Let $n,r,a$ be positive integers with g.c.d.$(a,d)=1$ , does there exist integer $m$ relatively prime to $n$ such that $d|m-a$?

Question

Let $n,r,a$ be positive integers with g.c.d.$(a,d)=1$ . Does there exist integer $m$ such that $d|m-a$ and g.c.d.$(m,n)=1$ ?

Answer 1

Your problem can be done using Chinese remainder theorem: $$\begin{align}m &\equiv a\mod d\\ m&\equiv1 \mod \frac{n}{gcd(n,d)} \end{align}$$ you can prove that if $m$ is an integer satisfying these two equation then it is a solution for your problem

Answer 2

You just need $\text{gcd}(m,n)=1$ and $m \equiv a \pmod{d}$. But $a$ and $d$ are coprime by hypothesis, therefore it is sufficient to set $m \equiv 1 \pmod{n}$.

In few words, you are searching an integer $m$ of the form $hn+1$ and of the form $kd+a$. There exists exactly one for each $ad$ consecutive integers by the chinese remainder theorem.