Let $d,n\in\mathbb Z_{>0}$ with $d\mid n$. Let $a\in\mathbb Z$ such that $\gcd(a,d)=1$. Let $t$ be the product of prime numbers that divide $n$, but don’t divide $a$. Show: $\gcd(a+td,n)=1$.
Assume $r\mid a+td$ and $r\mid n$. We want to show that $r=1$. I am trying to show that $r\mid a+td\implies r\mid a$ and $r\mid d$, because this will yield $r=1$.
I'm not sure how to do this. I was thinking of assuming $r\nmid d$, which would hopefully lead to a contradiction, so I would have $r\mid a$. The only thing I can think of, is that we need $r\nmid t+d$. Is this the right way? Any help would be appreciated.