Number of Non negative integer solutions of $x+2y+5z=100$ Find Number of Non negative integer solutions of $x+2y+5z=100$
My attempt: 
we have $x+2y=100-5z$ 
Considering the polynomial $$f(u)=(1-u)^{-1}\times (1-u^2)^{-1}$$
$\implies$
$$f(u)=\frac{1}{(1-u)(1+u)}\times \frac{1}{1-u}=\frac{1}{2} \left(\frac{1}{1-u}+\frac{1}{1+u}\right)\frac{1}{1-u}=\frac{1}{2}\left((1-u)^{-2}+(1-u^2)^{-1}\right)$$ 
we need to collect coefficient of $100-5z$ in the above given by
$$C(z)=\frac{1}{2} \left((101-5z)+odd(z)\right)$$
Total number of solutions is
$$S(z)=\frac{1}{2} \sum_{z=0}^{20} 101-5z+\frac{1}{2} \sum_{z \in odd}1$$
$$S(z)=540.5$$
what went wrong in my analysis?
 A: An alternative way.
Given $x+2y+5z=100$ and it is clear that $0\le z\le20$.
For any possible values of $z$, $x+2y=100-5z$.
Let us take $p=100-5z\ge0$. Solving the equation $x+2y=p$,
$(-p,p)$ is a solution. The general solution of $(x,y)$ is $$x=-p+2q,\ \ y=p-q,\ \ q\in\mathbb{Z}$$
If $p=2k$, then $k=\dfrac p2\le q\le p=2k$.
So, there are $k+1=\dfrac p2+1$ solutions for $(x,y)$
So, we have the following numbers as follows
$$p=100,95,90,85,80,75,......,15,10,5,0$$and$$k+1=51,48,46,43,41,38,......,8,6,3,1$$
The total number of solutions are $$4(10+20+30+40)+5(8+6+3+1)+51=541$$
A: I will find number of solutions of equation $5x+2y+z=10 n$ in general:
clearly the positive solutions $x_0, y_0, z_0$ of this equation are corespondent to the solution $x_0+2,y_0, z_0$ of equation $5x+2y+z=10(n+1)$.Clearly for $x=>2$, finding the solutions of $5x+2y+z=10(n+1)$ will lead to finding the solution of first equation,provided we consider $x-2$ in first equation.
If the number of solutions of equation $5x+2y+z=10(n+1)$ is $\phi(n+1)$ and that of equation $5x+2y+z=10n$ is $\phi(n)$ the difference of  $\phi(n+1)$ and $\phi(n)$ is equal to the number of solutions of equation $5x+2y+z=10(n+1)$ for $x=0$ and $x=1$. But this equation has $5n+6$ solutions for $x=0$, (i.e. $0=<y=<5n+5)$ and it has $5n+3$ solutions for $x=1$, (i.e $0=<y=<5n+2)$. Therefore we have:
$\phi(n+1)-\phi(n)=10n+9$
We can also search and find  that $\phi(1)=10$, so we can write:
$\phi(1)=10$
$\phi(2)-\phi(1)=10\times 1+9$
$\phi(3)-\phi(2)=10\times 2+9$ 
.
.
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$\phi(n)-\phi(n-1)=10(n-1)+9$
Summing theses relations gives:
$\phi(n)=5n^2 +4n +1$
In your question $n=10$, therefore number of solutions is $\phi(10)=5.10^2+4.10+1=541$
A: Note that the number of non-negative integer solutions of the following equation
$$ x+y=n $$
is $n+1$. Here $n$ is a non-negative integer. Clearly $5|(x+2y)$. Let 
$$x+2y=5k\tag{1}$$ 
where $0\le k\le 20$. For (1), if $k$ is odd, then so is $x$, and if $k$ is even, then so is $x$.
Case 1: $k$ is odd. Let $k=2n-1$ and $x=2m-1$. Then $1\le n\le 10$ and (1) becomes
$$ m+y=5n-2 $$
whose number of non-negative integer solutions is $5n-1$.
Case 2: $k$ is even. Let $k=2n$ and $x=2m$. Then $0\le n\le 10$ and (1) becomes
$$ m+y=5n $$
whose number of non-negative integer solutions is $5n+1$.
Thus the number of non-negative integer solutions is
$$ \sum_{n=1}^{10}(5n-1)+\sum_{n=0}^{10}(5n+1)=551 $$
A: Given: $x+2y=100-5z$, tabulate:
$$\begin{array}{c|c|c}
z&x&\text{count}\\
\hline
0&100,98,\cdots, 0&\color{red}{51}\\
1&\ \ 95,93,\cdots, 1&\color{blue}{48}\\
2&\ \ 90,88,\cdots, 0&\color{red}{46}\\
3&\ \ 85,83,\cdots, 1&\color{blue}{43}\\
4&\ \ 80,78,\cdots, 0&\color{red}{41}\\
\vdots&\vdots&\vdots\\
17&15,13,\cdots,1&\color{blue}{8}\\
18&10,8,\cdots,0&\color{red}{6}\\
19&5,3,1&\color{blue}{3}\\
20&0&\color{red}{1}\\
\hline
&&\color{red}{286}+\color{blue}{255}=541
\end{array}$$
