Count the number of positive integer $n \leq 2019$ such that $11n+20$ relatively prime with $3+5n$

Question

Here is my approach.

Since $$11n+20 = 2(3+5n) + (n+14)$$ then $$1=(11n+20,3+5n)=(3+5n,n+14)$$

I don't how to continue, because for some $n$, $3+5n > n+14$. And for other $n$, $3+5n \leq n+14$.

Any advice?

Answer 1

You are very close. $(5n+3,n+14)=(-67,n+14)$

So I am counting the number of positive integers such that $67\nmid n+14$. There are 30 numbers that fail so there are 1989 numbers that work.