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[Math Revolution GMAT math practice question]

When A and B are positive integers, is AB a multiple of 4?

1) The greatest common divisor of A and B is 6
2) The least common multiple of A and B is 30

Target question: Is AB a multiple of 4?

Useful property: If N is divisible by d, we can say that N = dk for some integer k
For example, if N is divisible by 5, we can say that N = 5k for some integer k


Statement 1: The greatest common divisor of A and B is 6
This means that A is divisible by 6, and B is divisible by 6.
Applying the above property, we can write A = 6k for some integer k, and B = 6j for some integer j
So, AB = (6k)(6j) = 36kj = (4)(9)(kj)
Aha! We can see that AB is a multiple of 4
So, the answer to the target question is YES, AB IS a multiple of 4
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The least common multiple of A and B is 30
There are several values of A and y that satisfy statement B. Here are two:
Case a: A = 10 AND B = 30 (the LCM of 10 and 30 is 30). In this case, AB = (10)(30) = 300. So, the answer to the target question is YES, AB IS a multiple of 4
Case b: A = 1 AND B = 30 (the LCM of 1 and 30 is 30). In this case, AB = (1)(30) = 30. So, the answer to the target question is NO, AB is NOT a multiple of 4
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Asking if \(AB\) is a multiple of \(4\) is equivalent to asking if \(AB = 4k\) for some integer k.

Condition 1)
Since \(A = 6a = 2*3\)*a and \(B = 6b = 2*3*b\) for some integers \(a\) and \(b\), \(AB = 2^2*3^2*ab = 4*3^2ab.\)
Thus, \(AB\) is a multiple of \(4\) and condition 1) is sufficient.

Condition 2)
If \(A = 6\) and \(B = 5\), then \(lcm(A,B) = 30\), but \(AB = 30\) is not a multiple of \(4\), and the answer is ‘no’.
If \(A = 6\) and \(B = 10\), then \(lcm(A,B) = 30\), and \(AB = 60\) is a multiple of \(4\). The answer is ‘yes’.
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A
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