Find the smallest positive integer $X$ such that $478^{870} \equiv X \ (\text{mod} \ 273)$ Appreciate if one could advise if my solution is correct. Here is my attempt of the problem: 
Since $(273, 478) =1,$ by Euler's theorem, $478^{\phi(273)}=478^{144} \equiv 1 \ \ (\text{mod} \ 273) \implies 478^{864} \equiv 1 \ \ (\text{mod} \ 273).$ 
Next, $478^{2} \equiv 22 \ \ (\text{mod} \ 39) \implies 478^{6} \equiv 22^{3} \equiv 1 \  \ (\text{mod} \ 39)$ and $478^{\phi(7)} = 478^{6} \equiv 1 \ \ (\text{mod} \ 7) $
Hence, $478^{6} \equiv 1 \ \ (\text{mod} \ 273)$ and $478^{864+6}= 478^{870} \equiv 1 \ \ (\text{mod} \ 273)$

Your idea to use Chinese Remainder Theorem is good, but we can apply it more cleanly.  We have $273 = 3\cdot 7\cdot 13,$ so we localize at each prime:
$$X\equiv 478^{870} \equiv 1^{870} \equiv 1 \pmod{3}$$
$$X\equiv 478^{870} \equiv 2^{6\cdot 145} \equiv 1 \pmod{7}$$
$$X\equiv 478^{870} \equiv (-3)^{6+12\cdot 72} \equiv 3^6 \equiv 27^2 \equiv 1^2 \equiv 1 \pmod{13}$$
By Chinese Remainder Theorem $X\equiv 1 \pmod{273}.$
