Is positive integer n – 1 a multiple of 3?

In a Yes/No Data Sufficiency question, statement(s) is sufficient if the answer is “always yes” or “always no” while a statement(s) is insufficient if the answer is "sometimes yes" and "sometimes no".

The question asks whether n-1 is a multiple of 3, and from (2) we have a definite NO answer to this question, so this statement is sufficient.

Hope it's clear.

Bumping for review and further discussion.

(1): Pick numbers. If n=5 then 5³-5 = 120 = multiple of 3, but n-1 = 4 no multiple of 3. And 4³-4 = 60 = multiple of 3 and 4-1 = 3 which is a multiple of 3. Insufficient. This will stay IS so the answer will be B or E.

(2) This is a bit trickier. First, simplify the expression:
n³+2n²+n = n(n²+2n+1) = n(n+1)² --> Multiple of 3. For this to be a multiple of 3, EITHER n OR n+1 is a multiple of 3 (both is not possible since they are consecutive integers).
Now pick numbers again: if n=5, then n+1 = 6 = multiple of 3, which satisfies the equation.
If n=3, then n is a multiple of 3, which again satisfies the equation.
Note that in both cases n - 1 is NOT a multiple of 3, which answers the question is n-1 a multiple of 3?

The Answer is B

WKT 0 is a multiple of any number.
(3*0=0)
Fact (1) \(n^3-n=0\)
\(n(n^2-1)=0\)
n=0,1 or -1. When n=0 ans is NO; when n=1 ans is YES. Hence INSUFF
Fact (2) \(n^3+2n^2+n=0\)
\(n(n^2+2n+1)=0\)
\(n(n+1)(n+1)=0\)
n=0,-1. When n=0 ans is NO; when n=-1 ans is NO. Hence SUFF
Ans: B

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is positive integer n – 1 a multiple of 3?

(1) n^3 – n is a multiple of 3

(2) n^3 + 2n^2+ n is a multiple of 3


When you modify the original condition and the question, they become n-1=3t(t is a positive integer)? --> n=3t+1?. There is 1 variable(n), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), it becomes n^3-n=(n-1)n(n+1). The multiplication of three consecutive integers is always a multiple of 6. So, n=3 -> no, n=4 -> yes, which is not sufficient.
For 2), n^3 + 2n^2+ n=3k(k is a positive integer) → n(n+1)^2=3k. In n(n+1)^2=3k, either n=3k or n=3k-1 should be valid. So, it is always no and sufficient.
Therefore, the answer is B.


--> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

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