Integers: Remainder Problems

If $$n$$ is a positive integer, which of the following must be divisible by 3?

Incorrect. [[Snippet]] __Plug In__ $$n = 2$$. >$$n^2 \cdot (n−3)$$ >$$= 2^2 \cdot (2-3)$$ >$$= 4 \cdot -1$$ >$$= -4$$ Eliminate this answer choice because -4 is not divisible by 3, so $$n^2·(n-3)$$ does not HAVE to be divisible by 3.

Incorrect. [[Snippet]] __Plug In__ $$n = 1$$. >$$(n-2)(n+1)(n+3)$$ >$$= (1-2)(1+1)(1+3)$$ >$$=-1\cdot 2\cdot 4$$ >$$=-8$$ Eliminate this answer choice because 8 is not divisible by 3, so $$(n-2)(n+1)(n-3)$$ does not HAVE to be divisible by 3.

Incorrect. [[Snippet]] __Plug In__ $$n = 1$$. >$$n(n+1)^3$$ >$$= 1(1+1)^3$$ >$$= 1 \cdot (2)^3$$ >$$= 1 \cdot 8$$ >$$= 8$$ Eliminate this answer choice because 8 is not divisible by 3, so $$n (n+1)^3$$ does not HAVE to be divisible by 3.

Correct. [[Snippet]] __Plug In__ $$n = 2$$. >$$n (n^2 - 1)$$ >$$= 2 (2^2 - 1) $$ >$$=2 (4 - 1)$$ >$$=2 (3)$$ >$$= 6$$ Since 6 is divisible by 3, this answer choice cannot be eliminated. All other answer choices can be eliminated by __Plugging In__ 2 or 1, so this is the correct answer.

Incorrect. [[Snippet]] __Plug In__ $$n = 2$$. >$$(n−1) \cdot n \cdot (n+3) $$ >$$=(2−1) \cdot 2 \cdot (2+3)$$ >$$= 1\cdot 2 \cdot 5$$ >$$= 10$$ Eliminate this answer choice because 10 is not divisible by 3, so $$(n−1) \cdot n \cdot (n+3)$$ does not HAVE to be divisible by 3.

It's true that when $$n=1$$, the expression becomes $$1(1^2 - 1) = 0$$. However, this isn't a problem, since 0 is in fact divisible by 3. As a matter of fact, 0 is divisible by every number except itself. This is because the term _divisible_ means _divisible without a remainder_; when dividing 0 by an integer $$x$$ yields 0, with no remainder—hence, 0 is divisible by $$n$$.

$$n(n^2 -1)$$

$$(n - 1) \cdot n \cdot (n+3)$$

$$n(n+1)^3$$

$$(n-2)(n+1)(n+3)$$

$$n^2(n-3)$$

Got it!

But when plugging in $$n=1$$, the expression results in 0. Doesn't this prove that the expression is __not__ always divisible by 3?