Let $n\geq 1$ be an integer and let $d|n$ be a divisor. I want to prove the following statement:
For any $1\leq k<d$ with $\gcd(k,d)=1$, there exists a number $1\leq u<n$ with $\gcd(u,n)=1$ such that $u=k$ mod $d$.
Can someone provide a proof or a reference?
Hint $ $ We seek $u = k\!+\!jd\,$ with $\,\gcd(k\!+\!jd,n)=1.\,$ Such a $j$ exists by here and $\,\gcd(k,d)=1$