Let $r$ be the remainder of $a$ when dividing by $b$. Prove: $2^r -1$ is the remainder of $2^a -1$ when dividing by $2^b -1$

Question

I am having trouble proving this question. Can anyone help me out?

Answer 1

Let $b\equiv r\bmod a$. This implies $a \mid b-r$ and $r<a$. So $2^a-1\mid 2^{b-r}-1$ and hence $2^a-1\mid(2^{b}-1)-(2^r-1)$. Now observe $2^r-1<2^a-1$ and hence the remainder is $2^r-1$.

Answer 2

here is my solution sorry I don't know how to write in latex I am new I hope that my solution is helpful :