jmc

First notice that $a\equiv 8(\mathrm{mod}9)$ tells us that $a\equiv 2(\mathrm{mod}3)$, so once we satisfy the former, we have the latter. So, we focus on the final three congruences. We do so by rewriting them as $$\begin{array}{rl}a& \equiv -1(\mathrm{mod}5),\\ a& \equiv -1(\mathrm{mod}7),\\ a& \equiv -1(\mathrm{mod}9).\end{array}$$ Since $\gcd (5,7)=\gcd (7,9)=\gcd (9,5)=1$, the above congruences apply that $a\equiv -1(\mathrm{mod}5\cdot 7\cdot 9)$, or $a\equiv 314(\mathrm{mod}315)$. So $a$ is of the form $314+315n$ for an integer $n$. The smallest non-negative number of this form is $\boxed{314}$, which satisfies the original congruences.