Lucky Seven

First, the number of elements in $S$ can be calculated using the stars and bars method . We find that

$|S| = \dbinom{28 + 3 - 1}{3 - 1} = \dbinom{30}{2} = \dfrac{30 \times 29}{2} = 435$ .

Now as $7$ is prime, $xyz$ will be divisible by $7$ if and only if at least one of $x,y,z$ to be divisible by $7$ . If we let

• $N_{x}, N_{y}, N_{z}$ be the number of elements of $S$ in which $x,y,z$ , respectively, are divisible by $7$ ,

• $N_{xy}, N_{xz}, N_{yz}$ be the number of elements of $S$ in which both subscript variables are divisible by $7$ , and

• $N_{xyz}$ be the number of elements of $S$ in which all of $x,y,z$ are divisible by $7$ ,

then by inclusion-exclusion the number of elements $(x,y,z)$ of $S$ for which $7|xyz$ is

$N_{x} + N_{y} + N_{z} - N_{xy} - N_{xz} - N_{yz} + N_{xyz}$ .

Now by symmetry $N_{x} = N_{y} = N_{z}$ . Also, since the sum of the variables is $28$ , i.e., a multiple of $7$ , if any two of $x,y,z$ are divisible by $7$ then so must the third variable. By this fact and symmetry we have that $N_{xy} = N_{xz} = N_{yz} = N_{xyz}$ .

To calculate $N_{x}$ , we first note that $x \in \{0, 7, 14, 21, 28\}$ . This in turn requires us to find the number of non-negative integer solutions to $y + z = 28 - x$ for each of the 5 possible values for $x$ . By stars and bars the number of solutions will be $\dbinom{28 - x + 2 - 1}{1} = 29 - x$ , and thus

$N_{x} = 29 + 22 + 15 + 8 + 1 = 75$ .

Next, to calculate $N_{xyz}$ , (and thus also $N_{xy}, N_{xz}$ and $N_{yz}$ ), we need to count all permutations of the triples $(0,0,28), (0,7,21), (0,14,14)$ and $(7,7,14)$ . This adds up to $3 + 6 + 3 + 3 = 15$ .

So by inclusion-exclusion the number of suitable triples is $3 \times 75 - 2 \times 15 = 195$ , which gives us a probability of $\dfrac{195}{435} = \dfrac{13}{29}$ . So finally $a + b = 13 + 29 = \boxed{42}$ .