:06 pm
GMATinsight wrote:If j is a positive integer, is (j^3-27)^2(j^3+1)^3 odd?

(1) j + 2 is even

(2) 2j is even

Answer: Option A
Question : Is (j^3-27)^2(j^3+1)^3 odd?

To answer this question, all we need is to know whether j is even or Odd

Statement 1: j + 2 is even
i.e. j is even
i.e. (j^3-27)^2(j^3+1)^3 = (Even-27)^2(Even+1)^3 = Odd*Odd = Odd
SUFFICIENT

Statement 1: 2j is even
i.e. j may be Even or Odd
NOT SUFFICIENT

Answer: option A
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by Max@Math Revolution » Wed Oct 14, 2015 5:54 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If j is a positive integer, is (j^3-27)^2(j^3+1)^3 odd?

(1) j + 2 is even
(2) 2j is even

If we modify the original condition and the question according to the Variable Approach Method, we ultimately want to know whether j is an even number as odd x odd=odd, j^3 -27=odd, and j^3+1=odd?
For condition 1, j+2=even, j=even-2=even. So j is an even number. This condition is sufficient.
For condition 2, 2j=even=2m(m=integer), j=integer. So this does not tell whether j is even. This condition is insufficient, and the answer is (A).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

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