Find the remainder when $2^{47}$ is divided by $47$ So I am trying to work out how to find remainders in several different way.
I have a few very similar question,
1.) Find the remainder when $2^{47}$ is divided by 47
So i have a solution that says 
$$2^{24} \equiv 2$$
$$2^{48} \equiv 4$$
$$2^{47} \equiv 2$$ Since $(2,47)=1$
and 
2.)Find the remainder when $2^{32}$ is divided by $47$:
We have $$2^6 = 64 \equiv 17$$
$$2^{12} \equiv 17^2 = 289 \equiv 7$$
$$2^{24} \equiv 7^2=49  \equiv 2$$
$$2^8 \equiv 4*17=68  \equiv 21$$
$$2^{32}=2^{8}*2^{24}  \equiv21*2 \equiv42$$
Okay so although i have some solutions here, Im not 100 percent sure how they derived this. For example on the second question, why have they started with $2^{6}$ before finding the remainder? Is there some steps missing or am I just missing the point?
Is this called something in particular? There are not many problems similar to and i do not know if it has a special name to search for.
P.S Please do not use fermat's little theorem, i want to understand this method, thanks :)

I don't know if this constitutes a full answer, but I cannot leave comments yet.  
Notice that $2^5 = 32 < 47$, and $2^6 = 64 > 47$.  That is, we are starting with the first power of $2$ that is greater than the number we're dividing by.  If the number were less, the remainder would simply be the number we started with.  
As for a term, try looking at Modular Arithmetic.  
