What is the probability that $XYZ$ is divisible by $5$? A solution of $X + Y + Z = 20$ in non-negative integers is chosen at random. What is the probability that $XYZ$ is divisible by $5$?
Edit:
This happens to be an exam question. So I can't use calculators or computers and have to get the answer in less than 20 minutes while showing systematic workings. I appreciate the answers below, but can someone instruct me on solving the question given the mentioned constraints?

$5\mid  XYZ\implies 5 \mid  XY(20-X-Y) \implies 5\mid XY(X+Y)$
(1)If $5\mid X$ let $X=5a$ where $a$ is any integer, so $Y$ can have $1+20-5a$ values.
So, here the number possible  values of $X,Y$ are $21+16+11+6+1=55$
(2)If $5\mid Y$ let $Y=5b$  where $b$ is any integer, leads to another $55$ values.
(3)If $5\mid Y$ and $5\mid X\implies 5(a+b)≤20\implies 0≤a+b≤4$ which has $(5+4+3+2+1)$ values= $15$ values, for example for $Y=0, X$ can be one of $0,5,10,15,20$.
So, (1)+(2)-(3) leads to $2\cdot 55 -15 =95$ values.
(4)If $5\mid (X+Y),$ but $5∤XY$
If $X=5a+r,Y=5b-r$ where $1≤r≤4$ so,$0≤a+b≤4$ and $1≤b≤4$ and $0≤a≤3$.
For each $r$, these $a,b$ can be chosen in $4+3+2+1=10$ ways,for example for $a=0, b$ can be one of $0,1,2,3$.
There can be $4$ values of $r$ ,leading to $4\cdot 10=40$ possible values of $X,Y$
So, number of the possible values of $X,Y$ such that $ 5\mid XY(X+Y)$ is $55+55-15+40=135$
As, $0 ≤ X+Y ≤ 20$, the number possible  values of $X,Y$ are $(0+ 1+2+...+21)=231$, for example for $X=a, a ≤ Y ≤ 20\implies 20-a+1$ values of $Y$ .
So, the required probability $=\frac{135}{231}=\frac{45}{77}$
The answer can be validated using the following Java code:
  void process() {
    int all = 0;
    int count = 0;
    for (int x = 0; x <= 20; x++) {
        boolean xMod5 = (x % 5 == 0);
        for (int y = 0; y <= 20 - x; y++) {
            if (xMod5 || y % 5 == 0 || (x + y) % 5 == 0) {
                count++;
            }
            all++;
        }
    }
    System.out.println(String.format("%d/%d", count, all));
 }

