2.1: Problem 02: Exhibit a complete residue system module 17 composed entirely of multiples of 3.
Solution
Observe that multiples of 3 leave the following remainders
$$
\begin{align*}
3 \bmod 17 &= 3 \\
6 \bmod 17 &= 6 \\
9 \bmod 17 &= 9 \\
12 \bmod 17 &= 12 \\
15 \bmod 17 &= 15 \\
18 \bmod 17 &= 1 \\
21 \bmod 17 &= 4 \\
24 \bmod 17 &= 7 \\
27 \bmod 17 &= 10 \\
30 \bmod 17 &= 13 \\
\cdots
\end{align*}
$$
In fact, we can construct a complete system of 17 from the first 17 multiples. This is also because \((3,17) = 1\).