Let $n$ be a positive integer. Prove that there exists a positive integer $m > 1000$ 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 $1000k+007$.
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do any numbers always divide numbers of this form ? – Aug 05 '17 at 01:14
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Can't you just find a prime ending in $007$? Are there any other restrictions on $n$? – platty Aug 05 '17 at 01:15
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2Dirichlet's theorem? – Angina Seng Aug 05 '17 at 01:15
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@platty I can't just use a prime, because $n$ can be any integer so $m$ changes according to $n$. If $m$ was just a prime, then if $n$ was a multiple of $m$ it wouldn't work – Annie Smith Aug 05 '17 at 01:25
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if there aren't constant factors then just look for when the divisors of N can't divide into the arithmetic sequence. – Aug 05 '17 at 01:32
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2Primes Are Forever, even James Bond primes. – Joe Knapp Aug 05 '17 at 02:30
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$(1000,7)=1\Rightarrow (m,n) = (100k+7,n)=1,$ for infinitely many $,k,,$ by Stieljes method in the linked dupe. $\ \ $ – Bill Dubuque Oct 29 '24 at 18:14
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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.
ajotatxe
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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.
Ross Millikan
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