1. ## Division proof

Prove that $(1+2+\cdots+9)|1^5+2^5+\cdots+9^5$.

2. ## Ok

here is how you do it....I will do a general proof...ok you want to prove that $\sum_{i=0}^n{i}|\sum_{i=0}^n{i^5}$ to do this all you need to do is utilize summation formulas and see that $\sum_{i=0}^n{i}=\frac{n(n+1)}{2}$ and that $\sum_{i=0}^m{i^5}=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$ then we need to prove that $\frac{\sum_{i=0}^n{i^5}}{\sum_{i=0}^n{i}}$ we do this by using substitution..the former thing can be rewritten as $\frac{\frac{n^2(n+1)^2(2n^2+2n-1)}{12}}{\frac{n(n+1)}{2}}$ which simplifies to $\frac{n(n+1)(2n^2+2n-1)}{6}$ now imputing 9 we get 2685...therfore we have proved your task

3. It's not difficult to prove it for any odd exponent $
{2a + 1}$

We want to show that $
\left. {\sum\limits_{j = 1}^n j } \right|\sum\limits_{j = 1}^n {j^{2a + 1} }$

Case $
n = 2 \cdot k
$

$
\sum\limits_{j = 1}^{2k} {j^{2a + 1} } = \sum\limits_{j = 1}^k {j^{2a + 1} } + \sum\limits_{j = 0}^{k - 1} {\left( {2 \cdot k - j} \right)^{2a + 1} } \equiv \sum\limits_{j = 1}^k {j^{2a + 1} } + \sum\limits_{j = 0}^{k - 1} {\left( { - j} \right)^{2a + 1} }=k^{2a+1} \left( {\bmod .k} \right)$

$
\Rightarrow \left. k \right|\sum\limits_{j = 1}^{2k} {j^{2a+1} }
$

$
\sum\limits_{j = 1}^{2k} {j^{2a + 1} } = \sum\limits_{j = 1}^k {j^{2a + 1} } + \sum\limits_{j = 1}^k {\left( {2 \cdot k + 1 - j} \right)^{2a + 1} } \equiv \sum\limits_{j = 1}^k {j^{2a + 1} } + \sum\limits_{j = 1}^k {\left( { - j} \right)^{2a + 1} } \left( {\bmod .2k + 1} \right)
$

$
\Rightarrow \left. (2k+1) \right|\sum\limits_{j = 1}^{2k} {j^{2a+1} }
$

Thus: $
\left. {\sum\limits_{j = 1}^{2k} j } \right|\sum\limits_{j = 1}^{2k} {j^{2a + 1} }$
since $
{k \cdot \left( {2k + 1} \right) = \sum\limits_{j = 1}^{2k} j }$
and $
\left( {2k + 1;k} \right) = 1$

Case: $
n = 2 \cdot k + 1
$

$
\sum\limits_{j = 1}^{2k + 1} {j^{2a + 1} } = \sum\limits_{j = 1}^k {j^{2a + 1} } + \sum\limits_{j = 0}^k {\left( {2 \cdot k + 1 - j} \right)^{2a + 1} } \equiv \sum\limits_{j = 1}^k {j^{2a + 1} } + \sum\limits_{j = 0}^k {\left( { - j} \right)^{2a + 1} } \left( {\bmod .2k + 1} \right)
$

$
\Rightarrow \left. (2k+1) \right|\sum\limits_{j = 1}^{2k+1} {j^{2a+1} }
$

$
\sum\limits_{j = 1}^{2k + 1} {j^{2a + 1} } = \sum\limits_{j = 1}^k {j^{2a + 1} } + \sum\limits_{j = 1}^{k + 1} {\left( {2 \cdot \left( {k + 1} \right) - j} \right)^{2a + 1} }

$
and

$

\sum\limits_{j = 1}^k {j^{2a + 1} } + \sum\limits_{j = 1}^{k + 1} {\left( {2 \cdot \left( {k + 1} \right) - j} \right)^{2a + 1} } \equiv \sum\limits_{j = 1}^k {j^{2a + 1} } + \sum\limits_{j = 1}^{k + 1} {\left( { - j} \right)^{2a + 1} } \equiv 0\left( {\bmod .k + 1} \right)
$

$
\Rightarrow \left. (k+1) \right|\sum\limits_{j = 1}^{2k+1} {j^{2a+1} }
$

Thus: $
\left. {\sum\limits_{j = 1}^{2k+1} j } \right|\sum\limits_{j = 1}^{2k+1} {j^{2a + 1} }$
since $
{(k+1) \cdot \left( {2k + 1} \right) = \sum\limits_{j = 1}^{2k+1} j }$
and $
\left( {2k + 1;k+1} \right) = 1$

Therefore for all $
n,a \in \mathbb{Z}^ +
$
we have $
\left. {\sum\limits_{j = 1}^n j } \right|\sum\limits_{j = 1}^n {j^{2a + 1} }
$