Which is the fastest way to find the remainder when $2^{400}$ is divided by $400$? 
Which is the fastest way to find the remainder when $2^{400}$ is divided by $400$?

My approach  is to break up $400$ as $16 \times 25$ and then apply CRT, I was wondering if there is any other approach that gives the result faster than using CRT.
Thanks,

As a relatively fast but purely mechanical method (not a whole lot of thought involved), I'd first note that the exponent $$400_{10}=110010000_2=2^4+2^7+2^8$$ so that $$2^{400}=2^{(2^4+2^7+2^8)}=2^{2^4}\cdot2^{2^7}\cdot2^{2^8}.$$
Now, by successively squaring:
$$\begin{align}
2^{2^1}&\equiv 4\mod 400
\\
2^{2^2}\equiv 4^2&\equiv 16\mod 400\qquad(*)
\\
2^{2^3}\equiv 16^2\equiv 256&\equiv-144\mod 400
\\
2^{2^4}\equiv (-144)^2\equiv 336&\equiv-64\mod 400
\\
2^{2^5}\equiv (-64)^2&\equiv 96\mod 400
\\
2^{2^6}\equiv 96^2&\equiv 16\equiv2^{2^2}\mod 400\qquad(*)
\\
2^{2^7}\equiv2^{2^3}&\equiv-144\mod 400
\\
2^{2^8}\equiv2^{2^3}&\equiv-64\mod 400
\end{align}$$
(Because we get the same result on the two $(*)$ lines, we now have a repeating pattern.)
So, $$\begin{align}
2^{400}&\equiv2^{2^4}\cdot2^{2^7}\cdot2^{2^8}\mod400
\\
&\equiv(-64)(-144)(-64)\mod400
\\
&\equiv176\mod400.
\end{align}$$
