Determine the smallest non-negative integer $a$ that satisfies the congruences: \begin{align*}
&a\equiv 2\pmod 3,\\
&a\equiv 4\pmod 5,\\
&a\equiv 6\pmod 7,\\
&a\equiv 8\pmod 9.
\end{align*}
First notice that $a\equiv 8\pmod 9$ tells us that $a\equiv 2\pmod 3$, 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{align*}
a&\equiv -1\pmod 5,\\
a&\equiv -1\pmod 7,\\
a&\equiv -1\pmod 9.
\end{align*} Since $\gcd(5,7)=\gcd(7,9)=\gcd(9,5)=1$, the above congruences apply that $a\equiv -1\pmod{5\cdot 7\cdot 9}$, or $a\equiv 314\pmod{315}$. 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.