How many positive solutions are there (Positive 3 tuples)? I want to find how many positive solutions for the Diophantine equation $4x + 2y + 5z = 100$
I found a particular solution $(x,y,z) = (50,-50,0)$ then I found a general solution (basis) $s(-2,-1,2) + t(1,-2,0)$ for the homogeneous equation $4x + 2y +5z = 0$ and then I added that to the particular solution to get $(x,y,z) = (50-2s+t,-50-s-2t,2s)$ as the solution. However, Now I need to calculate how many positive solutions. So I have $$50-2s + t > 0$$ $$ -50 -s -2t > 0$$ $$2s > 0$$
I did some algebraic manipulation to get $t > -30$ and $s > 0 $ Now, I still don't understand how to count how many positive solutions are there ?
I went to wolfram alpha and it said $106$ but I have no clue how ? Any Suggestions
http://www.wolframalpha.com/input/?i=50-2s+%2B+t+%3E+0+%2C+-50+-+s+-+2t+%3E+0+%2C+2s+%3E+0+
 A: Since $5z = 100 - 4x - 2y$, $z$ must be even.  Since $x, y, z > 0$, there are only nine possible values of $z$, namely, $z = 2, 4, 6, 8, 10, 12, 14, 16, 18$.  By substituting these values for $z$, we can reduce the problem to determining the number of solutions of the equations
\begin{align*}
4x + 2y & = 10\\
4x + 2y & = 20\\
4x + 2y & = 30\\
4x + 2y & = 40\\
4x + 2y & = 50\\
4x + 2y & = 60\\
4x + 2y & = 70\\
4x + 2y & = 80\\
4x + 2y & = 90
\end{align*}
in the positive integers.
If we divide each equation by $2$, we obtain
\begin{align*}
2x + y & = 5\\
2x + y & = 10\\
2x + y & = 15\\
2x + y & = 20\\
2x + y & = 25\\
2x + y & = 30\\
2x + y & = 35\\
2x + y & = 40\\
2x + y & = 45
\end{align*}
Since the value of $y$ is determined by the value of $x$, the number of solutions of each equation is the number of positive even integers less than the number on the right hand side of the equation.  Thus, the number of solutions of the linear Diophantine equation $4x + 2y + 5z = 100$ in the positive integers is 
$$2 + 4 + 7 + 9 + 12 + 14 + 17 + 19 + 22 = 106$$
