Hey guys!

By simple inspection, k can take only the following values:

k =1, 2, 4 and 5, so that it has no solution in x.

Given that k  is a positive integer less than 6,

how many values can k take on such that

$3x \equiv k \pmod{6}$ has no solutions in x ?

Rewrite:

$\begin{array}{|lrcll|} \hline 3x \equiv k \pmod{6} \\ \text{rewrite...} \\ \begin{array}{|rcll|} \hline 3x-k &=& n\cdot 6 \quad &| \quad n \in \mathbf{Z} \qquad (n \text{ is a integer}) \\ 3x &=& n\cdot 6 +k \\ x &=& \frac{n\cdot 6 +k}{3} \\ x &=& 2n+\frac{k}{3} \quad &| \quad \frac{k}{3} \text{ is a integer, if } k = 0 \text{ or } k = 3 \\ & & \quad &| \quad \frac{k}{3} \text{ is not a integer, if } k = 1,\ k = 2,\ k=4,\ k= 5 \qquad 0 < k < 6 \\ \hline \end{array} \\ \hline \end{array}$

Thanks guys!