Is positive integer n - 1 a multiple of 3?
(1) n3 - n is a multiple of 3
(2) n3 + 2n2+ n is a multiple of 3
OA: B
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Multiple of 3?
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nipunranjan
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Case 1:
n^3-n=n(n^2-1)=n(n+1)(n-1)
From the above equation either n is a multiple of 3 or n-1 is a multiple of 3 or n+1 is a multiple of 3--->Insufficient
Case 2:
n^3+2n^2+n=n(n^2+2n+1)=n(n+1)^2=n(n+1)(n+1)
Either n or n+1 is a multiple of 3--->If n is a multiple of 3 n-1 is not a multiple of 3 and if n+1 is a multiple of 3 then n-1 is also not a multiple of 3 since multiple of 3s are 3 integers apart.
Hence statement 2 is sufficient. OA is B
Cheers
n^3-n=n(n^2-1)=n(n+1)(n-1)
From the above equation either n is a multiple of 3 or n-1 is a multiple of 3 or n+1 is a multiple of 3--->Insufficient
Case 2:
n^3+2n^2+n=n(n^2+2n+1)=n(n+1)^2=n(n+1)(n+1)
Either n or n+1 is a multiple of 3--->If n is a multiple of 3 n-1 is not a multiple of 3 and if n+1 is a multiple of 3 then n-1 is also not a multiple of 3 since multiple of 3s are 3 integers apart.
Hence statement 2 is sufficient. OA is B
Cheers
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Brent@GMATPrepNow
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Target question: Is positive integer n-1 a multiple of 3?nipunranjan 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
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, 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
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Brent
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S1::
n³ - n = n(n² - 1) = n(n+1)(n-1)
Since (n-1) is an integer, n and (n+1) are integers too. This n³ - n is the product of three consecutive integers, and hence ALWAYS divisible by 3. (If I'm multiplying three consecutive integers, ONE of those integers is a multiple of 3, so the product is also a multiple of 3.)
This not only isn't sufficient, it's worthless, as this is true for every integer (n - 1). Since this statement tells us nothing, the answer is either B or E.
S2::
n³ + 2n² + n = n(n² + 2n + 1) = n(n+1)²
If this is a multiple of 3, then either n or (n+1) is a multiple of 3.
If n is a multiple of 3, note that (n - 3) is also a multiple of 3. Since (n - 1) = (n - 3) + 2, (n - 1) is not a multiple of 3: it is 2 greater than a multiple of 3, so its remainder when divided by 3 will be 2.
If (n + 1) is a multiple of 3, so is (n + 1) - 3, or (n - 2). Since (n - 1) = (n - 2) + 1, (n - 1) isn't a multiple of 3: it is 1 greater than a multiple of 3, so its remainder when divided by 3 will be 1.
In either case, (n-1) isn't a multiple of 3, so statement 2 is sufficient.
n³ - n = n(n² - 1) = n(n+1)(n-1)
Since (n-1) is an integer, n and (n+1) are integers too. This n³ - n is the product of three consecutive integers, and hence ALWAYS divisible by 3. (If I'm multiplying three consecutive integers, ONE of those integers is a multiple of 3, so the product is also a multiple of 3.)
This not only isn't sufficient, it's worthless, as this is true for every integer (n - 1). Since this statement tells us nothing, the answer is either B or E.
S2::
n³ + 2n² + n = n(n² + 2n + 1) = n(n+1)²
If this is a multiple of 3, then either n or (n+1) is a multiple of 3.
If n is a multiple of 3, note that (n - 3) is also a multiple of 3. Since (n - 1) = (n - 3) + 2, (n - 1) is not a multiple of 3: it is 2 greater than a multiple of 3, so its remainder when divided by 3 will be 2.
If (n + 1) is a multiple of 3, so is (n + 1) - 3, or (n - 2). Since (n - 1) = (n - 2) + 1, (n - 1) isn't a multiple of 3: it is 1 greater than a multiple of 3, so its remainder when divided by 3 will be 1.
In either case, (n-1) isn't a multiple of 3, so statement 2 is sufficient.
