Divisibility: Is positive integer n-1 a multiple of 3 ?

[II]

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

[Stuart@KaplanGMAT]

Is positive integer n – 1 a multiple of 3?

(1) n3 – n is a multiple of 3

(2) n3 + 2n^2+ n is a multiple of 3

note: I'm assuming that "n3" = "n^3"

(1) n^3 - n can be simplified to:

n(n^2 - 1) = n(n + 1)(n - 1)

We know that n is an integer (since we're told that n-1 is a positive integer).

Let's rearrange our expression:

(n-1)(n)(n+1) is a multiple of 3. For this to be a multiple of 3, at least 1 of the terms must be a multiple of 3.

However, since n is an integer, (n-1), n and (n+1) are 3 consecutive integers. EVERY set of 3 consecutive integers will contain exactly one multiple of 3. Therefore, statement (1) gives us ABSOLUTELY no useful information.

Eliminate (a), (c) and (d) [we can eliminate (c) because statement (1) is completely useless].

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

So: n(n^2 + 2n + 1) is a multiple of 3.

We can simplify further to:

n(n+1)(n+1) is a multiple of 3.

Therefore, either n or (n+1) is a multiple of 3.

As we saw earlier, (n-1), n and (n+1) are consecutive. If we know that n or (n+1) is a multiple of 3, it's IMPOSSIBLE for (n-1) to also be a multiple of 3.

Therefore, statement (2) gives us a definite "no" answer: sufficient.

Choose (b).

[Edthesock]

If the question had been about n + 1 rather than n - 1, would the answer be E, since both n as well as n + 1 could have potentially been the term that was the multiple of 3?

[Sauravgmat]

Is positive integer n ? 1 a multiple of 3?

(1) n3 ? n is a multiple of 3

(2) n3 + 2n^2+ n is a multiple of 3

note: I'm assuming that "n3" = "n^3"

(1) n^3 - n can be simplified to:

n(n^2 - 1) = n(n + 1)(n - 1)

We know that n is an integer (since we're told that n-1 is a positive integer).

Let's rearrange our expression:

(n-1)(n)(n+1) is a multiple of 3. For this to be a multiple of 3, at least 1 of the terms must be a multiple of 3.

However, since n is an integer, (n-1), n and (n+1) are 3 consecutive integers. EVERY set of 3 consecutive integers will contain exactly one multiple of 3. Therefore, statement (1) gives us ABSOLUTELY no useful information.

Eliminate (a), (c) and (d) [we can eliminate (c) because statement (1) is completely useless].

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

So: n(n^2 + 2n + 1) is a multiple of 3.

We can simplify further to:

n(n+1)(n+1) is a multiple of 3.

Therefore, either n or (n+1) is a multiple of 3.

As we saw earlier, (n-1), n and (n+1) are consecutive. If we know that n or (n+1) is a multiple of 3, it's IMPOSSIBLE for (n-1) to also be a multiple of 3.

Therefore, statement (2) gives us a definite "no" answer: sufficient.

Choose (b).

If you plugin any mulitple of 3 in to the First equation, for example, 3(9-1) = 24 ;
then n-1 can never be mulitple of 3 ?

If so then, statement A will always be NO; hence sufficient..isn't?

[analyst218]

Is positive integer n ? 1 a multiple of 3?

(1) n3 ? n is a multiple of 3

(2) n3 + 2n^2+ n is a multiple of 3

note: I'm assuming that "n3" = "n^3"

(1) n^3 - n can be simplified to:

n(n^2 - 1) = n(n + 1)(n - 1)

We know that n is an integer (since we're told that n-1 is a positive integer).

Let's rearrange our expression:

(n-1)(n)(n+1) is a multiple of 3. For this to be a multiple of 3, at least 1 of the terms must be a multiple of 3.

However, since n is an integer, (n-1), n and (n+1) are 3 consecutive integers. EVERY set of 3 consecutive integers will contain exactly one multiple of 3. Therefore, statement (1) gives us ABSOLUTELY no useful information.

Eliminate (a), (c) and (d) [we can eliminate (c) because statement (1) is completely useless].

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

So: n(n^2 + 2n + 1) is a multiple of 3.

We can simplify further to:

n(n+1)(n+1) is a multiple of 3.

Therefore, either n or (n+1) is a multiple of 3.

As we saw earlier, (n-1), n and (n+1) are consecutive. If we know that n or (n+1) is a multiple of 3, it's IMPOSSIBLE for (n-1) to also be a multiple of 3.

Therefore, statement (2) gives us a definite "no" answer: sufficient.

Choose (b).

If you plugin any mulitple of 3 in to the First equation, for example, 3(9-1) = 24 ;
then n-1 can never be mulitple of 3 ?

If so then, statement A will always be NO; hence sufficient..isn't?

That's what I thought too..
n is an integer and for n^3-n= multilpe of three to be true
n has to be multiple of 3 itself..
thus n-1 will always NOT be a multiple of 3.

Stuart correct us where our flaws are.

[dxgamez]

If you plug in any multiple of 3 to be n in Statement 1, the question will always be FALSE, i.e n-1 is NOT a multiple of 3.

But the qn states positive integer. So n could be any number.

For example, take n = 4 and plug into statement 1:-

4(4^2-1) = 4 x 15 = 60 which is a multiple of 3, thus n - 1 is a multiple of 3. And ans to qn would be Yes. This is insufficient cos we had NO earlier (for n to be multiple of 3).

Stuart, your take pls.

[Stuart@KaplanGMAT]

If you plug in any multiple of 3 to be n in Statement 1, the question will always be FALSE, i.e n-1 is NOT a multiple of 3.

But the qn states positive integer. So n could be any number.

For example, take n = 4 and plug into statement 1:-

4(4^2-1) = 4 x 15 = 60 which is a multiple of 3, thus n - 1 is a multiple of 3. And ans to qn would be Yes. This is insufficient cos we had NO earlier (for n to be multiple of 3).

Stuart, your take pls.

Correct. Statement (1) is insufficient because, depending on what number we plug in, we can get both a "yes" and a "no" answer.

As I showed in my original post, statement 1 basically says that the product of 3 consecutive integers is a multiple of 3. Since we already knew this information, statement (1) provides no information and is, therefore, insufficient.

In other words, you could choose ANY integer for (n-1) and still satisfy statement (1); you could let (n-1=3) to get a "yes" answer or (n-1=4) to get a "no" answer.