note: I'm assuming that "n3" = "n^3"II wrote: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
If you plugin any mulitple of 3 in to the First equation, for example, 3(9-1) = 24 ;Stuart Kovinsky wrote:note: I'm assuming that "n3" = "n^3"II wrote: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
(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).
That's what I thought too..Sauravgmat wrote:If you plugin any mulitple of 3 in to the First equation, for example, 3(9-1) = 24 ;Stuart Kovinsky wrote:note: I'm assuming that "n3" = "n^3"II wrote: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
(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).
then n-1 can never be mulitple of 3 ?
If so then, statement A will always be NO; hence sufficient..isn't?
Correct. Statement (1) is insufficient because, depending on what number we plug in, we can get both a "yes" and a "no" answer.dxgamez wrote: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.
