MathRevolution wrote:
[GMAT math practice question]
(number properties) \(m\) and \(n\) are integers. What is the value \((-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}\)?
\(1) m = n + 1\)
\(2) m = 3\)
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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.
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Since we have \(2\) variables (\(m\) and \(n\)) and \(0\) equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
Since we have \(m = n + 1\) and \(m = 3\), we can substitute \(m = 3\) into \(m = n + 1\) to get \(3 = n + 1\) and \(n = 2.\)
\((-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}\)
\(=(-1)^{3-2} +(-1)^{3+2} +(-1)^{3*2} +(-1)^{2*2}\)
\(=(-1)^1 +(-1)^5 +(-1)^6 +(-1)^4\)
\(=(-1) + (-1) + 1 + 1 = 0\)
Since both conditions together yield a unique solution, they are sufficient.
Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.
Condition 1)
Since \(m = n + 1\), \(m\) and \(n\) are consecutive integers, they have different parities, which means that if \(m\) is an odd integer, then \(n\) is an even integer, and if \(m\) is an even integer, then \(n\) is an odd integer.
Case 1: \(m\) is an odd integer and \(n\) is an even integer.
Then, \(m+n\) is an odd integer, \(m – n\) is an odd integer, \(mn\) is an even integer and \(2n\) is an even integer.
\((-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}\)
\(=(-1)^{odd} +(-1)^{odd} +(-1)^{even} +(-1)^{even}\)
\(=(-1) + (-1) + 1 + 1 = 0\)
Case 2: \(m\) is an even integer and \(n\) is an odd integer.
Then, \(m+n\) is an odd integer, \(m – n\) is an odd integer, \(mn\) is an even integer and \(2n\) is an even integer.
\((-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}\)
\(=(-1)^{odd}+(-1)^{odd} +(-1)^{even} +(-1)^{even}\)
\(=(-1) + (-1) + 1 + 1 = 0\)
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
Case 1: \(n\) is an even integer.
Then, \(m+n\) is an odd integer, \(m – n\) is an odd integer, \(mn\) is an even integer and \(2n\) is an even integer, since \(m = 3.\)
\((-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}\)
\(=(-1)^{odd} +(-1)^{odd} +(-1)^{even} +(-1)^{even}\)
\(=(-1) + (-1) + 1 + 1 = 0\)
Case 2: \(n\) is an odd integer.
Then, \(m+n\) is an even integer, \(m – n\) is an even integer, \(mn\) is an odd integer and \(2n\) is an even integer.
\((-1)^{m-n} +(-1)^{m+n} +(-1)^{mn} +(-1)^{2n}\)
\(=(-1)^{even} +(-1)^{even} +(-1)^{odd} +(-1)^{even}\)
\(=1 + 1 + (-1) + 1 = 2\)
Since condition 2) does not yield a unique solution, it is not sufficient.
If the question has both C and A as its answer, then A is an answer rather than C by the definition of DS questions. Also, this question is a 50/51 level question and can be solved by using the Variable Approach and the relationship between Common Mistake Type 3 and 4 (A or B).
Therefore, A is the answer.
Answer: A
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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