A common domino game is composed of 28 pieces. Each piece is formed by two integers that vary from $0$ to $6$, inclusive. All possible combinations $(a, b)$, with $a \leq b$, are listed exactly once. Note that the piece $(4,2)$ is listed as the piece $(2,4)$, since $2 \leq 4$. Excluding the piece $(0,0)$, for each of the other 27 pieces $(a, b)$, with $a \leq b$, we write on a board the fraction $\frac{a}{b}$.

a) How many distinct values are written in the form of fractions on the board? (Note that the fractions $\frac{1}{2}$ and $\frac{2}{4}$ have the same value and should be counted only once.)

b) What is the sum of the distinct values found in the previous item?
Solution

a) It's enough to start counting from the largest denominators and not repeat when the smaller ones appear.

i) For $b=6$, we have

$$
\left(\frac{0}{6}, \frac{1}{6}, \frac{2}{6}, \frac{3}{6}, \frac{4}{6}, \frac{5}{6}, \frac{6}{6}\right)=\left(0, \frac{1}{6}, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, \frac{5}{6}, 1\right)
$$

ii) For $b=5$, we should not repeat $0=0 / 5$ and $1=5 / 5$, as they have already been counted, so we have

$$
\left(\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}\right)
$$

iii) For $b=4$, we can only add the irreducible fractions with a denominator of 4, as we have already counted those with denominators of 1 and 2 when $b=6$, so we have

$$
\left(\frac{1}{4}, \frac{3}{4}\right)
$$

iv) When $b$ is 1, 2, or 3, we will have fractions that have already been counted in the case of $b=6$. Verify!

Therefore, the number of distinct values is $7+4+2=13$.

b) A good way to sum the 13 fractions is to consider their reduced forms seen in the previous item, that is,

$$
\begin{aligned}
\frac{0}{6}+\frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{4}{6}+\frac{5}{6}+\frac{6}{6} & =\frac{21}{6} \\
& =\frac{7}{2} \\
\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5} & =\frac{10}{5} \\
& =2 \\
\frac{1}{4}+\frac{3}{4} & =1
\end{aligned}
$$

So the total sum is $\frac{7}{2}+2+1=\frac{13}{2}$.