MathRevolution wrote:
[GMAT math practice question]
(Number Properties) \(a, b, \)and \(c\) are \(3\) different unit numbers. What is the \(3\)-digit number \(abc\)?
1) The \(5\)-digit number \(ababc\) is a multiple of \(12\).
2) The \(2\)-digit number \(ab\) is equal to \(c^2\).
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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 \(3\) variables (\(a, b\), and \(c\)) and \(0\) equations, E 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 the \(5\)-digit number \(ababc\) is a multiple of \(12\), it is a multiple of both \(3\) and \(4\). It means we have \(a + b + a + b + c = 2a + 2b + c\), which is a multiple of \(3\), and \(10b + c\) is a multiple of \(4\) since we can check the multiplicity of \(3\) with the sum of all digits and the multiplicity of \(4\) with the last two digits.
Since the \(2\)-digit number ab is equal to \(c^2\), we have \(10a + b = c^2.\)
Then, the possible solutions of (\(a, b, c\)) are (\(1, 6, 4\)), (\(4, 9, 7\)), (\(6, 4, 8\)), (\(8, 1, 9\)) from condition 2) since \(a, b\), and \(c\) are different and \(a\) is not equal to \(0\).
When we apply condition 1), we get \(a = 1, b = 6\) and \(c = 4. \) This is because \(a + b + a + b + c\) = multiple of \(3\), and in this case \(1 + 6 + 1 + 6 + 4 = 18\), which is a multiple of \(3\).
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)
\(a = 1, b = 3, c = 2\) and \(a = 7, b = 3, c = 2\) are possible solutions.
Since condition 1) does not yield a unique solution, it is not sufficient.
Condition 2)
\(A = 1, b = 6, c = 4\). and \(a = 4, b = 9, c = 7\) are possible solutions from the above reasoning.
Since condition 2) does not yield a unique solution, it is not sufficient.
Therefore, C is the answer.
Answer: C
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.
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