How to evaluate $1234^{1234} \pmod{5379}$? 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. 

Well this is far from perfect,but it works if you have enough time or a calculator.
$$93\equiv -70\pmod{163}$$
$$\begin{align}
93^{100}&\equiv(-70)^{100}\\
&= 490^{50}\cdot10^{50}\\
&\equiv 10^{50}\\
&= 2^{50}\cdot 5^{50}\\
&= 1024^5\cdot 3125^{10}\\
&\equiv 46^5\cdot 28^{10}\\
&=2^{25}\cdot 23^5\cdot 7^{10}\\
&= 2^{25}\cdot (23\cdot 7)^5\cdot 7^5\\
&= 2^{25}\cdot (161)^5\cdot 7^5\\
&\equiv 2^{10}\cdot2^{10}\cdot 2^5\cdot (-2)^5\cdot 7^5\\
&=1024\cdot 1024\cdot (-1024)\cdot 7^5\\
&\equiv -46^3\cdot 7^5\\
&= -(46\cdot 7)^3\cdot 7^2\\
&= -(322)^3\cdot 7^2\\
&\equiv -(-4)^3\cdot 49\\
&= 49\cdot 64\\
&\equiv 39 
\end{align}$$
