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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)
Since a, b and c are prime numbers less than 6 with a < b < c, we have a = 2, b = 3 and c = 5.
Thus we have a + b + c = 10.
Since condition 2) provides a unique solution, it is sufficient.
Condition 1)
(ab - 1)(bc - 1)(ca - 1)
= (ab^2c - ab - bc + 1)(ca - 1)
= a^2b^2c^2 - ab^2c - a^2bc - abc^2 + ab + bc + ca - 1
= a^2b^2c^2 - abc(a + b + c) + (ab + bc + ca) - 1
= abc{abc - (a + b + c)} + (ab + bc + ca) - 1
Since (ab - 1)(bc - 1)(ca - 1) is divisible by abc, we notice that ab + bc + ca - 1 is divisible by abc.
Then we have ab + bc + ca - 1 = abc*n for some integer.
When we divide both sides of the equation by abc, we have n = 1/a + 1/b + 1/c - 1/abc < 1/2 + 1/2 + 1/2 = 3/2, since 2 ≤ a < b < c or 1/c > 1/b > 1/a ≥ 1/2.
Then the positive integer n equals 1.
When we divide both sides of the equation ab + bc + ca - 1 = abc by bc, we have a = 1 + a/c + a/b - 1/bc < 1 + 1 + 1 - 1/bc < 3.
Then we have a = 2.
When we substitute a with 2 in the equation ab + bc + ca - 1 = abc, we have 2b + bc + 2c - 1 = 2bc or bc - 2b - 2c + 1 = 0.
Then we have bc - 2b - 2c + 4 = 3 or (b - 2)(c - 2) = 3.
Then we have b - 2 = 1, c - 2 = 3 or b = 3, c = 5.
Thus, we have a + b + c = 10.
Since condition 2) yields a unique solution, it is sufficient.
Therefore, D is the answer.
Answer: D
This question is a CMT4 (B) question: condition 2) is easy to work with, and condition 1) is difficult to work with. For CMT4 (B) questions, D is most likely to be the answer.
