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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 > 1, is integer j a prime number?

(1) When j is divided by 3, the remainder is 1.

(2) When j is divided by 2, the remainder is 1.

In the original condition, there is 1 variable(x), which should match with the number of equation. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), j=7 -> yes, j=25 -> no, which is not sufficient.
For 2), =7-> yes, j=25 -> no, which is not sufficient.
When 1) & 2), j=7 -> yes, j=25 -> no, which is not sufficient. Therefore, the answer is E.


 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Statement 1 - J could be 4, 7, 10, 13, 16, 19, 22, 25, 28 (keep adding 3) . NOT SUFF.
Statement 2 - J could be 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 (keep adding 2). NOT SUFF.
Statement 1 and 2 - J could be 7, 13, 19 and 25 (which is not a prime number) . NOT SUFF.

Hope it helps.
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The general expression for J will be 6P+1 and it may be a prime or non prime hence E
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Statement 1
j could be : ,4,7,10,13,16,19,22,25
non-prime and prime

Statement 2:
J could be: , 3, 5, 7, 9,11,13,15,17,19,21,23,25
non-prime and prime

Combined, j could be 7 or 25 from what we know - prime or non prime

Insufficient
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What is confusing about this problem is that all primes, besides 2 & 3, will be a product of 6n+1 or 6n-1. However, given that E is the correct answer, we should notice that all numbers that are products of 6n+1 or 6n-1 will not always be prime.
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Possible trap associated with this ques :
Quote:
Necessary condition but NOT sufficient condition for a prime number: All primes must be able to be represented at 6n±1 (other than 2 or 3), but not all integers of the form 6n±1 are prime.
After combining we can say that j is of form 6k+1. However, it is not a sufficient condition to prove that j is a prime number.
shasadou
If j > 1, is integer j a prime number?

(1) When j is divided by 3, the remainder is 1.

(2) When j is divided by 2, the remainder is 1.
­
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