jmc

We can re-write the problem as the following three equations: $$\begin{gathered} k=17a+1 \\ k=6b+1 \\ k=2c+1 \end{gathered}$$Therefore, $k-1$ is divisible by $17,$ $6,$ and $2.$ The smallest positive value of $k-1$ is thus $$\text{lcm}[17,6,2]=\text{lcm}[17,6]=17\cdot 6=102,$$and so the smallest possible value of $k$ is $k=102+1=\boxed{103}.$