The following is the answer approach given for the below problem in my old book. I am skeptical about the brute trial approach suggested (though k is found after 2 trials in this case). Is there a proper and better method in which we can find the least positive k in this case such that (840k + 3) is divisible by 9? (Regret the low-quality image)
If $9\mid(840k+3)$ then we have $$840k+3\equiv0\mod{9}$$ $$3k+3\equiv0\mod{9}$$ $$3k\equiv-3\mod{9}$$ $$\therefore k\equiv-1\equiv2\mod{3}$$ Hence $k=2$ is the smallest such positive integer $k$. All solutions are of the form $k=2+3n$ where $n\in\mathbb{N_0}$.