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 August 11th, 2016, 07:09 PM #1 Newbie   Joined: Aug 2016 From: Hong Kong Posts: 5 Thanks: 0 All primes are in form of 10nk+c Prove: For positive integer n, and any positive integer c that 0 < c < 10n and c is coprime with 10n, all primes larger than 10n can be expressed in form of 10nk+c where k is a positive integer. Remark: c may have various values. For example, when n=2, possible values of c are: 1,3,7,9,11,13,17,19
 August 11th, 2016, 11:53 PM #2 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 Let $m = nk$. Then setting $n = 1$, it is clear that $m$ can be any natural number. It is a consequence of the division algorithm that any positive number (including primes) can be represented by $10m + c$, where $0 < c < 10$. This proves the result.
 August 12th, 2016, 08:31 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond But if $c$ is coprime to $10n$ how would you represent, say, 12? Thanks from Joppy
 August 12th, 2016, 08:59 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 If $c$ isn't coprime to $10n$, $10n + c$ isn't a prime. Thanks from greg1313 and Joppy
 August 12th, 2016, 03:26 PM #5 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 Ah. I didn't read the question very well it seems. It is still trivial however, since 1, 3, 7 and 9 are all coprime to $10n$. This means that all odd numbers not ending in 5 (and thus not divisible by 5) can be represented in this way. Thanks from greg1313 and Joppy

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