MathRevolution wrote:
[Math Revolution GMAT math practice question]
(number property) When \(n\) is a positive integer, is \(\frac{n}{4}\) an integer?
1) \(n - 1\) is not divisible by \(2\)
2) \(n + 1\) is not divisible by \(2\)
=>
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.
Modifying the question:
Asking whether \(\frac{n}{4}\) is an integer is equivalent to asking whether \(m\) is a multiple of \(4\).
Since we have \(1\) variable (\(n\)) and \(0\) equations, D is most likely to be the answer. So, we should consider each condition on its own first.
Condition 1)
Since \(n – 1\) is not divisible by \(2, n – 1\) is an odd number and \(n\) is an even number.
If \(n = 4\), then \(n\) is a multiple of \(4\) and the answer is ‘yes’.
If \(n = 2,\) then \(n\) is not a multiple of \(4\) and the answer is ‘no’.
Thus, condition 1) is not sufficient, since it does not yield a unique solution.
Condition 2)
Since \(n + 1\) is not divisible by \(2, n + 1\) is an odd number and \(n\) is an even number.
If \(n = 4\), then \(n\) is a multiple of \(4\) and the answer is ‘yes’.
If \(n = 2\), then \(n\) is not a multiple of \(4\) and the answer is ‘no’.
Thus, condition 2) is not sufficient, since it does not yield a unique solution.
Conditions 1) & 2):
If \(n = 4,\) then neither \(n – 1\) nor \(n + 1\) is divisible by \(2\), but \(n\) is a multiple of \(4\) and the answer is ‘yes’.
If \(n = 2\), then neither \(n – 1\) nor \(n + 1\) is divisible by \(2\), and \(n\) is not a multiple of \(4\) and the answer is ‘no’.
Thus, both conditions together are not sufficient, since they do not yield a unique solution.
Therefore, the answer is E.
Answer: E
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.