\section*{Problem 5 - 061245}

Determine for each natural number \(n\) the number \(A(n)\) of all non-negative integer solutions of the equation \(5 x+2 y+z=10 n\).
}

To solve is

\[
5 x+2 y+z=10 n \quad(\star)
\]

Let's consider the smallest natural number. Let \(n=1\). Then we are looking for the non-negative solutions for \(5 x+2 y+z=10 \Longleftrightarrow 2 y+z=10-5 x\).

Obviously, \(x\) can only take the values \(0,1,2\). For \(x=0\), \(z\) can take the values \(10,8,6, \ldots, 0\). For \(x=1\), \(z\) can take the values \(5,3,1\), and for \(x=2\), \(z\) can only take the value 0.

We thus obtain a total of \(6+3+1=10\) solutions.

Now let's set \(x \mapsto x-2\) in \((\star)\). We get:

\[
5(x-2)+2 y+z=10 n \Longleftrightarrow 5 x+2 y+z=10(n+1)
\]

Let \(A(n+1)\) be the number of non-negative integer solutions of ( \(\square\) ). Then \(A(n+1)-A(n)\) is equal to the number of non-negative integer solutions for \(x=0\) and \(x=1\) in ( \(\square\) ).

1. Case: Let \(x=0 ; 2 y+z=10 n+10\)

It holds that \(0 \leq y \leq 5 n+5\) and thus we get \(5 n+6\) solutions.

2. Case: Let \(x=1 ; 5+2 y+z=10 n+10 \Longleftrightarrow 2 y+z=10 n+5\)

It holds that \(0 \leq y \leq 5 n+2\) and thus we get \(5 n+3\) solutions.

Thus, \(A(n+1)-A(n)=10 n+9\) and \(A(1)=10\). Recursively, we can thus calculate \(A(n)\).

\[
\left(A(n)=10+\sum_{k=1}^{n-1}(10 k+9)=10+9(n-1)+10 \frac{(n-1) n}{2}=5 n^{2}+4 n+1\right.
\]

\section*{Solution by a Member of the Matheplanet}

2. Solution:

Obviously, for each \(0 \leq x \leq 2 n\) and each \(0 \leq y \leq \frac{10 n-5 x}{2}\), there is exactly one \(0 \leq z=10 n-5 x-2 y\).

Thus, the sum

\[
A(n)=\sum_{x=0}^{2 n}\left(1+\left[\frac{10 n-5 x}{2}\right]\right)=2 n+1+\sum_{x=0}^{2 n}\left[\frac{10 n-5 x}{2}\right]
\]

needs to be determined.

For \(x=2 n\), we get the value 0 in the second sum. All other summands are obtained by setting \(x\) to either \(2 k\) or \(2 k+1\), where \(k\) runs through the natural numbers from 0 to \(n-1\). Thus,

\[
\begin{gathered}
A(n)=2 n+1+\sum_{k=0}^{n-1}\left(\left[\frac{10 n-5 \cdot 2 k}{2}\right]+\left[\frac{10 n-5 \cdot(2 k+1)}{2}\right]\right)= \\
=2 n+1+\sum_{k=0}^{n-1}(5(n-k)+5(n-k)-3)=2 n+1-3 n+10 \sum_{k=0}^{n-1}(n-k)= \\
=1-n+10 \sum_{\ell=1}^{n} \ell=1-n+10 \cdot \frac{n \cdot(n+1)}{2}=1-n+5 n^{2}+5 n=5 n^{2}+4 n+1
\end{gathered}
\]

Solution by cyrix