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

1. j+2 is even => j is even
suppose j=4
(j^3-27)(j^3+1)^3
=(64-27)(64+1)^3
=37(65)^3
=Odd x Odd = Odd

2. 2j is even
J may be even or odd
2j=6
j=3

2j=8
j=4

Answer A

IMO A is the answer.

We need to find if J is even or not.

Statement 1 states that j+2 is even so j is also even

Statement 2 states 2j is even where by j could be either even or odd.

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.