Find the value of k where 31k2 is divisible by 6

Dear student,
Applying the divisibility of 6 i.e a number should pass the test of 2 and 3
i.e it must be an even number and also the sum of digit is divisible by 3
$$\begin{gathered} 31k2 \\ \sin ce the unit digit is even therefore it satisfies the divisibility rue of 2 \\ also, 3+1+k+2=6+k \\ when k=0, 6+0=6 \\ it satisfies the divisibility rue of 3 \\ \Rightarrow k=0 \end{gathered}$$

Regards