How to find ord$[a]_m$ What are the steps to find minimum $k$ such that $\left([a]_m\right)^k=[1]_m$? In short, how to find ord$[a]_m?$ 
Specifically I need to find mod$\left( 173^{607},1147\right)$ but since $\varphi(1147)=1080>607$ I can't use Euler-Fermat's Th. on this one, and I don't feel like square-and-multiplying.

Since $1147=31\cdot 37$, by Chinese Remainder theorem, you have a ring isomorphim
$$\psi:\mathbb Z/1147\mathbb Z\longrightarrow\mathbb Z/31\mathbb Z\times\mathbb Z/37\mathbb Z$$
Thus 
$$\psi(173^{607}+1147\mathbb Z)=(173^{607}+31\mathbb Z,173^{607}+37\mathbb Z)$$
But
\begin{align*}
173^{607}
&\equiv 18^7\\
&=2^7\cdot 3^{14}\\
&\equiv 2^2\cdot (3^3)^4\cdot 3^2\\
&\equiv 2^2\cdot (-4)^4\cdot 3^2\\
&\equiv 2^{10}\cdot 3^2\\
&\equiv 9\pmod{31}
\end{align*}
because $2^5\equiv 1\pmod{31}$.
On the other hand:
\begin{align*}
173^{607}
&\equiv 25^{31}\\
&\equiv 3^{-31}\\
&\equiv 3^5\\
&\equiv 21\pmod{37}
\end{align*}
Consequently, $$173^{607}\equiv 9\cdot 37\cdot (-5)+21\cdot 31\cdot 6\pmod{1147}$$ because $-5\equiv 6^{-1}\equiv 37^{-1}\pmod{31}$ and $6\equiv (-6)^{-1}\equiv 31^{-1}\pmod{37}$.
Finally we get $173^{607}\equiv -45\cdot 37+126\cdot 31\equiv 1094\pmod{1147}$.
