MathRevolution wrote:
[GMAT math practice question]
(number properties) If \(m\) and \(n\) are prime numbers, is \(m^2 + n^2\) an even number?
\(1) m > 10\)
\(2) n > 20\)
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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.
In order for \(m^2 + n^2\) to be even, either both \(m\) and \(n\) must be even or both \(m\) and \(n\) must be odd. Since \(m\) and \(n\) are primes, we must have both \(m\) and \(n\) odd numbers as they are both greater than \(10\). Thus, C is the answer.
Since this question is a statistics 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)
If \(m = 11\) and \(n = 3\), then \(m^2+n^2 = 11^2+3^2 = 121 + 9 = 130\) is an even number, and the answer is “yes”.
If \(m = 11\) and \(n = 2\), then \(m^2+n^2 = 11^2+2^2 = 121 + 4 = 125\) is an odd number, and the answer is “no”.
Thus, condition 1) is not sufficient since it does not yield a unique solution.
Condition 2)
If \(m = 3\) and \(n = 23\), then \(m^2+n^2 = 3^2+23^2 = 9 + 529 = 538\) is an even number, and the answer is “yes”.
If \(m = 2\) and \(n = 23\), then \(m^2+n^2 = 2^2+23^2 = 4 + 529 = 533\) is an odd number, and the answer is “no”.
Thus, condition 2) is not sufficient since it does not yield a unique solution.
Therefore, C is the answer.
Answer: C
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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