Is positive integer n – 1 a multiple of 3?

(1) n^3 – n is a multiple of 3 --> \(n^3-n=n(n^2-1)=(n-1)n(n+1)=3q\). Now, \(n-1\), \(n\), and \(n+1\) are 3 consecutive integers and one of them must be multiple of 3, so no wonder that their product is a multiple of 3. However we don't know which one is a multiple of 3. Not sufficient.

(2) \(n^3 + 2n^2+ n\) is a multiple of 3 --> \(n^3 + 2n^2+ n=n(n^2+2n+1)=n(n+1)^2=3p\) --> so either \(n\) or \(n+1\) is a multiple of 3, as out of 3 consecutive integers \(n-1\), \(n\), and \(n+1\) only one is a multiple of 3 then knowing that it's either \(n\) or \(n+1\) tells us that \(n-1\) IS NOT multiple of 3. Sufficient.

Answer: B.
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(1) \(n^3-n = n(n^2-1) = n(n+1)(n-1)\)
This number will always be a multiple of three ... so not sufficient to answer our question

(2) \(n^3 + 2n^2+ n = n(n+1)^2\)
If this is a multiple of 3, then either n is a multiple of 3 or (n+1) is a multiple of 3. In either case, n-1 cannot be a multiple of 3.
Hence this is sufficient

Answer should be (b)

In your solution above, notice that for all the n's in case (2) n-1 is not a multiple of 3.
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But n-1 isn't mentioned in the 2) so how can we say something about it? how come that if n or n+1 is div by 3 then automatically n-1 is not?..

Out of any 3 consecutive integers only 1 is a multiple of 3. We know it's either n or n + 1, therefore it's not n-1.
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ahhhhh okok! now got it.. omg this is a tricky one....

Thank you Bunuel!

Hello,

What is the approximate level of this question on 800 scale?

Regards.

Check the original post for tags and other stats. It's a 700-level question.
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Target question: Is positive integer n – 1 a multiple of 3?

Statement 1: n³ – n is a multiple of 3
Let's do some FACTORING
n³ – n = n(n² - 1) = n(n + 1)(n - 1)
So, statement 1 tells us that n(n + 1)(n - 1) is a multiple of 3
Notice that (n-1), n and (n+1) are 3 CONSECUTIVE integers, and we know that 1 out of every 3 consecutive integers is a multiple of 3 (e.g., 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, etc.
So, ONE of the following n, (n + 1), or (n - 1) a multiple of 3, but WHICH ONE??
It COULD be the case that (n - 1) IS a multiple of 3, OR it could be the case that (n - 1) is NOT a multiple of 3
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: n³ + 2n² + n is a multiple of 3
More FACTORING
n³ + 2n² + n = n(n² + 2n + 1)
= n(n + 1)(n + 1)
So, statement 2 tells us that n(n + 1)(n + 1) is a multiple of 3
This means that EITHER n is a multiple of 3 OR (n + 1) is a multiple of 3
Let's examine both cases:

Case a: n is a multiple of 3
If n is a multiple of 3, then n - 1 CANNOT be a multiple of 3.
Why not?
Well, we already know that 1 out of every 3 consecutive integers is a multiple of 3
So, if n is a multiple of 3, the n+3 is also a multiple of 3 AND n+6 is a multiple of 3, AND n+9 is a multiple of 3, etc
Likewise, n-3 is a multiple of 3 AND n-6 is a multiple of 3, AND n-9 is a multiple of 3, etc.
Notice that n-1 is NOT among the possible multiples of 3
So, in this case, the answer to the target question is NO, n - 1 is NOT a multiple of 3

Case b: (n + 1) is a multiple of 3
If n+1 is a multiple of 3, then n - 1 CANNOT be a multiple of 3.
We'll use the same logic we used in case a above
1 out of every 3 consecutive integers is a multiple of 3
So, if n+1 is a multiple of 3, the n+1+3 (aka n+4) is also a multiple of 3 AND n+1+6 (aka n+7) is a multiple of 3, AND n+1+9 (aka n+10)is a multiple of 3, etc
Likewise, n+1-3 (aka n-2) is a multiple of 3 AND n+1-6 (aka n-5) is a multiple of 3, AND n+1-6 (aka n-5) is a multiple of 3, etc.
Notice that n-1 is NOT among the possible multiples of 3
So, in this case, the answer to the target question is NO, n - 1 is NOT a multiple of 3

IMPORTANT: for statement 2, there are only two possible cases, and in each case, the answer to the target question is the SAME: NO, n - 1 is NOT a multiple of 3
So, it MUST be the case that n - 1 is NOT a multiple of 3
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
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MANHATTAN PREP EXPLANATION

The question is in very simple form already; rephrasing the question isn't useful. The statements, however, can be rephrased.

Statement (1) gives the formula n3 - n. We can first factor out an n to get n(n2 – 1). Next, we can factor (n2 – 1) to get (n + 1)(n – 1). So, n3 - n factors to n(n + 1)(n – 1). Notice that the three factored terms represent consecutive integers: n – 1, n, and n + 1.

Now, let's rephrase statement (2). We first factor out an n to get n(n2 + 2n + 1). Next, we can factor (n2 + 2n + 1) to get (n + 1)(n + 1). So, n3 + 2n2 + n factors to n(n + 1)(n + 1). Notice that the factored terms represent two consecutive integers, with the larger of the two represented twice.

(1) INSUFFICIENT: (n – 1)(n)(n + 1) is a multiple of 3. Any three consecutive positive integers include exactly one multiple of 3 (if you don't remember this rule, try some real numbers and prove it to yourself). Of the three terms n – 1, n, and n + 1, one is a multiple of 3 but we have no way to determine which one.

(2) SUFFICIENT: n(n + 1)(n + 1) is a multiple of 3. This tells us that either n or n + 1 is a multiple of 3. The question asks whether the term n – 1 is a multiple of 3. Recall that n – 1, n, and n + 1 represent three consecutive integers and also recall that any three consecutive integers include exactly one multiple of 3. Note that we do not need to get this information from statement (1). If the multiple of 3 is either the n or the n + 1 term, then the n – 1 term cannot be a multiple of 3.

The correct answer is B.