"Stars and Bars" method when all variables are multiple of same number

Question

Find the number of non negative integral solutions of the equation $x + y + z + u = 100$, where $x, y, z$ and $u$ are multiple of $5.$

My approach using stars and bars:

Since all are multiples of $5,$ if have to find the values of $p,q,r$ and $s$ in:

$5p + 5q + 5r + 5s = 100$

Or, $p + q + r + s = 20$

Now, since they are multiples of $5$, each should have a minimum of $5 (x,y,z,u)$ or $1(p,q,r,s):$

$p + q + r + s = 16$

Applying stars and bars, answer should be $19C3 = 969$

What is wrong with my approach?

Answer 1

As you pointed out, the number of solutions of the original equation, with the multiples of $5$ constraint, is the number of non-negative solutions of $p+q+r+s=20$. This is $\binom{20+3}{3}$.

Remark: Note that $0$ is a multiple of $5$, for $0=(5)(0)$.