3. How many positive rational numbers less than $\pi$ have denominator at most 7 when written in lowest terms? (Integers have denominator 1.)
Solution: 54

We can simply list them. The table shows that there are $3+3+6+6+12+6+18=54$.
\begin{tabular}{r|ll} 
Denominator & \multicolumn{2}{|l}{ Values } \\
\hline 1 & $\frac{1}{1}, \frac{2}{1}, \frac{3}{1}$ \\
2 & $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}$ \\
3 & $\frac{1}{3}, \frac{2}{3}, \quad \frac{4}{3}, \frac{5}{3}, \quad \frac{7}{3}, \frac{8}{3}$ \\
4 & $\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \quad \frac{9}{4}, \frac{11}{4}$ \\
5 & $\frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \quad \frac{6}{5}, \frac{7}{5}, \frac{8}{5}, \frac{9}{5}, \quad \frac{11}{5}, \frac{12}{5}, \frac{13}{5}, \frac{14}{5}$ \\
6 & $\frac{1}{6}, \frac{5}{6}, \frac{7}{6}, \frac{11}{6}, \quad \frac{13}{6}, \frac{17}{6}$ \\
7 & $\frac{1}{7}, \frac{2}{7}, \ldots, \frac{6}{7}, \frac{8}{7}, \frac{9}{7}, \ldots, \frac{13}{7}, \quad \frac{15}{7}, \frac{16}{7}, \ldots, \frac{20}{7}$
\end{tabular}