I've been asked to prove this property of integers:
Let $\gamma$ be the remainder from the euclidean division of $n$ by $m$. Prove that the remainder from the division of $2^n-1$ by $2^m - 1$ is $2^\gamma - 1$.
She also asked me to find the quotient from this division. Can anyone help me? Tried a couple things here but couldn't figure it out myself. (I also already apologize for my poor math English)
We have $n = mr + \gamma$, $r\in\mathbb{Z}$, $m>\gamma$. Using Euclidean Division:
$$ \begin{align*} 2^n-1 &= 2^{mr + \gamma}-1 \\ &= 2^{mr}\cdot 2^\gamma - 1 \\ &= 2^\gamma(2^{mr}-1)+(2^\gamma - 1) \\ &= 2^\gamma[(2^m)^r-1]+(2^\gamma - 1) \\ &= 2^\gamma\cdot (2^m - 1)\cdot[2^{m(r-1)}+2^{m(r-2)} + \ldots 2^{2m} + 2^m + 1] + (2^\gamma-1) \\ &= (2^m - 1)\cdot q + (2^\gamma - 1) \end{align*} $$ where $q = 2^\gamma\cdot[2^{m(r-1)}+2^{m(r-2)} + \ldots 2^{2m} + 2^m + 1]$.
Notice that if $m > \gamma$, then $2^m - 1 > 2^\gamma - 1$.
If $2^n - 1 = (2^m - 1)\cdot q + (2^\gamma - 1)$ with $2^\gamma - 1 < 2^m - 1$, then $2^\gamma-1$ is the remainder of the asked division.