jmc

Note that the left-hand side is divisible by 3. Therefore, since 3 divides 6, the left-hand side will be equivalent to a multiple of 3. Therefore, if $k$ were to be 1, 2, 4, or 5, then the given congruence would have no solutions. On the other hand, if $k=0$ or $k=3$, then $x=0$ and $x=1$ (respectively) satisfy the given congruence. Therefore, there are $\boxed{4}$ values of $k$ such that $3x\equiv k(\mathrm{mod}6)$ has no solutions.