jmc

First we multiply $(7j+3)$ by 3 to get $21j+9$. Now we divide by 7 and get $$\frac{21j+9}{7}=3j+\frac{9}{7}=3j+1+\frac{2}{7}.$$ Since $j$ is an integer, we know that $3j+1$ is also an integer. We're left with the fraction $\frac{2}{7}$ when we divided by 7, which means the remainder is $\boxed{2}$.