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If n is an integer, is (n+1)(n+2)(n+3) divisible by 12? 1) n

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If n is an integer, is (n+1)(n+2)(n+3) divisible by 12? 1) n

by Max@Math Revolution » Mon Feb 25, 2019 11:20 pm

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If n is an integer, is (n+1)(n+2)(n+3) divisible by 12?

1) n is an even number.
2) n is a multiple of 4.
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Source: — Data Sufficiency |
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by Max@Math Revolution » Thu Feb 28, 2019 6:06 am

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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.

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

Since n+1, n+2 and n+3 are three consecutive integers, (n+1)(n+2)(n+3) is a multiple of 3.

Condition 2) tells us that n+1 and n+3 are odd integers, and n+2 is an even number which is not a multiple of 4. Thus, (n+1)(n+2)(n+3) is not a multiple of 4.
CMT(Common Mistake Type 1) states "no" is also an answer and a condition giving rise to the unique answer "no" is sufficient. Thus, condition 2) is sufficient.

Condition 1)
If n = 2, then (n+1)(n+2)(n+3) = 3*4*5 = 60 is a multiple of 12 and the answer is "yes".
If n = 4, then (n+1)(n+2)(n+3) = 5*6*7 = 210 is not a multiple of 12 and the answer is "no".
Thus, condition 1) is not sufficient, since it does not yield a unique solution.

Therefore, B is the answer.
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