jmc

The prime factorization of $210=2\cdot 3\cdot 5\cdot 7$. By the Chinese Remainder Theorem, it suffices to find the residues of $N$ modulo $5$, $6$, and $7$. Since the units digit of $N$ in base $6$ is equal to $0$, it follows that $N$ is divisible by $6$. Also, we note that $N$ is congruent modulo $b-1$ to the sum of its base $b$ digits. Indeed, if $N$ can be represented as $(\overline{a_k{a}_{k-1}\cdots a_0}{)}_b$, then $$\begin{array}{rl}N& \equiv a_k\cdot b^k+a_{k-1}\cdot b^{k-1}+\cdots +a_1\cdot b+a_0\\ & \equiv a_k\cdot ((b-1)+1{)}^k+\cdots +a_1\cdot ((b-1)+1)+a_0\\ & \equiv a_k+a_{k-1}+\cdots +a_1+a_0(\mathrm{mod}b-1).\end{array}$$It follows that $N\equiv 5+3+1+3+4+0\equiv 1(\mathrm{mod}5)$ and that $N\equiv 1+2+4+1+5+4\equiv 3(\mathrm{mod}7).$ By the Chinese Remainder Theorem and inspection, we determine that $N\equiv 31(\mathrm{mod}35)$, so that (by the Chinese Remainder Theorem again) $N\equiv \boxed{66}(\mathrm{mod}210)$.