What is the probability that $XYZ$ is divisible by $5$?

Solution 1:

There are two questions here:

1. How many triples of numbers $(X,Y,Z)$ add to 20? We can call this $A$.

2. How many of these are divisible by 5? We can call this $B$.

The answer will then be $B/A$.

First we calculate $A$. Let us write $C(x)$ for the total number of ways of choosing $(X,Y,Z)$ so that $X=x$ and $X+Y+Z=20$.

There are 21 possible choices for $X$. For each such $X$, there are $21-X$ possible choices for $Y$ that make $X+Y\le 20$, namely $\{0, 1, \ldots 20-X\}$. The other choices have $X+Y>20$ and therefore $X+Y+Z>20$. Once we have chosen $X$ and $Y$ at random, there is exactly one possible choice for $Z$ that makes $X+Y+Z=20$. So we have $C(x) = 21-x$.

We want the total of $C(x)$ for each $x$ between 0 and 20: $$\begin{eqnarray} A&=&\sum_{x=0}^{20} C(x) \\ &=& \sum_{x=0}^{20} (21-x)\\ &=& \sum_{x=0}^{20} 21 - \sum_{x=0}^{20} x \\ &=& 441 - 210 \\ &=& 231. \end{eqnarray}$$

Now we calculate $B$. 5 is prime, so $XYZ$ is divisible by 5 if and only if one of $X$, $Y$, or $Z$ is divisible by 5. We can use inclusion-exclusion: $B$ is the sum of the cases where (at least) $X$, $Y$, or $Z$ is divisible by 5, minus the cases where (at least) two are divisible by 5, plus the cases where all three are divisible by 5. That is, $$\begin{eqnarray}B&=&D_x + D_{y} + D_z \\ &&- D_{xy} - D_{xz} - D_{yz} \\ &&+ D_{xyz}\end{eqnarray}$$

Where $D_{xy}$ denotes the number of choices of $(X,Y,Z)$ where $5\mid X$ and $5\mid Y$, and similarly for the others.

By symmetry, $D_x = D_y = D_z$, and $D_{xy} = D_{xz} = D_{yz}$. Also, it is impossible to have two of $(X,Y,Z)$ divisible by 5 without the third also being divisible by 5, so $D_{xy} = D_{xyz}$. So the previous equation reduces to:

$$B = 3D_x - 2D_{xyz}$$

We calculate $D_x$: $X$ will be a multiple of 5 whenever $X\in\{0,5,10,15,20\}$, so we want $$\begin{eqnarray} &&\sum_{5\mid X} C(X) \\ &=& \sum_{X\in\{0,5,10,15,20\}} (21-X) \\ &=& 21\cdot5 - (0+5+10+15+20) \\ &=& 105 - 50 \\ &=& 55 \end{eqnarray}$$

To calculate $D_{xyz}$ is quick because there are very few such triples, and we can enumerate them by brute force: $(0,0,20)\ldots (0,20,0), (5,0,15)\ldots (5,15,0),\ldots (20,0,0)$. This is 5+4+3+2+1 = 15.

So we have $B = 3\cdot 55 - 2\cdot 15 = 135$.

Thus the answer is $B/A = 135/231 = 45/77$.

Solution 2:

Here's some GAP source code that finds the probability by exhaustively enumerating the triples $(x,y,z) \in \{0,1,\ldots,20\}^3$ for which $xyz$ is divisible by $5$. (Note: there are no solutions for $x+y+z=20$ if e.g. $x>20$, so we can assume $x \leq 20$, $y \leq 20$ and $z \leq 20$.)

n:=20;;
d:=5;;

div_by_d_count:=0;;
count:=0;;

for x in [0..n] do
  for y in [0..n-x] do
    z:=n-x-y;
    count:=count+1;
    if(x*y*z mod d=0) then
      div_by_d_count:=div_by_d_count+1;
    fi;
  od;
od;
Print([n,d]," ",div_by_d_count," ",count," ",div_by_d_count/count,"\n");

The output was:

[ 20, 5 ] 135 231 45/77

So the probability is $45/77$ (which concurs with Seyhmus Güngören's earlier result).

Solution 3:

Lets calculate the probability for $5|X$ and $Y\nmid5$ and as follows $Z\nmid5$.

Case 0: $X=0 \quad$16 possible solutions for $Y$, $Z$ follows.

Case 1: $X=5 \quad$there are 12 possible solutions

Case 2: $X=10 \quad$8 possible solutions

Case 3: $X=15 \quad$4 possible solutions

So we get $16+12+8+4=40$ solutions.

Now we can apply the same thing to Y and Z. $40*3=120$ solutions.

Now we have the only the case $5|X$ and $5|Y$ and as follows $5|Z$ left. The case can look like this: $(0,0,20), (0,5,15), (0,10,10)$ or $(5,5,10)$. Calculating all those permutations: $3+6+3+3=15$. So we get $15+120=135$ solutions.

Now we have to calculate, how many different values $X,Y,Z$ can take. $X$ can be any value between $0$ and $20$, so $21$ different values. $Y$ can have $21-X$ different values and $Z$ follows. Therefore we get: $\sum_{i=0}^{20}{21-i}=21*21-\frac{20*21}{2}=441-210=231$

So the probability is: $$\frac{135}{231}$$