Number Theory: m>1000, m ends in 007, m is relatively prime to n Let $n$ be a positive integer. Prove that there exists a positive integer $m &gt; 1000$ with the following two properties: $m$'s last 3 digits are 007, and $m$ is relatively prime to $n$.
I know that I should first solve the case that $n$ is relatively prime to 1000 and then try to reduce the second case to the first but I don't really know how to do that. 
ALso I know that I can express $m$ as $1000k+007$.
You can use the Chinese Remainder Theorem.  Having the last three digits  of $m$ be $007$ means $m \equiv 7 \pmod {1000}$  As long as $n$ has no factors of $2$ or $5$ you can just form a set of congruences $m \equiv 1 \pmod p$ where $p$ ranges over the primes that divide $n$.  That set of congruences along with $m \equiv 7 \pmod {1000}$ has a solution and you are done.  I leave patching up the cases where $2$ or $5$ divides $n$ to you.
As Lord Shark has pointed, Dirichlet's theorem implies that there are infinitely many primes that end with $007$. Choose one greater than $n$.

Actually, the series
$$\sum_{p\text{ prime that ends with $007$}}\frac 1p$$
diverges.