TITLE: How to evaluate $1234^{1234} \pmod{5379}$?
QUESTION [6 upvotes]: Note: $5379 = 3 \times 11 \times 163$.
I tried Chinese Remainder Theorem and Fermat's Little Theorem, got as far as:
$$
1234^{1234} = 1 \pmod{3} \\
1234^{1234} = 5 \pmod{11}
$$
With a bit more work:
$$1234^{1234} = 93^{100} \pmod{163}$$
But $93^{100}$ doesn't really help?
WolframAlpha tells me that $\phi(5379)=3240>1234$
So I can't use Euler's Theorem?
N.B This appeared on a 1st year undergrad problem sheet. So presumably, not too much technology is required.

REPLY [3 votes]: A little noodling produced the following, where all congruences are mod $163$:
$$93\equiv256=2^8\implies2^{10}\cdot93^{100}\equiv2^{810}=2^{162\cdot5}\equiv1$$
Noting that $2^{10}=1024\equiv46$, it remains to compute $46^{-1}$ mod $163$.  This could be done by a straightforward Euclidean algorithm, but I found it easy enough to argue as follows:
$$\begin{align}
23\cdot7=161\equiv-2
&\implies46\cdot7\equiv-4\\
&\implies46\cdot7\cdot(-41)\equiv164\equiv1\\
&\implies46^{-1}\equiv-287\equiv39
\end{align}$$
All in all, we have
$$93^{100}\equiv39\mod163$$
Remark:  The one computation I had to do off to the side was 
$$1024-6\cdot163=1024-978=46$$
Everything else I could do easily in my head.