Trying to find a modulo for $2^{24}$ I was trying to figure out which modulo $n$ would make $2^{24}$ congruent to $1  \bmod n$. One answer is $241$, and I wonder if there is a good way to find it.
I tried to use Fermat's little theorem or Chinese Remainder theorem, but $24$ is not a prime and $241$ is not composite.

We want to find factors of 
\begin{align}(2^{24}-1)&=(2^{12}-1)(2^{12}+1)\\
&= (2^6-1)(2^6+1)(2^{12}+1)
\end{align}
Hence we can for example, pick our $n$ to be $2^6-1, 2^6+1$ or $2^{12}+1$. Notice that $241$ is a factor of $2^{12}+1$.
Also note that $$2^{24}=(2^3)^8=(7+1)^8$$
Hence we can pick $n=7$.
Also $$2^{24}=(2^8)^3=((2^8-1)+1)^3$$
Hence, we can pick $n=2^8-1$ or its factors and so on.
