MathRevolution wrote:
[GMAT math practice question]
(Number Property) \(a, b,\) and \(c\) are integers. Is \(2(a^4 + b^4 + c^4)\) a perfect square?
1) \(a = 1, b = 1\), and \(c = -2\)
2) \(a + b + c = 0\)
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Condition 1)
Since we have \(a = 1, b = 1,\) and \(c = 2\), we have
\(2(a^4 + b^4 + c^4) = 2(1^4 + 1^4 + 2^4) \)
\(= 2(1 + 1 + 16) \)
\(= 2*18 \)
\(= 36.\)
\(2(a^4 + b^4 + c^4) = 36\) is a perfect square and the answer is ‘yes’.
Since condition 1) yields a unique solution, it is sufficient.
Condition 2)
Since \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca), \)
we have \(a^2 + b^2 + c^2 = -2(ab + bc + ca).\)
Then, rearranging the second formula gives us:
\((a^2 + b^2 + c^2)^2 = (-2(ab + bc + ca))^2 \)
\(= 4(ab + bc + ca)(ab + bc + ca)\)
\(= 4(a^2b^2 + ab^2c + a^2bc + ab^2c + b^2c^2 + abc^2 + a^2bc + abc^2 + a^2c^2)\)
\(= 4((a^2b^2 + b^2c^2 + a^2c^2 + 2ab^2c + 2a^2bc + 2abc^2) \)
\(= 4(a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c))\)
\(= 4(a^2b^2 + b^2c^2 + c^2a^2), \)since \(a + b + c = 0\)
Following the pattern in the first equation gives us:
\((a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2 + c^2a^2)\)
We now have two equations:
\( (a^2 + b^2 + c^2)^2 = a^4 + b^4 + c^4 + 2(a^2b^2 +b^2c^2+ c^2a^2)\)
\((a^2 + b^2 + c^2)^2 = 4(a^2b^2 + b^2c^2 + c^2a^2)\)
Combining the two equations gives us:
\(a^4 + b^4 + c^4 + 2(a^2b^2 + b^2c^2+ c^2a^2) = 4(a^2b^2 + b^2c^2 + c^2a^2)\)
\(a^4 + b^4 + c^4 = 2(a^2b^2 + b^2c^2 + c^2a^2)\)
\(2(a^4 + b^4 + c^4) = 4(a^2b^2 + b^2c^2 + c^2a^2) = (a^2 + b^2 + c^2)^2.\)
Thus, \(2(a^4 + b^4 + c^4)\) is a perfect square.
Since condition 2) yields a unique solution, it is sufficient.
Therefore, D is the answer.
Answer: D
This question is a CMT 4(B) question: condition 1) is easy to work with, and condition 2) is difficult to work with. For CMT 4(B) questions, D is most likely the answer.
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