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_ka_{k-1}\cdots a_0})_b $, then \begin{align*}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 \pmod{b-1}.\end{align*} It follows that $ N \equiv 5+3+1+3+4+0 \equiv 1 \pmod{5} $ and that $ N \equiv 1 + 2 + 4 + 1 + 5 + 4 \equiv 3 \pmod{7} $. By the Chinese Remainder Theorem and inspection, we determine that $ N \equiv 31 \pmod{35} $, so that (by the Chinese Remainder Theorem again) $ N \equiv \boxed{66} \pmod{210} $.