[GMAT math practice question]
(number properties) p and q are different positive integers. What is the remainder when p^2 + q^2 is divided by 4?
1) p and q are prime numbers.
2) p and q are not consecutive integers.
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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.
Since we have 2 variables (p and q) and 0 equations, C is most likely to be 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 p and q are prime numbers, which are not consecutive integers, p and q are odd integers.
So, both p^2 and q^2 have remainder 1 when they are divided by 4.
Thus, p^2 + q^2 has remainder 2 when it is divided by 4.
Since conditions 1) & 2) yield a unique solution, when they are applied together, 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)
If p = 2 and q = 3, then p^2 + q^2 = 4 + 9 = 13, which has remainder 1 when it is divided by 4.
If p = 3 and q = 5, then p^2 + q^2 = 9 + 25 = 34, which has remainder 2 when it is divided by 4.
Condition 1) is not sufficient since it doesn’t yield a unique solution.
Condition 2)
If p = 3 and q = 5, then p^2 + q^2 = 9 + 25 = 34, which has remainder 2 when it is divided by 4.
If p = 3 and q = 6, then p^2 + q^2 = 9 + 36 = 45, which has remainder 1 when it is divided by 4.
Condition 2) is not sufficient since it doesn’t 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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