If $m$ is an odd integer and $n \in \mathbb N$, prove that the system of congruence
$2x \equiv 2n (mod\, m)$
$x \equiv m(mod \, 2^n) $
has exactly one integer solution $x$ with $0 \le x \lt 2^nm$
To start off, I know that $x \equiv n(mod \, m)$ since $gcd(m,2)=1$
I thought about using the Chinese Remainder Theorem since $gcd(m,2^n)=1$ as well but I am not exactly sure how to approach this. Does anyone know how I can solve this? Thanks in advance!
"I thought about using the Chinese Remainder Theorem since $\gcd(m,2^n)=1$" That's exactly what you should do. The approach is simple: First off, snice $m$ is odd you may rewrite $2x\equiv 2n$ to $x \equiv n$.
You now have two congruences: $$ x \equiv n \mod m\\ x \equiv m \mod 2^n $$ and the Chinese remainder theorem says that because $\gcd(m, 2^n) = 1$, there is exactly one solution $\mod m2^n$.