The Ultimate Q51 Guide [Expert Level]

[GMAT math practice question]

(Number Properties) a, b and c are positive integers. Is 2(a^4+b^4+c^4) a perfect square?

1) a + b + c = 0
2) abc = 6

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

Condition 2)
Condition 2) tells us that abc = 6. For three numbers to multiply to 6, the numbers must be 1, 2, and 3 and the order does not matter. We can substitute these numbers into the given equation to get:
2(a^4 + b^4 + c^4) = 2(1^4 + 2^4 + 3^4) = 2(1 + 16 + 81) = 2*98 = 196 = 14^2, which is a perfect square.
Thus condition 2) is sufficient.

Condition 1)
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
Then, we have a^2 + b^2 + c^2 = -2(ab + bc + ca) since a + b + c = 0.
We have (a^2 + b^2 + c^2)^2 = (-2(ab + bc + ca))2 when we square both sides.
Its left-hand side is
(a^2 + b^2 + c^2)^2
= (a^2 + b^2 + c^2) (a^2 + b^2 + c^2)
= a^4 + a^2b^2 +a^2c^2 + a^2b^2 + b^4 + b^2c^2 + a^2c^2 + b^2c^2 + c^4
= a^4 + b^4 + c^4 + 2a^2b^2 + 2b^2c^2 + 2c^2a^2.

Its right-hand side is
(-2(ab + bc + -ca))^2
= 4(ab + bc + ca)(ab + bc + ca)
= 4(a^2b^2+ ab^2c + a^2bc + ab^2c + b^2c^2 + abc^2 + a^2bc + abc^2 + c^2a^2
= 4(a^2b^2 + b^2c^2 + c^2a^2 + 2a^2bc + 2ab^2c + 2abc^2)
= 4(a^2b^2 + b^2c^2 + c^2a^2) + 2abc(a + b + c))
= 4(a^2b^2 + b^2c^2 + c^2a^2), since a + b + c = 0
We have a^4 + b^4 + c^4 = 2(a^2b^2 + b^2c^2 + c^2a^2).
Then 2(a^4 + b^4 + c^4) = 4(a^2b^2 + b^2c^2 + c^2a^2) = (-2(ab + bc + ca))^2
Thus 2(a^4 + b^4 + c^4) is a square of -2(ab + bc + ca).
We realize we have only used condition 1), but haven’t used condition 2).
It means condition 1) alone is sufficient.
Thus, condition 1) is sufficient.

Therefore, D is the answer.
Answer: D

This question is a CMT 4(B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C, or D.