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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 $1^3 + 2^3 + \cdots + (n-1)^3$.

Leucippus
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Marlow37
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    Hint. $k^3+(2005-k)^3$ is divisible by $2005$. See also https://math.stackexchange.com/questions/62171/proving-13-23-cdots-n3-left-fracnn12-right2-using-induct – Robert Z Dec 06 '22 at 21:12
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    The sum of the first $n-1$ cubes is $1^3 + 2^3 + \cdots + (n-1)^3 = ((n-1)^2 n^2)/4$ - see Robert Z's link. However that sum is not always divisible by $n$ - for example check $n = 6$. It might be interesting to look at when that sum is divisible by $n$. – Michael Lugo Dec 06 '22 at 21:32
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    @MichaelLugo It appears to be the case whenever $\neg (n \equiv 2 \mod 4)$. The case of $n$ odd is fairly straightforward - see my answer. If $n$ is even, we can show that the sum equals $(n/2)^3 \mod n$ by pairing up everything except the middle element using Robert Z’s comment. By factoring $n = m 2^{k+1}$ where $m$ is odd, we see $(n/2)^3 \equiv 0 \mod n$ if and only if $k > 0$; that is, iff $n$ is divisible by 4. – Mark Saving Dec 06 '22 at 22:29
  • See Gauss / Wilson reflection. The answers below are special cases of this. – Bill Dubuque Dec 07 '22 at 03:10

2 Answers2

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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.

Mark Saving
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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.

ConMan
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