Prove that $1^3 + 2^3 + ... + 2004^3$ is divisible by $2005$. This is a first year high school problem. There's no series or any other more "advanced" math involved. There should just be a way to factor $2005$ out of the sum.
Judging by similar problems I assume that it is, in general, true that, for a natural number $n>2$, $n$ divides $1^3 + 2^3 + \cdots + (n-1)^3$.
 A: One nice proof goes as follows.
Let $S$ be the sum $1^3 + \cdots + 2004^3$. Consider the sum $S’ = (1 - 2005)^3 + \cdots + (2004 - 2005)^3$. Note that $S \equiv S’ \mod 2005$. Note that $S’ = (-2004)^3 + \cdots + (-1)^3 = (-1)^3 + \cdots + (-2004)^3 = - (1^3 + \cdots + 2004^3) = -S$. Then $S \equiv -S \mod 2005$. That is, $2S \equiv 0 \mod 2005$. Since $2$ and $2005$ are relatively prime, we have $S \equiv 0 \mod 2005$. That is, $S$ is divisible by 2005.
Note that the only thing special about $2005$ is that it’s odd, so we can generalise this greatly.
A: Here's maybe a more 9th-grade-friendly approach, assuming you know how to factorise a sum of two cubes:
$\begin{eqnarray} S & = & 1^3 + 2^3 + \ldots + 2003^3 + 2004^3 \\
& = & (1^3 + 2004^3) + (2^3 + 2003^3) + \ldots + (1002^3 + 1003^3) \\
& = & (1 + 2004)(1^2 + 1 \times 2004 + 2004^2) + (2 + 2003)(2^2 + 2 \times 2003 + 2003^2) + \ldots + (1002 + 1003)(1002^2 + 1002 \times 1003 + 1003^2) \\
& = & 2005\left(1^2 + 1 \times 2004 + 2004^2 + 2^2 + \ldots) \right) \end{eqnarray}$
Note that this trick only works because we have an even number of terms so we can pair them up, so for example $1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 225$ which is not divisible by 6.
