Krdytkyu

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?

An alternative way.

Given $x+2y+5z=100$ and it is clear that $0le zle20$.

For any possible values of $z$, $x+2y=100-5z$.

Let us take $p=100-5zge0$. Solving the equation $x+2y=p$,
$(-p,p)$ is a solution. The general solution of $(x,y)$ is $$x=-p+2q, y=p-q, qinmathbb{Z}$$

If $p=2k$, then $k=dfrac p2le qle p=2k$.

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 $$4(10+20+30+40)+5(8+6+3+1)+51=541$$

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:

$phi(1)=10$

$phi(2)-phi(1)=10times 1+9$

$phi(3)-phi(2)=10times 2+9$
.

.

.

$phi(n)-phi(n-1)=10(n-1)+9$

Summing theses relations gives:

$phi(n)=5n^2 +4n +1$

In your question $n=10$, therefore number of solutions is $phi(10)=5.10^2+4.10+1=541$

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
$$x+2y=5ktag{1}$$
where $0le kle 20$. 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 $1le nle 10$ 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 $0le nle 10$ 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
$$ sum_{n=1}^{10}(5n-1)+sum_{n=0}^{10}(5n+1)=551 $$

Given: $x+2y=100-5z$, tabulate:
$$begin{array}{c|c|c}
z&x&text{count}\
hline
0&100,98,cdots, 0&color{red}{51}\
1& 95,93,cdots, 1&color{blue}{48}\
2& 90,88,cdots, 0&color{red}{46}\
3& 85,83,cdots, 1&color{blue}{43}\
4& 80,78,cdots, 0&color{red}{41}\
vdots&vdots&vdots\
17&15,13,cdots,1&color{blue}{8}\
18&10,8,cdots,0&color{red}{6}\
19&5,3,1&color{blue}{3}\
20&0&color{red}{1}\
hline
&&color{red}{286}+color{blue}{255}=541
end{array}$$