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
(1) n^3 - n is a multiple of 3
(2) n^3 + 2n^2+ n is a multiple of 3
Please don't include the answer to the question in the subject line. This will allow others to attempt the question without seeing the final answer.[email protected] wrote: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
Target question: Is positive integer n-1 a multiple of 3?
Statement 1: n^3 - n is a multiple of 3
Factor: n^3 - n = n(n^2 - 1) = n(n-1)(n+1) = (n-1)(n)(n+1)
Notice that n-1, n and n+1 are three consecutive numbers.
IMPORTANT: Statement 1 is simply telling us that the product of 3 consecutive integers is divisible by 3. This is not new information. The product of any 3 consecutive integers will always be divisible by 3. In fact, there's a rule that says, "The product of n consecutive integers is divisible by n, n-1, n-2, . . . 2, and 1"
Since statement 1 is just some rule that already exists in mathematics, we already knew this information before we even examined statement 1. So, there's no way that statement 1 could possibly add any information to help us answer the target question.
As such, statement 1 is NOT SUFFICIENT
Statement 2: n^3 + 2n^2+ n is a multiple of 3
Factor: n^3 + 2n^2+ n = n(n^2 + 2n + 1) = n(n+1)(n+1)
This means that EITHER n is a multiple of 3 OR n+1 is a multiple of 3.
Let's examine both possible cases:
case a: If n is a multiple of 3, then we can find other multiples of 3 by adding or subtracting multiples of 3 to n. So, for example, n+3 and n+6 will be also be multiples of 3. Likewise, n-3 and n-6 will be also be multiples of 3. Since n-1 is just 1 less than n, n-1 cannot be a multiple of 3 .
case b: If n+1 is a multiple of 3, then n-1 cannot be a multiple of 3 , Since n-1 is just 2 less than n+1.
Since both possible cases yielded the same answer to the target question, statement 2 is SUFFICIENT
Answer = B
Cheers,
Brent
