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 $\frac{7}{2}+2+1=\frac{13}{2}$.
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?