Cruising the right way

We're looking for $a,b$ such that $\overline{30a0b03}$ is divisible by 13. To do this, we'll look at powers of 10 mod 13.

$\begin{aligned} &10^0\equiv 1\mod{13}\\&10^1\equiv 10\mod{13}\equiv{-3}\mod{13}\\&10^2\equiv -30\mod{13}\equiv{-4}\mod{13}\\&10^3\equiv -40\mod{13}\equiv{-1}\mod{13}\\&10^4\equiv -10\mod{13}\equiv{3}\mod{13}\\&10^5\equiv 30\mod{13}\equiv{4}\mod{13}\\&10^6\equiv 40\mod{13}\equiv{1}\mod{13}\\\end{aligned}$

Now we can use this to get conditions on $a$ and $b$ . Specifically, we can reduce the number mod 13 using what we got above to $3*1+0*(-3)+b*(-4)+0*(-1)+a*3+0*4+3*1 = \boxed{3a-4b+6}.$ Our condition then is that $3a-4b+6\equiv 0\mod{13}$ . Solving for $b$ (using the fact that $4^{-1} = 10 \mod{13}$ ), we get $b \equiv 4^{-1}(3a+6) \equiv 10(3a+6) \equiv 4a+8\mod{13}.$

This tells us that for each $a$ , there is exactly one $b$ that works. But we have to be careful, because $b$ is an integer mod 13, so it might not be a single digit. So now we explicitly calculate each solution pair when $a$ is a single digit to get

$\begin{aligned}&(0,8)\\&(1,12)\\&(2,3)\\&(3,7)\\&(4,11)\\&(5,2)\\&(6,6)\\&(7,10)\\&(8,1)\\&(9,5)\\\end{aligned}$

Of these, 7 are single digit values for $a$ , so the answer is $\boxed{7}$ .